1 Affine Varieties

We will begin following Kempf’s Algebraic Varieties, and eventually will dothings more like in Hartshorne. We will also use various sources for commutativealgebra.

What is algebraic geometry? Classically, it is the study of the zero sets ofpolynomials.

We will now fix some notation. k will be some fixed algebraically closed field,any ring is commutative with identity, ring homomorphisms preserve identity,and a k-algebra is a ring R which contains k (i.e., we have a ring homomorphismι : k → R).

P ⊆ R an ideal is prime iff R/P is an integral domain.Algebraic SetsWe define affine n-space, An = kn = (a1, . . . , an) : ai ∈ k.Any f = f(x1, . . . , xn) ∈ k[x1, . . . , xn] defines a function f : An → k :

(a1, . . . , an) 7→ f(a1, . . . , an).Exercise If f, g ∈ k[x1, . . . , xn] define the same function then f = g as

polynomials.

Definition 1.1 (Algebraic Sets). Let S ⊆ k[x1, . . . , xn] be any subset. ThenV (S) = a ∈ An : f(a) = 0 for all f ∈ S.

A subset of An is called algebraic if it is of this form.

e.g., a point (a1, . . . , an) = V (x1 − a1, . . . , xn − an).Exercises

1. I = (S) is the ideal generated by S. Then V (S) = V (I).

2. I ⊆ J ⇒ V (J) ⊆ V (I).

3. V (∪αIα) = V (∑Iα) = ∩V (Iα).

4. V (I ∩ J) = V (I · J) = V (I) ∪ V (J).

Definition 1.2 (Zariski Topology). We can define a topology on An by definingthe closed subsets to be the algebraic subsets. U ⊆ An is open iff An \U = V (S)for some S ⊆ k[x1, . . . , xn].

Exercises 3 and 4 imply that this is a topology.

The closed subsets of A1 are the finite subsets and A1 itself.

Definition 1.3 (Ideal of a Subset). If W ⊂ An is any subset, then I(W ) =f ∈ k[x1, . . . , xn] : f(a) = 0 for all a ∈W

Facts/Exercises

1. V ⊆W ⇒ I(W ) ⊆ I(V )

2. I(∅) = (1) = k[x1, . . . , xn]

3. I(An) = (0).

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Definition 1.4 (Affine Coordinate Ring). W ⊂ An is algebraic. Then A(W ) =k[W ] = k[x1, . . . , xn]/I(W )

We can think of this as the ring of all polynomial functions f : W → k.

Definition 1.5 (Radical Ideal). Let R be a ring and I ⊆ R be an ideal, thenthe radical of I is the ideal

√I = f ∈ R : f i ∈ I for some i ∈ N

We call I a radical ideal if I =√I.

ExerciseIf I is an ideal, then

√I is a radical ideal.

Proposition 1.1. W ⊆ An any subset, then I(W ) is a radical ideal.

Proof. We have that I(W ) ⊆√I(W ).

Suppose f ∈√I(W ). Then f i ∈ I for some i. That is, for all a ∈ W ,

f i(a) = 0. Thus, f(a)m = 0 = f(a). And so, f(a) ∈ I.

Exercises

1. S ⊆ k[x1, . . . , xn], then S ⊆ I(V (S)).

2. W ⊆ An then W ⊆ V (I(W )).

3. W ⊆ An is an algebraic subset, then W = V (I(W )).

4. I ⊆ k[x1, . . . , xn] is any ideal, then V (I) = V (√I) and

√I ⊆ I(V (I))

Theorem 1.2 (Nullstellensatz). Let k be an algebraically closed field, and I ⊆k[x1, . . . , xn] is an ideal, then

√I = I(V (I)).

Corollary 1.3. k[V (I)] = k[x1, . . . , xn]/√I.

To prove the Nullstellensatz, we will need the following:

Theorem 1.4 (Nother’s Normalization Theorem). If R is any finitely gener-ated k-algebra (k can be any field), then there exist y1, . . . , ym ∈ R such thaty1, . . . , ym are algebraically independent over k and R is an integral extensionof the subring k[y1, . . . , ym].

Proof is in Eisenbud and other Commutative Algebra texts.

Theorem 1.5 (Weak Nullstellensatz). Let k be an algebraically closed field,and I ( k[x1, . . . , xn] any proper ideal, then V (I) 6= ∅.

Proof. We may assume without loss of generality that I is actually a maximalideal. Then R = k[x1, . . . , xn]/I is a field. R is also a finitely generated k-algebra, and so by Normalization, ∃y1, . . . , ym ∈ R such that y1, . . . , ym arealgebraically independent over k and that R is integral over k[y1, . . . , ym].

Claim: m = 0. Otherwise, y−11 ∈ R is integral over k[y1, . . . , ym], and so then

y−p1 +y1−p1 f1+. . .+y−1

1 fp−1+fp = 0 for fi ∈ k[y1, . . . , ym]. Multiplying through

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by yp1 gives 1 = −(y1f1 + . . . + yp−11 fp−1 + yp1fp) ∈ (y1), which contradicts the

algebraic independence.Thus, the field R is algebraic over k. As k is algebraically closed, R = k.

k ⊆ k[x1, . . . , xn]→ R = kLet ai = the image in k of xi. Then xi−ai ∈ I. Thus, the ideal generated by

(x1−a1, . . . , xn−an) ⊆ I ( k[x1, . . . , xn], and so they I = (x1−a1, . . . , xn−an),as it is a maximal ideal.

V (I) = V (x1 − a1, . . . , xn − an) = (a1, . . . , an) 6= ∅

Note: Any maximal ideal of k[x1, . . . , xn] is of the form (x1−a1, . . . , xn−an)with ai ∈ k.

This is NOT true over R, look at the ideal (x2 + 1) ⊆ R[x]. It is, in fact,maximal.

Now, we can prove the Nullstellensatz.

Proof. Let I ⊆ k[x1, . . . , xn] be any ideal. We will prove that I(V (I)) =√I.

It was an exercise that√I ⊆ I(V (I)).

Let f ∈ I(V (I)). We must show that f ∈√I.

Looking at An+1, we have the variables, x1, . . . , xn, y. Set J = (I, 1− yf) ⊆k[x1, . . . , xn, y].

Claim: V (J) = ∅ ⊂ An+1. This is as, if p = (a1, . . . , an, p) ∈ V (J), then(a1, . . . , an) ∈ V (I), then (1− yf)(p) = 1− bf(a1, . . . , an). But f(a1, . . . , an) =0, so (1− yf)(p) = 1, and so p /∈ V (J).

By the Weak Nullstellensatz, J = k[x1, . . . , xn, y]. Thus 1 = h1g1 + . . . +hmgm + q(1− yf) where g1, . . . , gm ∈ I and h1, . . . , hm, q ∈ k[x1, . . . , xn, y].

Set y = f−1, and multiply by some big power of f to get a polynomialequation once more.

Then fN = h1g1 + . . .+ hmgm where the hi = fNhi(x1, . . . , xn).And so, we have fN ∈ I, and thus, f ∈

√I, by definition.

exercise: V (I(W )) = W in the Zariski Topology.Irreducible Algebraic SetsRecall: V (y2 − xy − x2y + x3) = V (y − x) ∪ V (y − x2), and V (xz, yz) =

V (x, y) ∪ V (z).

Definition 1.6 (Reducible Subsets). A Zariski Closed subset W ⊆ An is calledreducible if W = W1 ∪W2 where Wi ( W and Wi closed.

Otherwise, we say that W is irreducible.

Proposition 1.6. Let W ⊆ An be closed. Then W is irreducible iff I(W ) is aprime ideal.

Proof. ⇒: Suppose I(W ) is not prime. Then ∃f1, f2 /∈ I(W ) such that f1f2 ∈W . Set W1 = W ∩ V (f1) and W2 = W ∩ V (f2).

As fi /∈ I(W ), W 6⊆ V (fi), and so Wi ( W . Now we must show that W =W1 ∪W2. Let a ∈W . Assume a /∈W1. Then f1(a) 6= 0, but f1(a)f2(a) = 0, sof2(a) = 0, thus a ∈W2.⇐: Exercise

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This gives us the beginning of an algebra-geometry dictionary.Algebra Geometry

k[x1, . . . , xn] Anradical ideals closed subsetsprime ideals irreducible closed subsetsmaximal ideals points

In fact, this is an order reversing correspondence. So I ⊆ J ⇐⇒ V (I) ⊇V (J), but this requires I, J to be radical.

Definition 1.7 (Notherian Ring). A ring R is called Notherian if every idealI ⊆ R is finitely generated.

Exercise A ring R is Notherian iff every ascending chain of ideals I1 ⊆ I2 ⊆. . . stabilizes, that is, ∃N such that IN = IN+1 = . . ..

Theorem 1.7 (Hilbert’s Basis Theorem). If R is Notherian, then R[x] isNotherian.

Corollary 1.8. k[x1, . . . , xn] is Notherian.

Definition 1.8 (Notherian Topological Space). A topological space X is Notherianif every descending chain of closed subsets stabilizes.

Corollary 1.9. An is Notherian.

W1 ⊇W2 ⊇ . . . closed in An, then I(W1) ⊆ I(W2) ⊆ . . . ideals in k[x1, . . . , xn],and so must stabilize.

Theorem 1.10. Any closed subset of a Notherian Topological Space X is aunion of finitely many irreducible closed subsets.

Proof. Assume the result is false. ∃W a closed subset of X which is not theunion of finitely many irreducible closed sets.

As X is Notherian, we may assume that W is a minimal counterexample. Wis not irreducible, and so W = W1∪W2, where Wi ( W and Wi closed. The Wi

can’t be counterexamples, as W is a minimal one, but then W = W1 ∪W2 andeach Wi is the union of finitely many irreducible closed sets. Thus, W cannotbe a counterexample.

Corollary 1.11. Every closed W ⊆ An is union of finitely many irreducibleclosed subsets.

Example: V (xy) = V (x) ∪ V (y) ∪ V (x− 1, y).Recall: X is a topological space, then if Y ⊆ X is any subset, it has the

subspace topology, that is, U ⊆ Y is open iff ∃U ′ ⊆ X open such that U =U ′ ∩ Y .

Note:

1. W ⊆ Y is closed iff W = W ∩ Y , where the closure is in X.

2. X is Notherian implies that Y is Notherian in the subspace topology.

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Definition 1.9 (Zariski Topology on X ⊆ An). If X ⊆ An, then the ZariskiTopology on X is the subspace topology.

Definition 1.10 (Components of X). If X is any Notherian Topological Space,then the maximal irreducible closed subsets of X are called the (irreducible)components of X.

Exercises

1. X has finitely many components.

2. X = the union of its irreducible components.

3. X 6= union of any proper subset of its components.

4. A topological space is Notherian if and only if every subset is quasi-compact.

5. A Notherian Hausdorff space is finite.

Recall: X ⊆ An closed. Then A(X) = k[x1, . . . , xn]/I(X).

Definition 1.11. If f ∈ A(X), set D(f) = a ∈ X : f(a) 6= 0.

Proposition 1.12. The sets D(f) form a basis for the Zariski Topology on X.

Proof. Let p ∈ U ⊆ X, U open. Show that p ∈ D(f) ⊆ U for some f ∈ A(X).Z = X\U a closed subset of X, and Z ( Z∪p implies that I(Z) ) I(Z∪p).

Take any f ∈ I(Z) \ I(Z ∪ p). Then f vanishes on Z but not at p, sop ∈ D(f).

Regular FunctionsLet X ⊆ An be an algebraic subset, and U ⊆ X is a relatively open subset

of X.

Definition 1.12 (Regular Function). A function f : U → k is called regular if fis locally rational. That is, ∃ open cover U = ∪αUα and functions pα, qα ∈ A(X)such that ∀a ∈ Uα, qα(a) 6= 0 and f(a) = pα(a)/qα(a).

We define k[U ] to be the set of regular functions from U to k.

Note:

1. k[U ] is a k-algebra.

2. A(X) ⊆ k[X].

ExampleLet X = V (xy − zw) ⊆ A4. f : U → k can be defined by f = x/w on D(w)

and f = z/y on D(y). Thus, f ∈ k[U ].Exercise; 6 ∃p, q ∈ A(X) such that q(a) 6= 0 and f(a) = p(a)/q(a) for all

a ∈ U .

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Lemma 1.13. Let q1, . . . , qn ∈ A(X). Then D(q1) ∪ . . . ∪ D(qn) = X iff(q1, . . . , qm) = (1) = A(X).

Proof. ⇐: 1 =∑hiqi, hi ∈ A(X), then the qi cannot all vanish at any point,

and so we are done.⇒: Take Qi ∈ k[x1, . . . , xn] such that qi = Qi ∈ A(X). D(q1)∪. . .∪D(qm) =

X, so X ∩ V (Q1) ∩ . . . ∩ V (Qm) = ∅ = V (I(X), Q1, . . . , Qm) = ∅, and so,by the weak nullstellensatz, (I(X), Q1, . . . , Qm) = (1) ⊆ k[x1, . . . , xn], and so(q1, . . . , qm) = (1) = A(X).

Theorem 1.14. Let X ⊆ An be an algebraic set. Then k[X] = A(X).

Proof. Let f ∈ k[X]. Then X = U1 ∪ . . .∪Um and there are pi, qi ∈ A(X) suchthat qi 6= 0 and f = pi/qi on Ui.

We can refine the open cover such that each Ui = D(gi) for some gi. Note:f = pi/qi = pigi

qigion Ui = D(gi) = D(giqi). We can replace pi with pigi and qi

by qigi.Then we can assume that Ui is D(qi) = D(q2

i ). Thus, X = D(q21) ∪ . . . ∪

D(q2m). By the lemma, we know that 1 =

∑mi=1 hiq

2i , hi ∈ A(X). Note q2

i f =qipi on Ui, and qi = 0 outside of Ui.

f = 1f =∑mi=1 hiq

2i f =

∑mi=1 hiqipi, so f ∈ A(X).

Definition 1.13 (Spaces With Functions). A space with functions (SWF) isa topological space X together with an assignment to each open U ⊆ X of ak−algebra k[U ] consisting of functions U → k. These are called regular func-tions. It must also satisfy the following:

1. If U = ∪Uα is an open cover and f : U → k any function, then f isregular on U iff f |Uα is regular on Uα for all α.

2. If U ⊆ X is open, f ∈ k[U ], then D(f) = a ∈ U : f(a) 6= 0 is open and1f ∈ k[D(f)].

Note: OX(U) = k[U ] is another common notation.Examples:

1. Algebraic sets. These are called Affine Algebraic Varieties

2. M is a differentiable manifold, k = R, k[U ] = C∞ functions U → R.

3. X is a SWF, U ⊆ X open subset, then U is a SWF. OU (V ) = OX(V ).

Definition 1.14 (Morphism of SWFs). Let X,Y be SWFs, then a morphismϕ : X → Y is a continuous map which pulls back regular functions to regularfunctions. i.e., if V ⊆ Y is open, f ∈ OY (V ), then ϕ∗(f) ∈ OX(ϕ−1(V )),ϕ∗(f) = f ϕ.

Definition 1.15 (Isomorphism). ϕ : X → Y is an isomorphism if ϕ is amorphism and ∃ a morphism ψ : Y → X such that ϕψ = idY and ψϕ = idX .

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Exercises

1. The id function of a SWF is a morphism.

2. Compositions of morphisms are morphisms.

3. Let X be any SWF and Y ⊆ An closed, that is, an affine variety. Thenf = (f1, . . . , fn) : X → Y , is a morphism iff fi ∈ k[X] for all i.

Example: A1 \ 0 is (isomorphic to) an affine variety defined by V (1− xy).LocalizationLet R be a ring and S ⊆ R multiplicatively closed subset. That is, s, t ∈

S ⇒ st ∈ S and 1 ∈ S.We define S−1R = f/s : f ∈ R, s ∈ S. We consider f/s to be the same

element as g/t iff there exists u ∈ S such that u(ft− sg) = 0. This is a ring byfsgt = fg

st and fs + g

t = ft+gsst .

Exercise: Check these assertions.Special Case: If f ∈ R, then Rf = S−1R where S = fn : n ∈ N. In fact,

Rf ∼= R[y]/(1− fy).

Definition 1.16 (Reduced Ring). R is a reduced ring iff fn = 0 implies f = 0for all f ∈ R. Equivalently, (0) =

√(0).

Facts:

1. R reduced implies S−1R is reduced

2. R/I reduced iff I =√I

Proposition 1.15. Let X ⊆ An be a closed affine variety and f ∈ A(X). ThenD(f) if an affine variety, with affine coordinate ring A(X)f .

Proof. Let I = I(X) ⊆ k[x1, . . . , xn] define J = (I, yf − 1) ⊆ k[x1, . . . , xn, y].Let φ : D(f)→ V (J) ⊆ An+1 by (a1, . . . , an) 7→ (a1, . . . , an, f(a1, . . . , an)−1).

Note: φ is an isomorphism.What remains is to compute the coordinate ring. A(X) is reduced, and

so A(X)f is reduced. A(X)f = k[x1, . . . , xn, y]/J , so J is a radical ideal, soJ =

√J = I(V (J)). Therefore, k[D(f)] = k[V (J)] = k[x1, . . . , xn, y]/J =

A(X)f .

Definition 1.17 (Prevariety). A prevariety is a space with functions X suchthat X has a finite open cover X = U1 ∪ . . .∪Um where Ui is an affine variety.

Example: Any affine variety is a prevariety.Exercise: Any prevariety is a Notherian Topological SpaceExample: An open subset of a prevariety is a prevariety. This follows from

the previous proposition and the fact that principle open sets are a basis for thetopology of any affine variety.

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Proposition 1.16. X is a space with functions, Y ⊆ An an affine variety, wehave a 1-1 correspondence:morphisms X → Y ⇐⇒ k-algebra homomorphism A(Y ) → k[X] by

ϕ ⇐⇒ ϕ∗

Proof. Note φ 7→ φ∗ is a well defined map. Write A(An) = k[y1, . . . , yn], thenI(Y ) ⊆ k[y1, . . . , yn]. Then yi is the image of yi in A(Y ). Assume that α :A(Y )→ k[X] is a k-algebra homomorphism. We define φ : X → An by φ(x) =(φ1(x), . . . , φn(x)).

If f ∈ I(Y ) then f(y1, . . . , yn) = 0, so f(φ1, . . . , φn) = α(f(y1, . . . , yn)) = 0,and so φ(X) ⊆ Y . Note, φ∗(yi) = yi φ = φi = α(yi). Thus, φ∗ = α.

If φ : X → Y is a morphism, then φi = yi φ = φ∗(yi)Thus, φ is the morphism that we construct from φ∗.

Corollary 1.17. Two affine varieties are isomorphic iff their affine coordinaterings are isomorphic as k-algebras

Exercise: An \ (0, . . . , 0) is not affine for n ≥ 2.

Proposition 1.18. We have a one-to-one correspondence between affine vari-eties and reduced finitely generated k-algebras, up to isomorphism, by X 7→ k[X].

Proof. Last time, we proved that two affine varieties are isomorphic iff theircoordinate rings are isomorphic. Thus, X 7→ k[X] is injective.

Let R be a finitely generated reduced k-algebra generated by r1, . . . , rn ∈R. There is a k-alg homomorphism φk[x1, . . . , xn] → R by xi 7→ ri which issurjective. Set I = kerφ and let X = V (I) ⊆ An.

I is radical, as R is reduced, so k[X] = k[x1, . . . , xn]/I ' R.

Note: Assume m ⊆ R is a maximal ideal, then φ : k[x1, . . . , xn] → R asin proof, then M = φ−1(m) is maximal, and M = (x1 − a1, . . . , xn − an).R/m = k[x1, . . . , xn]/M = k.

Canonical ConstructionLet R be a finitely generated reduced k-algebra. Then define Spec−m(R) =

m ⊆ R max ideals.The topology will be that the closed sets V (I) = m ⊇ I|I ⊆ R and ideal .Let f ∈ R. We define f : Spec−m(R) → k by f(m) =image of f in

R/m = k. So f is a function from Spec−m to k.I.E., f(m) ∈ k ⊆ R is the unique element such that f − f(m) ∈ m.Finally, if U ⊆ Spec−m is open, f : U → k is some function, then f is

regular if f is locally of the form f(m) = p(m)/q(m) where p, q ∈ R.Exercise: Spec−m(R) ∼= X, where X is the affine variety with coordinate

ring R, as spaces with functions.Subspaces of SWFsLet X be any space with functions, and Y ⊆ X any subset. Then give Y an

”inherited” SWF structure as follows:We give Y the subspace topology, and if U ⊆ Y is open and f : U → k is a

function, then f is regular iff f can be locally extended to a regular function on

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X. That is, for every point y ∈ U , there is an open subset U ′ ⊆ X containingy and F ∈ OX(U ′) such that f(x) = F (x) for all x ∈ U ∩ U ′.

Exercises

1. Y is a SWF

2. i : Y → X the inclusion map is a morphism.

3. Let Z be a SWF, φ : Z → Y function. Then φ is a morphism iff i φ is amorphism.

4. The SWF structure on Y is uniquely determined by (2) and (3) together.

5. Let Z ⊆ Y ⊆ X. Then Z inherits the same structure from Y and X.

Example: X ⊂ An an algebraic set inherits structure from An. If Y ⊆ X isclosed, then Y inherits structure from X (or An).

Proposition 1.19. A closed subset of a prevariety is a prevariety.

Proof. Let X be a prevariety, and Y ⊆ X a closed subset. X = U1 ∪ . . . ∪ Umwhere Ui are open affine subsets of X.

Ui ∩ Y is a closed subset of Ui, which implies that Ui ∩ Y is affine, and so Yhas the open cover (U1 ∩ Y ) ∪ . . . ∪ (Un ∩ Y ), and so is a prevariety.

Projective Space

Theorem 1.20. Two distinct lines in the place intersect in exactly one point.(except when parallel)

Theorem 1.21. A line meets a parabola in exactly two points. (except whenfalse)

We need projective space to remove the bad cases.

Definition 1.18 (Projective Space). Define an equivalence relation on An+1 \0 by

(a0, . . . , an) ∼ (λa0, . . . , λan) where λ ∈ k∗ = k \ 0.So Pn = (An+1 \ 0)/ ∼ and π : An+1 \ 0 → Pn the projection.

There is a topology on Pn given by U ⊆ Pn is open iff π−1(U) ⊆ An+1 isopen.

The regular functions on Pn are f : U → k such that π∗(f) = π f :π−1(U)→ k is regular.

Thus, Pn is a SWF called Projective Space. Note: Pn = lines through theorigin in An+1, and this method of thinking is often very helpful.

We will use the notation (a0 : . . . : an) for the image π(a0, . . . , an) ∈Pn. If f ∈ k[x0, . . . , xn] is a homogeneous polynomial of total degree d, thenf(λa0, . . . , λan) = λdf(a0, . . . , an). Thus it is well-defined to ask if f(a0 : . . . :an) = 0 or not.

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Definition 1.19. D+(f) = (a0 : . . . : an) ∈ Pn : f(a0 : . . . : an) 6= 0.

Theorem 1.22. Pn is a prevariety.

Proof. Let Ui = D+(xi) ⊆ Pn for 0 ≤ i ≤ n.Claim: Ui ' An.

φ : An → Ui : (a0, . . . , ai−1, ai+1, . . . , an) 7→ (a0 : . . . : ai−1 : 1 : ai+1 : . . . : an)

ψ : Ui → An : (a0 : . . . : an) 7→(a0

ai, . . . ,

aiai, . . . ,

anai

)

Note:Pn = D+(x0)∐V+(x0) = An

∐Pn−1, that is, An = (1 : a1 : . . . an)

the usual n-space and Pn−1 = 0 : a1 : . . . : an points at ∞.The points at ∞ correspond to lines through the origin in An, that is, we

can think of them as being directions.Example: P2 has ”homogeneous coordinate ring” k[x, y, z]. We can think of

that as A2 = D+(z) ⊂ P2 and we know that k[A2] = k[x/z, y/z]. We want tointersect a parabola with a line.

The vertical line is V (x/z − 1) = V+(x−z) and the parabola is V (y/z − (x/z)2) =V+(yz − x2).

And so, V+(x− z) ∩ V+(yz − x2) = (1 : 1 : 1), (0 : 1 : 0), where (0 : 1 : 0)is the point at infinity in the direction ”up”.

Exercise: k[Pn] = k.Exercise: X a SWF, φ : Pn → X a function, then φ is a morphism iff

φ π : An+1 \ 0 → X is a morphism.

Definition 1.20 (Projective Coordinate Ring of Pn). We define k[x0, x1, . . . , xn]to be the coordinate ring of Pn. An ideal I ⊆ k[x0, . . . , xn] is homogeneous if itis generated by homogeneous polynomials. Equivalently, f ∈ I iff each homoge-neous component is in I.

Definition 1.21. If W ⊆ Pn is a subset, then I(W ) = I(π−1(W )) ⊆ k[x0, . . . , xn].Notice that I(W ) is homogeneous. Let f = f0 + . . . + fd ∈ I(W ), fi a form ofdegree i, then (a0 : . . . : an) ∈ W , so 0 = f(λa0, . . . , λan) = f0(a0, . . . , an) +. . .+ λdfd(a0, . . . , an). As this is true for all λ, fi(a0, . . . , an) = 0 for all i, andso fi ∈ I(W ).

Definition 1.22. If I ⊆ k[x0, . . . , xn] is a homogeneous ideal, then defineV+(I) = (a0 : . . . , an) ∈ Pn : f(a0, . . . , an) = 0 for all f ∈ I

Theorem 1.23 (Projective Nullstellensatz). If I ⊆ k[x0, . . . , xn] is a homoge-neous ideal, then

1. V+(I) = ∅ ⇒ (x0, . . . , xn)N ⊆ I for some N > 0. That is,√I = (1) or

(x0, . . . , xn).

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2. V+(I) 6= ∅ then I(V+(I)) =√I.

Proof. 1. V+(I) = ∅ ⇐⇒ V (I) = ∅ or V (I) = 0. By the regular nullstel-lensatz,

√I = I(V (I)) = (1) or (x0, . . . , xn).

2. V+(I) 6= ∅. Then π−1(V+(I)) = π−1(V+(I)) ∪ 0 = V (I) ⊆ An+1. SoI(V+(I)) = I(V (I)) =

√I.

This gives us a 1-1 correspondence between closed subsets of Pn and radicalhomogeneous ideals in k[x0, . . . , xn] except for (x0, . . . , xn). This ideal is oftencalled the irrelevant ideal.

Definition 1.23 (Locally Closed subset). X is a topological space, W ⊆ X asubset is locally closed if it is the intersection of an open set in X and a closedset in X.

Note: A locally closed subset of a prevariety is a prevariety.Terminology: a projective variety is any closed subset of Pn considered as

a space with functions. A Quasi-projective variety is a locally closed subset ofPn. An affine variety is a closed subset of An. A quasi-affine variety is a locallyclosed subset of An.

We notice that anything affine is also quasi-affine and anything quasi-affineis quasi-projective. Something that is projective will also be quasi-projective.

Exercise: Pn is not quasi-affine for n ≥ 1. Later: If X is both projective andquasi-affine, then X is finite.

Definition 1.24 (Projective Coordinate Ring). X ⊆ Pn is a closed projectivevariety, then the projective coordinate ring of X = k[x0, . . . , xn]/I(X).

Warning: This definition depends on the embedding of X in Pn.Example: φ : P1 → P2 by φ(a : b) = (a2 : ab : b2). This is a morphism. In

fact, it is an isomorphism of P1 and V+(xz − y2), but the coordinate ring of P1

is k[s, t] and the coordinate ring of V+(xz − y2) is k[x, y, z]/(xz − y2). Thesetwo rings are NOT isomorphic as k-algebras.

Definition 1.25 (Projective Closure of an affine variety). X ⊆ An is affine.Then we know that An = D+(x0) ⊆ Pn, and X ⊂ An ⊆ Pn makes X a quasi-projective variety, so we take X = the closure of X in Pn.

I = I(X) ⊆ k[x1, . . . , xn] if f = f0 + . . . + fd ∈ k[x1, . . . , xn] where fi is aform of degree i. Then we define f∗ = xd0f0 + xd−1

0 f1 + . . .+ fd ∈ k[x0, . . . , xn].And I∗ is the ideal generated by f∗ : f ∈ I in k[x0, . . . , xn].

Exercise: I(X) = I(X)∗.Example: I = (y−x2, z−x2) ⊆ k[x, y, z]. Then X = V (I) ⊆ A3 = D+(w) ⊂

Pn. I(X) = I∗ = (yw − x2, y − z) ⊇ (wy − x2, wx− x2)So V+(wy − x2, wz − x2) = X ∪ V+(x,w).We now recall that a graded ring is a ring R with decomposition R = ⊕d≥0Rd

as an abelian group such that Rd ·Re ⊆ Rd+e.

11

e.g., R = k[x0, . . . , xn]. f ∈ Rd ⇒ Rf is a Z-graded ring g ∈ Rp implies thatg/fm ∈ Rf is homogeneous of degree p−md.

Definition 1.26. R(f) = homogeneous elements of degree zero = (Rf )0 =g/fm : g ∈ Rdm.

Exercise: f ∈ k[x0, . . . , xn] homogeneous implies that k[x0, . . . , xn](f) is afinitely generated reduced k-algebra.

Theorem 1.24. f /∈ k ⇒ D+(f) ⊆ Pn is affine and in fact k[D+(f)] =k[x0, . . . , xn](f).

Proof. k[D+(f)] = h ∈ k[D(f)] : h(λx) = h(x),∀λ ∈ k∗, x ∈ D(f).If h ∈ k[D(f)] = k[x0, . . . , xn]f , h = g/fm, g ∈ k[x0, . . . , xn].

g

fm(λa0, . . . , λan) =

g

fm(a0, . . . , an) ⇐⇒ g homogeneous of degree md

Therefore, k[D+(f)] = k[x0, . . . , xn](f).The identity map k[D+(f)] → k[D+(f)] gives a morphism φ : D+(f) →

Spec−m(k[D+(f)]) by φ(x) = Mx where Mx = I(x) ⊆ k[D+(f)]Observe that if x, y ∈ D+(f), x 6= y then ∃ homogeneous g ∈ k[x0, . . . , xn]

such that deg(g) = d and g(x) = 0 with g(y) 6= 0. So Mx 6= My. gf ∈Mx, /∈My.

Thus, φ is injective. Set hi = xdif ∈ k[D+(f)] for 0 ≤ i ≤ n.

Ui = D(hi) ⊆ D+(f), Vi = D(hi) ⊆ Spec−m(k[D+(f)]) = m 63 hi.Now we must check thatD+(f) = ∪ni=0Ui and Spec−m(k[D+(f)]) = ∪ni=0Vi.It is enough to prove that φ : Ui → Vi is an isomorphism for all i.D+(xi) ⊆ Pn is affine. k[D+(xi)] = k[x0/xi, . . . , xn/xi] = k[x0, . . . , xn](xi).Thus, Ui = D(f/xdi ) ⊆ D+(xi) is affine, so k[Ui] = (k[x0, . . . , xn](xi))f/xdi ,

which is k[x0, . . . , xn](x,f) = k[D+(f)]hi = k[Vi]. Thus, Ui ' Vi.

Example: f = xz − y2 ∈ k[x, y, z]. X = D+(f) ⊆ P2, R = k[x, y, z](f). Ris generated by A = x2/f , B = y2/f , C = z2/f , D = xy/f , E = yz/f , andF = xz/f .

So X ' V (AB −D2, AC − F 2, BC − E2, F − B − 1) ⊆ A6 by (x : y : z) 7→(A,B,C,D,E, F ).

Exercise:X ⊆ Pn a projective variety f ∈ R = k[x0, . . . , xn]/I(X) is homo-geneous, then D+(f) ⊆ X is affine with affine coordinate ring k[D+(f)] = R(f).

2 Algebraic Varieties

ProductsLet X,Y be two sets. Then X×Y , the cartesian product, is the set (x, y) :

x ∈ X, y ∈ Y .What is X×Y , really? Well, it is a set with projection πX : X×Y → X and

πY : X × Y → Y . This set with the projections satisfies a universal property inthe category of sets.

12

For any set Z with arbitrary functions f : Z → X and g : Z → y, thereexists a unique function φ : Z → X × Y such that f = πX φ and g = πY φ.

X

Y

Z X × Y......................................................................πX

.................................................................. ............πY

..................................................

..................................................

..................................................

................................. ............

f

....................................................................................................................................................................................... ............

g............. ............. ............. .................... ............∃!

Definition 2.1 (Product of SWFs). Let X,Y be spaces with functions. Aproduct of X and Y is a SWF called X × Y with morphism πX : X × Y → Xand πY : X × Y → Y which satisfies the above universal property except with”morphisms” rather than ”functions”.

Exercise: Assume that (P, πX , πY ) and (P ′, π′X , π′Y ) are two products of X

and Y . Then they are isomorphic by unique isomorphism. (See homeworkproblem)

Example: A1 × A1 = A2. NOTE: A2 does not have the product topology!General Construction: X,Y spaces with functions. Then X × Y = (x, y) :

x ∈ X, y ∈ Y is the point set. If U ⊂ X and V ⊂ Y are open, then U × V ⊂X × Y is open, as it is π−1

X (U) ∩ π−1Y (V ).

Let g1, . . . , gu ∈ OX(U) and h1, . . . , hn ∈ OY (V ) set f(u, v) =∑ni=1 gi(u)hi(v).

Then f : U × V → k must be regular. Thus, DU×V (f) = (u, v) ∈ U × V :f(u, v) 6= 0 must be open in X × Y . And so, we define our topology byS ⊆ X × Y is open iff it is a union of sets DU×V (f). The regular functionsF : S → k are the functions that can locally be written as f ′(u, v)/f(u, v) onsome DU×V (f).

That is, ∀(x, y) ∈ S ∃U ⊆ X open and V ⊆ Y open and f(u, v) =∑ni=1 gi(u)hi(v) and f ′(u, v) =

∑mj=1 g

′j(u)h′j(u) with gi, g′j ∈ OX(U) and hi, h′j ∈

OY (V ) such that (x, y) ∈ DU×V (f) ⊆ S and F (u, v) = f ′(u, v)/f(u, v) for all(u, v) ∈ DU×V (f).

Exercises (for X × Y above)

1. X × Y is an SWF

2. πX : X × Y → X and πY : X × Y → Y are morphisms.

3. X × Y is the product of X and Y .

Remark: X,Y SWFs, and U ⊆ X, V ⊆ Y are arbitrary subsets, U, V haveinherited structure as SWFs. Then U ×V has the product space with functionsstructure and subspace SWF structure U × V ⊆ X × Y . These are in fact thesame, due to the universal properties.

For now, we call U × V the product. We obtain the following diagram.

U × V

V

U

X × Y

Y

X

Z

................................................................................................................. ............

................................................................................................................. ............

.................................................................. ............πV

......................................................................πU

.................................................................. ............πY

......................................................................πX

............. ............. ......................... ............i............................................................................................... ............

φ

13

φ is a morphism iff iφ is one, and so we see that the two structures are thesame.

Definition 2.2 (Separated SWF). A SWF X is separated if ∀ SWFs Y andmorphisms f, g : Y → X the set y ∈ Y : f(y) = g(y) ⊆ Y is closed.

Example: Let X = (A1 \ 0) ∪ O1, O2. We can define φi : A1 → X by

taking a 7→

a a 6= 0Oi a = 0

We define a topology by U ⊆ X is open iff φ−1i (U) ⊆ A1 is open for all i. A

function f : U → k is regular iff φ∗i (f) = f φi : φ−1i (U)→ k is regular for all i.

X is a prevariety as X = φ1(A1) ∪ φ2(A1) and φi(A1) ' A1. However, it isnot separated, as a ∈ A1 : φ1(a) = φ2(a) = A1 \ 0 is not closed in A1.

Definition 2.3 (Algebraic Variety). An algebraic variety is a separated preva-riety.

Exercise:

1. Any subspace of a separated SWF is separated.

2. A product of separated SWFs is separated.

Remark: If X is any SWF, then ∆ : X → X×X : x 7→ (x, x) is a morphism.Now we set ∆X = ∆(X) ⊆ X ×X. Then ∆ : X → ∆X is an isomorphism.

Lemma 2.1. X is separated iff ∆X ⊆ X ×X is closed.

Proof. ⇒: πi : X ×X → X be the projections. ∆X = z ∈ X ×X : π1(z) =π2(z) is closed.⇐: Let Y be a SWF, f, g : Y → X maps. Define φ : Y → X × X by

φ(y) = (f(y), g(y)) is a morphism. Now y ∈ Y : f(y) = g(y) ⇒ φ−1(∆X) isclosed.

Exercise: A topological space X is Hausdorff iff ∆X ⊆ X ×X is closed.NB: Product topology!Exercise: An × Am = An+m.

Proposition 2.2. All affine varieties are varieties.

Proof. Enough to show that An itself is separated.∆An ⊆ An × An = A2n is closed, as k[A2n] = [x1, . . . , xn, y1, . . . , yn], so

∆An = V (xi − yi).

Products of Affine Varieties

Lemma 2.3. Let A,B be finitely generated reduced k-algebras. Then A ⊗k Bis a finitely generated reducted k-algebra and k algebraically closed.

Recall: ring structure on A⊗B by (a1 ⊗ b1)(a2 ⊗ b2) = a1a2 ⊗ b1b2.

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Proof. If A is generated by a1, . . . , an and B is generated by b1, . . . , bm thenA ⊗ B is generated by a1 ⊗ 1, . . . , an ⊗ 1, 1 ⊗ b1, . . . , 1 ⊗ bm. For example,k[x1, . . . , xn]⊗ k[y1, . . . , ym] = k[x1, . . . , xn, y1, . . . , ym].

Let X,Y be affine varieties such that k[X] = A, k[Y ] = B. We defineφ : A⊗kB →the set of all functions X×Y → k by f ⊗g 7→

[(x, y) 7→ f(x)g(x)

]is a k-algebra homomorphism.

φ is injective (which implies thatA⊗B is reduced): Suppose φ (∑ni=1 fi ⊗ gi) =

0. WLOG we can assume g1, . . . , gn are linearly independent. Let x ∈ X. Then∑ni=1 fi(x)gi = 0 ∈ B, but gi is linearly independent so fi(x) = 0 for all i

and x, thus fi = 0 ∈ A.

Theorem 2.4. 1. If X,Y are affine then X × Y is affine and k[X × Y ] =k[X]⊗k k[Y ].

2. A product of prevarieties is a prevariety.

Proof. 1 ⇒ 2: X,Y prevarieties, X = ∪Ui, Y = ∪Vj with Ui, Vj affine. ThenX × Y = ∪i,jUi × Vj affine.

Now, we must only prove 1. Set P = Spec-m(k[X]⊗k[Y ]), We have k-algebrahomomorphisms k[X]→ k[X]⊗k[Y ] : f 7→ f ⊗ 1 and k[Y ]→ k[X]⊗k[Y ] : g 7→1⊗ g. These give morphism πX : P → X and πY : P → Y . Claim: P = X ×Y .Let Z be a SWF, p : Z → X and q : Z → Y .

Define k-alg homomorphism k[X] 7→ k[Y ] → k[Z] by f ⊗ g 7→ p∗(f)q∗(g).This gives us a morphism φ : Z → P demonstrating that P is the product.

Remark: If Y is affine and X ⊆ Y is closed, then k[X] ' k[Y ]/I(X).On the other hand, assume X,Y are affine, ϕ : X → Y is a morphism andϕ∗ : k[Y ]→ k[X] is surjective. Then set I = ker(ϕ∗) ⊆ k[Y ], we get

k[Y ]/I

k[Y ] k[X]

V (I)

Y X.................................................................................................... ............ϕ∗

................................................... ............

...............................................................'

......................

.............................................

inj

............................................................................................................................. ϕ ...................................................................... '

Therefore ϕ is an embedding of X as a closed subset of Y .Recall: ∆ : X → X ×X : x 7→ (x, x) gives X ' ∆X = ∆(X) = (x, x) : x ∈

X.

Proposition 2.5. A prevariety X is separated iff ∀ open affine U, V ⊆ X, U∩Vis affine and k[U × V ] = k[U ]⊗ k[V ]→ k[U ∩ V ] = k[∆U∩V ] is surjective.

Proof. ⇒: U ∩ V ' ∆U∩V = ∆X ∩ (U × V ) ⊆ U × V is closed, thus U ∩ V isaffine and k[U × V ]→ k[U ∩ V ] is surjective.⇐: If U, V, U ∩ V are affine and k[U × V ] → k[U ∩ V ] is surjective, then

∆ : U∩V → U×V is an inclusion of closed subsets. Thus, ∆X∩(U×V ) ⊆ U×Vclosed. So if X×X = ∪U ×V is an open cover then ∆X ⊆ X×X is closed.

Exercises

1. X is a prevariety such that ∀x, y ∈ X there is an open affine U ⊆ X suchthat x, y ∈ U . Then X is separated.

15

2. Pn has this property.

Corollary 2.6. Quasi-projective varieties are separated.

We want to show that the products of projective varieties are again projec-tive.

Let X ⊆ Pn and Y ⊆ Pm are closed. Then X × Y ⊆ Pn × Pm is closed. Itis enough to show that Pn × Pm is projective, that is, Pn × Pm ⊂ PN is closed.

Segre Map: If N = (n + 1)(m + 1) − 1 = nm + n + m, then we can defines : Pn × Pm → PN : (x0 : . . . : xn)× (y0 : . . . : ym) 7→ (x0y0 : x0y1 : . . . : x0ym :x1y0 : . . . : xnym).

We call the projective coordinates on PN as zij for 0 ≤ i ≤ n, 0 ≤ j ≤ m.Exercise: s : Pn × Pm → V+(zijzpq − ziqzpj) ⊆ PN

Note: Pn × Pn ⊆ Pn2+2n is closed.Exercise: ∆Pn = V+(zij − zji) ⊆ PN .Complete Varieties: analogues of compact manifolds

Definition 2.4 (Complete). A variety X is complete if for any variety Y , theprojection πY : X × Y → Y is closed. (i.e.: Z ⊆ X × Y is closed implies thatπY (Z) ⊆ Y is closed)

Note: 1) closed subsets of complete varieties are complete.2) Products of complete varieties are complete.Examples:Points are complete.A1 is not complete, as Z = V (xy− 1) ⊆ A1×A1 = A2 = X ×Y is such that

πY (Z) is not closed, as it is A1 \ 0

Proposition 2.7. Let ϕ : X → Y be a morphism of varieties. If X is complete,then ϕ(X) is closed in Y and is complete.

Proof. Γ(ϕ) = (x, ϕ(x)) ∈ X × Y : x ∈ X ⊆ X × Y = (ϕ× 1)−1(∆Y )As Y is separated, Γ(ϕ) ⊆ X × Y is closed. X is complete implies that

ϕ(X) = πY (Γ(ϕ)) ⊆ Y is closed.Now, let Z ⊆ ϕ(X)× Y ′ be closed. Then

X × Y ′

Y ′

ϕ(X)× Y ′.................................................. ............ϕ× 1

............................................................................................................................................................................ ............

πY ′

.............................................................................................................................

πY ′

W = (ϕ×1)−1(Z) ⊆ X×Y ′ is closed, πY ′(Z) = πY ′((ϕ×1)(W )) = πY ′(W )is closed.

Exercise: ϕ : X → Y is a continuous map of topological spaces then X isirreducible implies that ϕ(X) is irreducible.

Proposition 2.8. If X is an irreducible complete variety then k[X] = k.

16

Proof. Let f ∈ k[X]. f : X → A1 a morphism. As X is complete, f(X) mustbe is irreducible, closed and complete. Thus, it must be a point. Thus, f isconstant.

Proposition 2.9. A complete quasi-affine variety is finite.

Proof. X is such a variety, without loss of generality X is irreducible. X ⊆ An islocally closed, then xi : X ⊆ An → k must be constant, and so x is a point.

Theorem 2.10. Pn is complete.

Note: I ⊆ S = k[x0, . . . , xn] a homogeneous ideal, then V+(I) is not emptyiff Id ( Sd for all d ∈ N.

Proof. Let Y be a variety and Z ⊆ Pn × Y be closed. Show that πY (Z) ⊆ Y isclosed.

Y = ∪Yi and open affine cover. It is enough to show that πY (Z) ∩ Yi ⊆ Yiis closed, that is, πYi(Z ∩ (Pn × Yi)) is closed. So we take πYi : Pn × Yi → Yi.

Thus, WLOG, we assume Y is affine. Then let C(Z) = (π × id)−1(Z) ⊆An+1 × Y . k[An+1 × Y ] = S ⊗ k[Y ] = k[Y ][x0, . . . , xn] = ⊕d≥0Sd ⊗k k[Y ], so itis a graded ring.

Note that (y, (a0, . . . , an)) ∈ C(Z) implies that (y, (λa0, . . . , λan)) ∈ C(Z)for λ ∈ k.

Thus, I(C(Z)) ⊆ k[An+1 × Y ] is a homogeneous ideal. We write I(C(Z)) =(f1, . . . , fm) with fi ∈ Sdi ⊗ k[Y ].

For y ∈ Y , fi(y) = fi(−, y) ∈ Sdi . We observe that y ∈ πY (Z) iff ∃x ∈ Pnsuch that (x, y) ∈ Z. This happens iff V+(f1(y), . . . , fm(y)) 6= ∅ ⊆ Pn. This istrue iff (f1(y), . . . , fm(y))d 6= Sd for all d ≥ 0.

Fix d ≥ 0, then y ∈ Y defines a linear map ΦY : ⊕mi=1Sd−di → Sd :(g1, . . . , gm) 7→

∑mi=1 fi(y)gi.

Note: Every entry of the matrix ΦY is a regular function of Y .

Now, (f1(y), . . . , fm(y))d 6= Sd iff rank(ΦY ) < dim(Sd) =(n+ dn

)which

holds iff all minors in ΦY of size(n+ dn

)vanish. Therefore, Wd = y ∈ Y :

(d1(y), . . . , fm(y))d 6= Sd ⊆ Y is closed. Finally, πY (Z) = ∩d≥0Wd, and so isclosed.

Challenge: Find a complete variety that is not projective.Exercise:

1. Let X be a topological space and W ⊆ X is a subset. W = X iff W∩U 6= ∅for all nonempty open sets U ⊆ X.

2. f : X → Y continuous and W = X and f(X) = Y then f(W ) = Y

3. X is irreducible and ∅ 6= U ⊆ X is open. Then U = X and U is irreducible.

17

Rational Map: X and Y are irreducible varieties. The ideal is that a mor-phism f : X → y is uniquely determined by restriction to any non-empty opensubset of X.

Consider pairs (U, f) where U ⊆ X nonempty and open and f : U → Y is amorphism. Relation: (U, f) ∼ (V, g) iff f = g on U ∩ V because Y is separatedand X is irreducible, this is an equivalence relation. Checking this is an exercise.

Definition 2.5 (Rational Map). A rational map f : X 99K Y is an equivalenceclass for ∼.

X irreducible implies that U ∩ W ⊇ U ∩ V ∩ W dense, and Y separatedimplies that f = h on a closed subset of U ∩W .

Remark: If f : X 99K Y then there is a maximal open U ⊆ X where f isdefined as a morphism. U = ∪(V,g)∼fV ⊂ X.

Example: f : A2 99K A2 : (x, y) 7→ (x/y, y/x2) defined as a morphism ofD(xy).

Exercise: f : A2 99K A2 ⊆ P2. Find the max open where f : A2 99K P2 isdefined.

Definition 2.6 (Rational Function). A rational function on X is a rationalmap f : X 99K A1 = k. f is given by a regular function f : U → k where∅ 6= U ⊆ X open.

k(X) = f : X 99K k is the field of rational functions on X.

Note: If (U, f), (V, g) ∈ k(X) then f + g, f − g, fg : U ∩ V → k definerational functions on X. If f 6= 0 in k[U ] then ∅ 6= D(f) ⊆ U is open, and1/f : D(f)→ k is regular. Thus, 1/f = (D(f), 1/f) ∈ k(X).

Examples: k(An) = k(x1, . . . , xn). k(Pn) = k(x1/x0, . . . , xn/x0).

Proposition 2.11. Let X be an irreducible variety

1. If ∅ 6= U ⊆ X open, then k(X) = k(U).

2. If X is affine, then k(X) = k[X]0 =field of fractions of k[X].

Proof. 1. k(X)→ k(U) : (V, g)→ (V ∩ U, g|V ∩U ) is isomorphism.

2. Define k[X]0 → k(X) : f/g 7→ (D(g), f/g).

Injective: As this is a homomorphism of fields, it is enough to say thatit is not identically zero, and it maps 1 to (X, 1), which is not the zerofunction.

Surjective: If f : U → k is regular, ∅ 6= U ⊆ X open. Find 0 6= g ∈ k[X]such that ∅ 6= D(g) ⊆ U . Then f ∈ k[D(g)] = k[X]g ⊆ k[X]x0 .

Definition 2.7 (Dominant). (U, f) : X 99K Y is dominant if f(U) = Y .

18

Indep. of rep.: If ∅ 6= V ⊆ U open, then f(V ) = Y by the homework.Suppose (U, f) : X 99K Y is dominant and (V, g) : Y 99K Z is any rational

map, then we can compose g f : X 99K Z, as f(U) = Y so f(U) ∩ V 6= ∅, sof−1(V ) 6= ∅ ⊆ U .

g f = (f−1(V ), g f).Exercise: If f, g both dominant, then g f dominant.

Proposition 2.12. X,Y irreducible varieties, then there is a one to one corre-spondence between φ : X → Y with φ dominant and field extensions k(Y ) ⊆k(X) by φ 7→ φ∗ = [h 7→ h φ]

Proof. WLOG X,Y are affine. The map φ 7→ φ∗ is:Injective: So we let ψ : X 99K Y dominant and ψ∗ = φ∗. Take D(h) ⊆ X

such that φ and ψ are both defined on D(h).

k[Y ]

k(Y )

k[X]h

k(X)

........

........

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⊆

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⊆

............................................................................................... ............

............................................................................................... ............φ∗ = ψ∗

This diagram commutes, and so φ = ψ on D(h).Surjective: Let α : k(Y ) → k(X) be a k-algebra homomorphism. k[Y ] is

generated by f1, . . . , fn. α(fi) = gi/hi, gi, hi ∈ k[X]. Let h = h1 . . . hn ∈ k[X].So α : k[Y ] → k[X]h is a k-alg hom. This gives us a morphism φ : D(h) → Yand φ∗ = α.

Definition 2.8 (Birational). Let f : X 99K Y be a rational map. It is birationalif f is dominant and there is a dominant g : Y 99K X such that f g = idY andg f = idX as rational maps.

Definition 2.9 (Birationally Equivalent). X and Y are birationally equivalent(often X and Y are birational) written X ≈ Y if there exists f : X 99K Y abirational map.

Examples: A2 ≈ P2 ≈ P1 × P1.If U, V ⊆ X open and X irred., then U ≈ Y .

Theorem 2.13. The following are equivalent

1. X ≈ Y

2. k(X) ' k(Y ) as k-algebras

3. ∃∅ 6= U ⊂ X,V ⊂ Y open such that U ' V as varieties.

Proof. 3⇒ 2 is clear from the first prop on this topic.2⇒ 1 is clear from the second prop on this topic.1 ⇒ 3: Let (U, f) : X 99K Y and (V, g) : Y 99K X be inverses. Set

U0 = f−1(V ) ⊆ U . Then g f : U0 → V → X must be the inclusion of U0 into

19

X. Thus, g(f(U0)) ⊆ U0, so f(U0) ⊆ g−1(U0). Set V0 = g−1(U0) ⊆ V . ThenU0 ' V0 as varieties by f, g.

Definition 2.10 (Rational Variety). An irreducible variety is rational if it isbirational to An for some n.

Examples: Any curve C ⊆ P2 of degree 2 is rational.Let C = V (y2−x3−x2) ⊆ A2 is rational. Let φ : C 99K A1 by φ(x, y) = y/x.

The inverse should be ψ : A1 → C by ψ(t) = (1 − t2, t − t3). So φ ψ(t) =(t− t3)/(1− t2) = t = idA1 , and the opposite is also an identity (Exercise, showthis).

Challenge: Show that E = V (y2 − x3 + x) ⊆ A2 is not rational.Big Challenge: If C is any irreducible variety and there exists a dominant

rational map A1 99K C then C is rational.Transcendence DegreeLet k ⊆ L a field extension. Then L is algebraic over k if for all f ∈ L, there

is a polynomial equation fn + a1fn−1 + . . .+ an = 0 for ai ∈ k.

S ⊆ L subseteq, then S is algebraically independent over k if for all s1, . . . , sn ∈S with si 6= sj for i 6= j, then k[x1, . . . , xn]→ L by xi 7→ si is injective.

Definition 2.11 (Transcendence Basis). A transcendence basis for L over k isa set B ⊆ L such that B is algebraically independent over k and L is algebraicover k(B).

Theorem 2.14. 1. All transcendence bases have the same cardinality.

2. If S ⊆ Γ ⊆ L subsets such that S is alg indep over k and L is alg overk(Γ), then there exists a transcendence basis B for L over k such thatS ⊆ B ⊆ Γ.

Proof. The idea is as ”any vector space has a basis.”

Exercise: Prove where L is a finitely generated extension of k.Lang’s Algebra contains the proof.

Definition 2.12 (Transcendence Degree). The transcendence degree tr degk(L) =tr deg(L) = the number of elements in any transcendence basis for L over k.

Definition 2.13 (Dimension). Let X be an irreducible variety. Then definedim(X) = tr deg(k(X)).

Examples:

1. dim(An) = tr degk k(x1, . . . , xn) = n

2. If X is irreducible and dim(X) = 0, then k ⊆ k(X) is algebraic extension,then k(X) = k. Thus, X is a point.

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Some terminology: a curve is a variety of dimension 1, a surface is a varietyof dimension 2, and an n-fold is a variety of dimension n.

Notation: IfR is a finitely generated domain over k, then we write tr deg(R) =tr deg(R0).

We will state the following without proof.

Theorem 2.15 (Principle Ideal Theorem). If R is a finitely generated domainover k and 0 6= f ∈ R and P ⊆ R is a minimal prime, then P is a minimalprime containing f . Then tr deg(R/P ) = tr deg(R)− 1.

Geometric Statement: If X is any irreducible variety and 0 6= f ∈ k[X] andif Z ⊆ V (f) is an irreducible component then dimZ = dimX − 1.

Proof. Take U ⊆ X open affine such that U ∩Z 6= ∅. Then Z ∩U = V (P ) ⊆ U ,P ⊆ k[U ] prime ideal. Z is a component of V (f) ⇒ P is minimum over(f) ⊆ k[U ].

Thus, dim(Z) = tr deg(k[U ]/P ) = tr deg k[U ]− 1 = dimX − 1.

Theorem 2.16. Let X be an irreducible variety, and let ∅ 6= X0 ( X1 ( . . . (Xn = X be a maximal chain of irreducible closed subsets. Then dim(X) = n.

Proof. WLOG, X is affine. Take 0 6= f ∈ I(Xn−1). Then Xn−1 ⊆ V (f) is acomponent. PIT says that dimXn−1 = dimX − 1.

Induction implies that dimXn−1 = n− 1, so dimX = n.

Definition 2.14. If X is any variety, set dim(X) =the supremum of all n suchthat ∃ a chain ∅ 6= X1 ( X1 ( . . . ( Xn ⊆ X where Xi ⊆ irreducible and closedfor all i.

Exercises

1. X = X1 ∪ . . . ∪Xm and Xi ⊆ X closed, then dimX = max dim(Xi)

2. dim(X × Y ) = dimX + dimY

Recall: If R is a ring, then dim(R) is the supremum of all n such that ∃Pn ( Pn−1 ( . . . ( P0 ⊆ R where Pi is a prime ideal.

Note: If X is affine then dimX = dim k[X].

Theorem 2.17 (PIT For Several Equations). If X is an irreducible varietyand f1, . . . , fr ∈ k[X] and Z ⊆ V (f1, . . . , fr) are components, then dimZ ≥dimX − r.

Proof. Enough to show that if W ⊆ X is a closed subset and each componentof W has dim ≥ d, then each component of W ∩ V (f) has dim ≥ d − 1 for allf ∈ k[X].

Let Z ⊆W be a component. If f |Z = 0 then V (f)∩Z = Z. If f |Z 6= 0 thenevery component of Z ∩ V (f) has dim= dim(Z)− 1 ≥ d− 1.

Therefore W ∩ V (f) = union of finitely many irreducible closed subsets ofdim ≥ d− 1.

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Lemma 2.18 (Prime Avoidance). X is an affine variety, Z ⊆ X an irreducibleclosed subset and X1, . . . , Xm ⊆ X are also irreducible closed subsets, then ifXi 6⊆ Z then ∃f ∈ I(Z) such that f /∈ I(Xi).

Proof. Induction on m.If m = 1, then X1 6⊆ Z ⇒ I(Z) 6⊆ I(X1)⇒ ∃f ∈ I(Z) \ I(X1).For m ≥ 2, take fi ∈ I(Z) such that fi /∈ I(Xj) for j 6= i. If any fi /∈ I(Xj),

then done. Take f = fi.If fi ∈ I(Xi) for all i, then f = f1 + f2f3 . . . fm ∈ I(Z).

Definition 2.15 (Codimension). If X is any variety, Z ⊆ X closed and irre-ducible, let X1, . . . , Xm be the components of X containing Z. Set codim(Z;X) =dim(X1 ∪ . . . ∪Xm)− dimZ.

E.g. X is the union of a line and a plane, Z is a single point of X. Thencodim(Z;X) is 2 if it is a point in the plane, 1 otherwise.

Theorem 2.19 (Reverse PIT). X affine, Z ⊆ X irreducible closed and c =codim(Z;X). Then ∃f1, . . . , fc ∈ k[X] such that Z ⊆ V (f1, . . . , fc) irreduciblecomponent.

Proof. If Z is a component of Z, then c = 0.Otherwise, no components of X are contained in Z, so the lemma implies

that there exists f1 ∈ k[X] such that f1 ∈ I(Z) and f1 does not vanish on anycomponent of X. PIT implies that codim(Z;V (f)) < c.

Induction on c gives us that there are f2, . . . , fc ∈ I(Z) such that Z is acomponent of V (f1, . . . , fc).

ResultantsLet K be an arbitrary field, and f(T ) = anT

n + . . .+ a1T + a0 and g(T ) =bmT

m + . . .+ b1T + b0 ∈ K[T ].Q: Do f(T ) and g(T ) have a common factor?

Set A =

an a2 a1 a0 0

an a2 a1 a0

bm b1 b0bm b1 b0

bm b1 b0

.

Definition 2.16 (Resultant). We define Res(f, g) = detA ∈ K.

Let ~v = (cm−1, c0, dn−1, d1, d0) ∈ Kn+m, then ~v ·A = (rn+m−1, . . . , r1, r0) ∈Kn+m. Then (cm−1

m−1 + . . .+ c1T + c0)f(T ) + (dn−1Tn−1 + . . .+d1T +d0)g(T ) =

rm+n−1Tm+n−1 + . . .+ r1T + r0.

Proposition 2.20. Suppose an 6= 0, then Res(f, g) 6= 0 ⇐⇒ (f, g) = 1 ∈K[T ].

Proof. Res(f, g) = 0 iff ∃~v ∈ Km+n such that ~v · A = 0 iff ∃p(T ), q(T ) of deg≤ m− 1, n− 1 such that p(T )f(T ) = q(T )g(T ), iff (f, g) 6= 1.

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If f(T ) =∑ni=0 aiT

i then we allow formal differentiation, that is, f ′(T ) =∑ni=1 iaiT

i−1 ∈ K[T ].Note that (fg)′ = f ′g + fg′, and similar rules still hold.

Corollary 2.21. If f(T ) = anTn + . . . + a1T + a0 then f(T ) has n different

roots in K iff Res(f, f ′) 6= 0.

Proof. f(T ) = an∏ri=1(T − αi)di , where αi 6= αj for i 6= j. Then f ′(T ) =

an∑ri=1 di(T − αi)di−1

∏j 6=i(T − αj)dj .

Res(f, f ′) = 0 iff (f, f ′) 6= 1 iff f ′(α`) = 0 for some ` iff d` ≥ 2 for some`.

Definition 2.17 (Discriminant). The discriminant of f(T ) is Res(f, f ′).

Exercise: a 6= 0 and f(T ) = aT 2 + bT + c then discriminant= −a(b2 − 4ac).Remark: If char(K) = 0 and if f(T ) ∈ k[T ] is an irreducible polynomial,

then (f(T ), f ′(T )) = 1 so Res(f, f ′) 6= 0. In char(K) = p, then f ′(T ) may bezero, for example (T p + 1)′ = 0.

Remark: ϕ : X → Y is a morphism, then dim(ϕ(X)) ≤ dim(X). This is ask(ϕ(X)) ⊆ k(X).

Theorem 2.22. φ : X → Y is a dominant morphism of irreducible varieties,such that k(Y ) ⊆ k(X) is a finite extension of degree d. Suppose that char(k) = 0or k(X)/k(Y ) is separable. Then ∃ dense open V ⊆ Y such that |φ−1(y)| = dfor all y ∈ V .

Proof. Assume X,Y affine and k[X] = k[Y ][f ]. Let P (T ) = adTd + . . . +

a1T + a0 ∈ k(Y )[T ] be the minimum polynomial for f ∈ k(X) over k(Y ). ieP (f) = 0 ∈ k(X).

WLOG, ai ∈ k[Y ] for all i and we can replace Y with D(ad) and X withφ−1(D(ad)) ⊆ X. We may assume that ad = 1. Now k[X] = k[Y ][T ]/(P (T )).This implies that X ' V (P ) ⊆ Y ×A1 πY→ Y , and φ : X → Y goes through thispath.

If (y, t) ∈ Y × A1 then we set Py(T ) =∑di=0 ai(y)T i ∈ k[T ]. (y, t) ∈ X iff

Py(t) = 0. Let ∆ = Res(P, P ′) ∈ k[Y ]. P (T ) irreducible and char(k) = 0 implythat ∆ 6= 0. Note that Res(Py, P ′y) = ∆(y).

Thus, if y ∈ D(∆), then Py(t) = 0 has exactly d solutions.Now the general case: X and Y are irreducible varieties and φ : X → Y dom-

inant. Let V ⊆ Y and U ⊆ φ−1(V ) ⊆ X be open affines. Then dim(φ(X \ U)) ≤dim(X \ U) < dim(X) = dim(Y ). Thus φ(X \ U) ( Y , and so ∃h ∈ k[V ] suchthat D(h) ∩ φ(X \ U) = ∅.

We can replace X with D(φ∗h) and Y with D(h). And so, WLOG, X,Yaffine.

φ dominant implies that k[Y ] ⊆ k[X] and k[X] generated by f1, . . . , fn.Then k[Y ] ⊆ k[Y ][f1] ⊆ . . . ⊆ k[Y ][f1, . . . , fn] = k[X] gives X = Xn → Xn−1 →. . .→ X1

ψ→ Y a sequence of dominant maps.Induction on n: ∃ a dense open U ⊆ X1 such that all points of U are hit by

d1 = [k(X) : k(X1)] pts of X.

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As above: ψ(X1 \ U) ( Y implies ∃h ∈ k[Y ] such that D(h)∩ψ(X1\U) = ∅,and so ψ−1(D(h)) = D(φ∗h) ⊆ U . ψ : D(φ∗h) → D(h) gives k[D(φ∗h)] =k[X]φ∗h = k[X]h. Thus, k[Y ][f1]h = k[Y ]h[f1] = k[D(h)][f1].

Thus, the first case implies that ∃∅ 6= V ⊆ D(h) ⊆ Y open such that|ψ−1(y)| = [k(X1) : k(Y )]. Since ψ−1(D(h)) ⊆ U we have |φ−1(y)| = [k(X1) :k(Y )] · d1 = d.

Exercise: πY : X × Y → Y is an open map. That is, if U ⊆ X × Y is openthen πY (U) is open in Y .

Corollary 2.23. φ : X → Y is a dominant morphism of irreducible varieties.Then φ(X) contains a dense open subset of Y .

Proof. We can assume that X,Y are affine. Choose B = f1, . . . , fn ⊆ k[X]such that B is a transcendence basis for k(X)/k(Y ).

Then k[Y ] ⊆ k[Y ][f1, . . . , fn] ⊆ k[X] gives Xψ→ Y × An πY→ Y and φ is the

composition.The theorem says that there is a open subset U ⊆ Y × An such that U ⊆

ψ(X). As πY is an open mapping, πY (U) is open.

Definition 2.18. Let X be a variety.

1. W ⊆ X is locally closed if W = open ∩ closed.

2. W ⊆ X is constructible if W =the union of finitely many locally closedsubsets.

Example: W = D(xy)∪0 ⊆ A2. Notice: φ : A2 → A2 : (x, y) 7→ (x2y, xy),then φ(A2) = W .

Exercise: φ : X → Y is an arbitrary morphism of varieties, then φ(X) isconstructible.

3 Nonsingular Varieties

Local Rings

Definition 3.1 (Local Ring at a point). If X is an irreducible variety and x ∈ Xthen OX,x = f ∈ k(X) : f(x) defined. This is a local ring with maximal idealmx = f ∈ OX,x : f(x) = 0.

Note: If U ⊆ X is any open subset, then k[U ] =⋂x∈U OX,x ⊆ k(X).

Let U ⊆ X open affine, x ∈ U then M = I(x) ⊆ k[U ]. f ∈ OX,x then f isdefined on D(h) ⊆ U for some h ∈ k[U ] \M , so f = g/hn where g ∈ k[U ]. Andso, OX,x = g/h : g, h ∈ k[U ], h(x) 6= 0 = k[U ]M .

Remark: X not irreducible implies that OX,x = lim−→U3x OX(U). If U ⊆ X

open affine, x ∈ U we still have that OX,x = k[U ]I(x).Note: If X is irreducible then dim(X) = dim OX,x for any x ∈ X. If we let

X be any variety, then dim(X) = maxx∈X dim OX,x.

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Definition 3.2 (Regular Local Ring). If (R,m) is a local ring, then F = R/mis a field called the residue field of R and m/m2 is an F -vector space.

R is a regular local ring if dimF (m/m2) = dimR.

Exercise: If R is a Notherian Local Ring then dimF (m/m2) =min numberof generators for m ≥ dim(R). The equality is from Nakayama, the inequalityfrom PIT.

Definition 3.3. Let X be a variety and x ∈ X.

1. X is nonsingular at x if OX,x is a regular local ring.

2. Otherwise, X is singular at x.

3. X is nonsingular if all points x ∈ X are nonsingular.

Exercise: S is a commutative ring, and M ⊆ S is a maximal ideal, thenset R = SM a local ring. Then unique maximal ideal m = MSM . Show thatS/M ' R/m and M/M2 = m/m2.

Example: Let C = V (f) ⊆ A2 a curve. Let P = (0, 0) ∈ C. So f = ax+by+HOT . And k[C] = k[x, y]/(f). Let M = I(P) = (x, y) = (x, y)/(f) ⊆ k[C].

mP /m2P = M/M2 = (x, y)/(x2, xy, y2, f) = (x, y)/(x2, xy, y2, ax + by). So

P ∈ C is nonsing iff dimk(M/M2) = dim(C) = 1. This is iff ax + by 6= 0 ∈k[x, y], note that ax + by 6= 0 implies that V (f) looks like V (ax + by) close tothe point.

Exercise: Find all singular points of V (y2 − x3 − x2), V (y2 − x3).X ⊂ An a closed affine variety. Then I = I(X) = (f1, . . . , ft) ⊆ S =

k[x1, . . . , xn]. Idea: X is nonsing at P ∈ X iff X has a tangent space at P .

Definition 3.4 (Jacobi Matrix). Let JP =[∂fi∂xj

(P )]

be a t× n matrix, we callthis the Jacobi matrix.

Note: If ~v ∈ kn then Jp · ~v ∈ kt is the partial derivative of (f1, . . . , ft) at pin the direction ~v.

ker(JP ) = ~v ∈ kn : Jp · ~v = ~0 is a candidate for a tangent space.

Lemma 3.1. Let P ∈ X ⊆ An, then rank(Jp) + dimk(mP /m2P ) = n.

Proof. SetM = I(P) ⊆ S. Define d : M → kn by d(f) = ( ∂f∂x1(P ), . . . , ∂f∂xn (P )).

d is surjective as d(xi − pi) = ei. Note: f, g ∈ S that d(fg) = f(P )d(g) +d(f)g(P ). So d(M2) = 0. Thus, d : M/M2 → kn is an isomorphism if it isinjective. It is injective as the two vector spaces are of the same dimension andit is surjective.

d(fi) = ith row of JP and d(∑gifi) =

∑gi(P )d(fi) so d(I) =row span of

JP in kn. Thus d : I + M2/M2 → row span of JP is an isomorphism. Thusrank(Jp) + dim(M/I + M2) = dim(M/M2) = n. Finally OX,P = (S/I)M andmP /m

2P = (M/I)/(M/I)2 = M/I +M2.

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Theorem 3.2. P ∈ X ⊆ An, then rank(JP ) ≤ n − dim(OX,P ) and rank JP =n− dim OX,P ⇐⇒ P nonsingular.

Proof. rank(Jp) = n− dim(mP /m2P ) ≤ n− dim OX,P .

We have equality iff dimk(mP /m2P ) = dim OX,P .

Example: X = V (z2 − x2y2) ⊆ A3 where char(k) 6= 2, 3. Then J =[−2xy2,−2x2y, 3z2]. p ∈ X is a nonsingular point iff rank(Jp) = 3 − x = 1,so Xsing = V (z3 − x2y2, xy2, x2y, z2) = V (xy, z).

Exercise: X is a variety and p ∈ X. Then OX,P is a domain iff p is in onlyone component.

Theorem 3.3. Any Notherian regular local ring is a domain. (in fact, a UFD,and even Macaulay)

Conclude: The points on an intersection of two components are singular.

Proposition 3.4. Xsing ⊆ X is a closed subset of X.

Proof. Let X = X1 ∪ . . . ∪ Xm be the components of X. Then Xsing =⋃mi=1(Xi)sing ∪

⋃i6=j Xi ∩ Xj . The latter are closed, so without loss of gen-

erality, X irreducible and affine.X ⊆ An closed, I(X) = (f1, . . . , ft) ⊆ k[x1, . . . , xn], P ∈ Xsing iff p ∈ X

and rank(Jp) < n− dim OX,p = n− dim(X).Let m1, . . . ,mN be all of the minors of size n−dim(X) in J = [ ∂fi∂xj

]. Xsing =X ∩ V (m1, . . . ,mN ).

Fact: Xsing 6= X.

Lemma 3.5. If p ∈ X = V (g1, . . . , gr) ⊆ An and if rank Jp(g1, . . . , fr) = rthen OX,p is regular local of dimension n− r.

Proof. PIT implies that dim OX,p ≥ n− r.I(X) = (f1, . . . , ft) ⊇ (g1, . . . , gr). Thus, row span Jp(f1, . . . , ft) ⊇ row span

Jp(g1, . . . , gr), so r = rank Jp(g1, . . . , gr) ≤ rank Jp(f1, . . . , ft) ≤ n−dim OX,p ≤r.

Theorem 3.6 (Implicit Function Theorem). If f1, . . . , fc are holomorphic func-tions in a classical nbhd of p ∈ Cn. Suppose det

(∂fi∂xj

(p))

1≤i,j≤c6= 0. Then ∃

holomorphic functions w1, . . . , wc on classical open subset of Cn−c and classicalopen subset V ⊆ Cn such that p ∈ V and so that for all z ∈ V , f1(z) = . . . =fc(z) = 0 iff z − i = wi(zc+1, . . . , zn) for all 1 ≤ i ≤ c.

Theorem 3.7. X ⊆ Cn a complex affine variety. p ∈ X a nonsingular point.Then a classical neighborhood of p in X is holomorphic to a classical open subsetof Cd where d = dim OX,p.

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Proof. WLOG, X is irreducible. I(X) = (f1, . . . , ft) and rank Jp(f1, . . . , ft) =n − d = c. WLOG, det( ∂fi∂xj

(p)) 6= 0. Set Y = V (f1, . . . , fc) ⊆ Cn, thenrank Jp(f1, . . . , fc) = c implies that p is a nonsingular point of Y . So OY,p isregular local of dimension d.

Now, only one component of Y contains p, p ∈ X and X ⊆ Y . dimXis the same as the dimension of the component of Y containing p, and as Xis irreducible, X is the component of Y containing p. Then there exists openU ⊆ An such that X ∩ U = Y ∩ U = V (f1, . . . , fc) ∩ U . We apply the IFT, letV ⊂ U , p ∈ V , and w1, . . . , wc be as in the IFT.

Define π : Cn → Cd by π(z) = (zc+1, . . . , zn). Then π : X∩V → π(X∩V ) ⊆Cd is an holomorphism, as we can get an inverse map π−1 : π(X ∩ V )→ X ∩ Vby (zc+1, . . . , zn) 7→ (w1(z), . . . , wc(z), zc+1, . . . , zn).

Corollary 3.8. Every nonsingular complex variety variety is a complex mani-fold.

If X is an affine variety, recall that pts in X are in 1-1 correspondence withmax ideals in k[X].

So X an irreducible variety, p ∈ X corresponds to local rings OX,P = f ∈k(X) : f(P ) defined .

Lemma 3.9. X an irred var, x, y ∈ X, if OX,x ⊆ OX,y then x = y.

Proof. Take open affine x ∈ U ⊆ X, y ∈ V ⊆ X. X is separated, so U ∩ V isaffine and k[X]⊗ k[Y ]→ k[U ∩ V ] is surjective, and k[U ] ⊆ OX,x ⊆ OX,y withk[V ] ⊆ OX,y, thus, k[U ∩ V ] ⊆ OX,y.

k[V ] ⊆ k[U ∩ V ] ⊆ OX,y proves that y ∈ U ∩ V . Then k[U ∩ V ] ∩ my is amax ideal in k[U ∩ V ] so it corresponds to a point in U ∩ V which maps to yunder U ∩ V → V the inclusion.

Thus, x, y ∈ U ⊆ X. U is affine. If x 6= y, then there is an f ∈ k[U ] suchthat f(x) 6= 0, f(y) = 0. Then 1

f ∈ OX,x and 1f /∈ OX,y.

Nonsingular Curves

Definition 3.5 (Curves). A curve is an irreducible variety of dimension one.

Example: C = A1. k[C] = k[t] and k(C) = k(t). Let 0 6= f ∈ k(t). What isthe order of vanishing of f at 0?

f = p/q, p, q ∈ k[t], we can write p = tnp0 and q = tmq0 with p0(0) 6= 0 andq0(0) 6= 0. f = (p0/q0)tn−m, so v(f) = n−m is the order of vanishing.

Note that OC,0 = k[t](t) with m0 = (t) ⊆ OC,0 is a maximal ideal. Thenf = utn−m where u = p0/q0 is a unit in OC,0.

Definition 3.6 (Discrete Valuation Ring). A discrete valuation ring, or DVR,is a Notherian regular local ring of dimension 1.

Examples are OC,P where C is a curve and P a nonsingular point.Let (R,m) be a DVR, m = (t), t is a uniformizing parameter, and K = R0

is the field of fractions of R.

27

Claim: Any f ∈ K∗ = K \ 0 can be written f = utn where u ∈ R \ m aunit in R and n ∈ Z. WLOG, f ∈ R \ 0.

Assume that the claim is false, then choose a counterexample such that (f)is maximal. f is not a unit implies that f ∈ m, so f = gt for some g ∈ R.

(f) ( (g) as (f) = (g)⇒ g = hf ⇒ f = thf ⇒ th = 1, contradiction.So g = utn, u ∈ R a unit implies that f = utn+1.Check that if f = utn, u a unit, then u, n are unique. n = minp ∈ Z : f ∈

(tp) ⊆ K.

Definition 3.7 (Valuation Map). v : K∗ → Z is a valuation map v(f) = n iff = utn with u ∈ R \m.

Note that R = f ∈ K∗ : v(f) ≥ 0∪0 and m = f ∈ K∗ : v(f) > 0∪0.Rules: v(fg) = v(f) + v(g). v(f + g) ≥ min(v(f), v(g)) if f, g, f + g ∈ K∗.

If f = utn, g = vtm and n ≤ m, then f + g = (u+ vtm−n)tn = u′trtn.Example: C a curve, p ∈ C nonsing, then R = OC,p is a DVR with K =

R0 = k(C). vp : K(C)∗ → Z is a valuation, f ∈ k(C)∗ vp(f) =the order ofvanishing of f at p. vp(f) > 0 iff f ∈ mp, vp(f) = 0 iff f(p) 6= 0 and vp(f) < 0iff f is not defined at p.

Lemma 3.10. Let R be a DVR, R0 = K and S any ring such that R ⊆ S ⊆ K.Then S = R or S = K.

Proof. If S 6= R, then take f ∈ S, f /∈ R. m = (t), f = utn, n < 0, S ⊇ R[f ] =K.

Recall: A domain R is called integrally closed iff for all f ∈ R0, f is integralover R implies that f ∈ R.

Theorem 3.11. If R is any Notherian Local Domain of dimension one, thenR is regular iff R is integrally closed.

Exercise: Prove ⇒ (easy, as DV R ⇒ PID ⇒ UFD ⇒ int closed, so provethe last one)

Proposition 3.12. Let A be a domain. A is integrally closed iff AP is integrallyclosed for all maximal ideals P ⊆ A.

Proof. ⇒: Easy⇐: A = ∩P⊆AAP over maximal ideals P .

Note: A a f.g. domain over k, X = Spec−m(A), then A = k[X] =∩p∈XOX,p = ∩Ap.

Definition 3.8 (Dedekind Domain). A Dedekind Domain is an integrally closedNotherian domain of dimension 1.

Note: If X is an irred affine variety, X nonsingular curve iff k[X] is aDedekind domain.

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OX,p a DVR for all p ∈ X iff k[X]p integrally closed for all max ideals p bythe theorem, each of these is integrally closed iff k[X] is, and that is the def ofa Dedekind domain.

Finiteness of Integral ClosureLet R be a finitely generated domain over k, K = R0, and K ⊆ L a finite

field extension. Then R is defined to be the integral closure of R in L. That is,f ∈ L : f integral over R. Fact R is an integrally closed domain with field offractions L.

Let f ∈ L. Then fn + a1fn−1 + . . .+ an = 0 with ai ∈ K. Take b ∈ R such

that bai ∈ R for all i. Then (bf)n + ba1(bf)n−1 + . . .+ bnan = 0. Thus bf ∈ R.b ∈ R, so f ∈ (R)0.

Theorem 3.13 (Finiteness of Integral Closure). R is a finitely generated R-module.

In particular: R is a finitely generated K algebra.

Definition 3.9 (DVR of K/k). k ⊆ K is a field extension, a DVR of K/k is asubring R ⊆ K such that:

1. R is a DVR

2. R0 = K

3. k ⊆ R.

Let K be a function field of dimension 1 over k.i.e., k ⊆ K is finitely generated as a field extension and it has transcendence

degree 1.Q: ∃ a nonsingular curve C such that k(C) = K?Key construction: Let f ∈ K \ k. Then k(f) ⊆ K is a finite field extension.

It is finitely generated by assumption, and f must be a transcendence basisfor K/k, thus it is algebraic, and so it is a finite field extension.

Set B = k[f ] ⊆ K.Finiteness of integral closure says that B is a finitely generated k-algebra,

and so B is a Dedekind domain with B0 = K.

Proposition 3.14. X = Spec−m(B) is a nonsingular curve.

1. Points in X are in 1-1 correspondence with DVRs R of K/k such thatf ∈ R.

2. Points in V (f) correspond to the DVRs R of K/k such that f ∈ mR.

Proof. X → DVRs R 3 f is well-defined and injective.Let R be a DVR of K/k with f ∈ R.k[f ] ⊆ R⇒ B ⊆ R.Set M = B ∩mR ⊆ B a prime ideal. Then BM ⊆ R 6= K ⇒ M 6= 0 and so

BM is a DVR of K/k.And so, the lemma implies that BM = R.Now we prove (b). M ∈ V (f) ⇐⇒ f ∈M ⇐⇒ f ∈MBM = m.

29

Corollary 3.15. Every DVR R of K/k is the local ring of a nonsingular curveat some point. In particular, R/mR = K.

Proof. Let f ∈ R \ k, B = k[f ] ⊆ K.Then R = BM for some maximal ideal M ∈ Spec−m(B). So R = OX,p.

Corollary 3.16. Given f ∈ K∗, there are only finitely many DVRs R of K/ksuch that f ∈ mR.

Also have finitely many R such that f /∈ R.

Proof. WLOG, f /∈ k. Set B = k[f ] ⊆ K. R : f ∈ mR is in correspondencewith V (f) ⊆ Spec−m(B). PIT implies that dimV (f) = 0. Thus, V (f) is afinite set.

Note: f /∈ R iff 1/f ∈ mR.

Definition 3.10. CK = DVRs of K/kElements of CK will be called ”points” P . DVR given by P is RP with

maximal ideal mP .

We define a topology on CK to have as closed sets the finite sets and all ofCK . We also let f ∈ K and P ∈ CK , assume f ∈ RP .

Definition 3.11. f(P ) is defined to be the image of f by RP → RP /mP = k.i.e. f(P ) ∼= f mod mP .

The regular functions on a nonempty open subset U ⊆ CK are then the setk[U ] = ∩p∈URP ⊆ K.

This makes CK a SWF.Note: If U ⊆ CK open, f ∈ k[U ] then D(f) = P ∈ U : f(P ) 6= 0 = P ∈

U : f /∈ mP is open by the second corollary.Example Let K = k(t). The DVRs of K/k are k[t](t−a) for a ∈ k and

k[1/t](1/t).Then CK and P1 are in 1-1 correspondence, with k[t](t−a) corresponding to

a ∈ A1 and the other point corresponding to the point at infinity.Note: If f ∈ k[t](t−a) then f(t) ∼= f(a) mod (t− a), f(k[t](t−a)) = f(a).

Theorem 3.17. CK is a non-singular curve and k(CK) = K.

Proof. Take any f ∈ K \ k. B = k[f ] ⊆ K. U = p ∈ CK : f ∈ RP ⊆ CK isopen.

We define φ : Spec−m(B)→ U by M 7→ BM . The prop implies that this isbijective.

φ is a homeomorphism, as the closed sets are the finite sets in both. Itremains to show that this is a morphism of spaces with functions.

For V ⊆ Spec−m(B) open, then k[V ] = ∩M∈VBM = ∩P∈φ(V )RP =k[φ(V )].

Thus, φ : Spec−m(B)→ U is an isomorphism of spaces with functions.Note: If P ∈ CK then f ∈ RP or 1/f ∈ RP . Thus, Ck = Spec−m(k[f ]) ∪

Spec−m(k[f−1]). This is an open affine cover, and so Ck is a prevariety.

30

Let P,Q ∈ CK . Enough to find an open affine U ⊆ CK such that P,Q ∈ U .Take f ∈ RP \mP and f /∈ k. (f = 1 + t, (t) = mP ).

If f ∈ RQ then P,Q are both in Spec−m(k[f ]).Otherwise, 1/f ∈ RQ, so both are in Spec−m(k[1/f ]).Thus CK is a variety. It is nonsingular and of dimension one by construction.

We must show that it is irreducible.CK is, in fact, irreducible because its proper closed sets are finite and CK is

infinite.

Proposition 3.18. Let C be an irreducible curve and P ∈ C a nonsingularpoint. Let Y be any projective variety and φ : C \ P → Y is any morphism ofvarieties. Then ∃! extension φ : C → Y .

Note: No points in Y are ”missing”.

Proof. Y ⊆ Pn closed subset. It is enough to make φ : C → Pn.WLOG φ(C \ p) 6⊆ V+(xi) for all i. Set U = D(x0, x1, . . . , xn) ⊆ Pn.

φ(C \ p) ∩ U 6= ∅.Set fij = xi/xj φ ∈ k(C). Defined on φ−1(U) 6= ∅.vP : k(C)∗ → Z is the valuation given by OC,P .Set ri = vP (fi0) for 0 ≤ i ≤ n. Choose j such that rj minimal.As xi

xj= xi/x0

xj/x0, fij = fi0/fj0, so vP (fij) = ri − rj ≥ 0. Thus, fij ∈ OC,P for

all i. Note that if Q ∈ φ−1(U), then φ(Q) = (f0j(Q) : f1j(Q) : . . . : fjj(Q) = 1 :. . . : fnj(Q)).

We can then extend φ to P by this expression. Then φ is a morphism onφ−1(U) ∪ p and so φ : C → Pn is a morphism.

Lemma 3.19. R ⊆ K is a local ring, k ⊆ R, R is not a field. Then R iscontained in some discrete valuation ring of K/k.

Proof. Set B = R ⊆ K. Lying over implies that there exists some maximalideal M ⊆ B such that M ∩R = mR.

Claim: BM is a DVR of K/k.Let 0 6= f ∈ mR. S = k[f ] is a Dedekind domain, S ⊆ B. M = M ∩ S is a

maximal ideal of S.Thus SM is a DVR of K/k. SM ⊆ BM ( K and a lemma from before says

that we have equality of SM and BM .

Theorem 3.20. CK is a projective curve.

Proof. Let f ∈ K \ k. U = Spec−m(k[f ]), V = Spec−m(k[f−1]), and CK =U ∪ V an open affine cover.

U ⊆ AN closed. U ⊆ PN projective closure.The proposition implies that the inclusion U → U extends to a morphism

ϕ1 : CK → U .Similarly, we take V to be the projective closure of V . Then V → V extends

to ϕ2 : CK → V .

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We now define ϕ : Ck → U × V by ϕ(P ) = (ϕ1(P ), ϕ2(P )). Set Y =ϕ(CK) ⊆ U × V .

Y is a projective variety.Claim: ϕ : CK → Y is an isomorphism.Note: ϕ(U) ⊆ U × V is a closed subset. Let ψ = ϕ2 × id : U × V → V × V .

ϕ(U) = (u, v) ∈ U × V : ϕ2(u) = v = ψ−1(∆V ) closed. Thus, ϕ(U) =Y ∩ (U ×V ), which implies that ϕ : U → Y ∩ (U ×V ) is bijective, isomorphism.

So πU : Y ∩ (U × V )→ U is the inverse.Similarly, ϕ : V → Y ∩ (U × V ) is an isomorphism.Note: k(Y ) = k(CK) = K, and forall P ∈ CK , OY,ϕ(P ) = RP ⊆ K. Thus, ϕ

is injective.For surjective, let y ∈ Y . Then k ⊆ OY,y ⊆ K is a local ring. By the lemma,

OY,y ⊆ RP for some P ∈ CK . OY,y ⊆ RP = OY,ϕ(P ) so y = ϕ(P ).

Corollary 3.21. Any curve is birational to some nonsingular projective curve.

Corollary 3.22. X is any nonsingular curve, then X ∼= some open subset ofCK , K = k(X).

Proof. ϕ : X → CK by ϕ(x) = P where P ∈ CK such that RP = OX,x ⊆ K.Injectivity is clear. Claim: ϕ(X) ⊆ CK is open. Take U ⊆ X open affine.

k[U ] is generated by f1, . . . , fn. P ∈ ϕ(U) iff k[U ] ⊆ RP , iff fi ∈ RP for all i.Thus, ϕ(U) = ∩ni=1 Spec−m(k[fi]) open. Thus ϕ(X) ⊆ CK is open.ϕ : X → ϕ(X) is a homeomorphism. To check that ϕ is an isomorphism,

then U ⊆ X open gives k[U ] ∩x∈U OX,x = ∩P∈ϕ(U)Rϕ(x) = k[ϕ(U)].

Exercise: Two nonsingular projective curves are isomorphic iff they have thesame function field.

Degree of Projective Varieties in PnBezout: f1, . . . , fn ∈ k[x0, . . . , xn] homogeneous of degrees d1, . . . , dn. Then

V+(f1, . . . , fn) has cardinality at most d1 . . . dn or is infinite. If it is finite andcounted with multiplicity, then it is equal to d1 . . . dn.

Classical Definition: X ⊆ Pn closed, then deg(X) = #(X∩V ) where V ⊆ Pnis a linear subspace with dimV + dimX = n.

e.g. f ∈ S = k[x0, . . . , xn] a square-free homogeneous polynomial. Then#(V+(f)∩ general line) = deg f .

Warning: V+(xz−y2), V+(x) ⊆ P2. These are isomorphic but have differentdegrees. So degree is not a property of a projective variety, but rather one ofthe embedding into projective space.

Example: f ∈ S is square-free, deg f = d. Then X = V+(f), I(X) = (f),R = S/(f) is the projective coordinate ring.

Note: dimk(Sm) =(m+ nn

), where Sm is all forms of degree m. This is

1n! (m+n)(m+n− 1) . . . (m+ 1), which is actually a polynomial in m of degreen with lead coefficient 1

n! .

32

Consider 0→ Sf→ S → R→ 0 implies that 0→ Sm−d → Sm → Rm → 0 is

exact. Then dimRm = dimSm−dimSm−d, which is(m+ nn

)−(m+ n− d

n

),

which is 1n! (m + n) . . . (m + 1) − 1

n! (m + n − d) . . . (m + 1 − d). Which is apolynomial of degree n− 1.

This has lead coefficient 1n! (∑i−∑

(i− d)) = ndn! = d

(n−1)! .Recall: A graded S-module is a module M with a decomposition M =

⊕d∈ZMd as abelian group such that SmMd ⊂Mm+d.

Definition 3.12. Ann(M) = f ∈ S : fM = 0 is a homogeneous ideal,Supp(M) = V+(Ann(M)) ⊆ Pn.

Reason for Supp is x ∈ Pn, P = I(x) ⊆ S is a homogeneous prime ideal.MP 6= 0 iff P ⊇ Ann(M), iff x ∈ Supp(M).

Example: X ⊆ Pn closed, then Ann(S/I(X)) = I(X), so Supp(S/I(X)) =V+(I(X)) = X.

Note: Md is a k-vector space for all d ∈ Z because S0Md ⊆ Md. If M is afinitely generated graded S-module, then dimkMd <∞ forall d.

Definition 3.13 (Hilbert Function). HM (d) = dimkMd is the Hilbert functionof M .

Theorem 3.23. If M is a finitely generated graded S-module then ∃!PM (z) ∈Q[z] such that PM (d) = dimk(Md) for all d sufficiently large. We call PM (z)the Hilbert Polynomial.

Definition 3.14 (Numerical Polynomial). P (x) ∈ Q[z] is a numerical polyno-mial if P (d) ∈ Z for all d sufficiently large in Z.

Example:(zm

)= 1

m!z(z − 1) . . . (z −m+ 1).

Note:(

zm

): m ∈ N

is a basis over Q for Q[z].

Lemma 3.24. P (z) = c0 + c1

(z1

)+ . . .+ cr

(zr

)∈ Q[z], ci ∈ Q. Then TFAE

1. P (z) is a numerical polynomial.

2. P (d) ∈ Z for all d ∈ Z.

3. ci ∈ Z.

Proof. 3⇒ 2⇒ 1 are easy, so we need 1⇒ 3.

We know that(z + 1m

)−(zm

)=(

zm− 1

). Thus P (z + 1) − P (z) =

c1 + c2

(z1

)+ . . .+ cr

(z

r − 1

).

We perform induction on r: P (z) is numeric, then P (z + 1) − P (z) is alsonumeric. Thus, c1, . . . , cr ∈ Z, and so c0 must also be an integer.

33

Theorem 3.25 (Affine Dimension Theorem). X,Y ⊆ An closed and irreducible,then Z ⊆ X ∩ Y component has dimZ ≥ dimX + dimY − n.

Proof. X ∩Y = (X×Y )∩∆An = (X×Y )∩V (x1−y1, . . . , xn−yn) ⊆ An×An,use PIT.

WARNING: Does not prevent X ∩ Y = ∅.

Theorem 3.26 (Projective Dimension Theorem). X,Y ⊆ Pn closed irreducible,Z ⊆ X∩Y a component, then dimZ ≥ dimX+dimY −n. If dimX+dimY −nis nonnegative, then X ∩ Y is nonempty.

Proof. First statement follows from ADT.Set s = dimX, t = dimY . C(X) = π−1(X) ⊆ An+1, then dimC(X) = s+1,

dimC(Y ) = t+ 1.Every component of C(X) ∩ C(Y ) has dimension ≥ s+ 1 + t+ 1− n− 1 =

s+ t− n+ 1 ≥ 1, and 0 ∈ C(X) ∩ C(Y ), so such a component exists.

Definition 3.15 (Twisted Module). Let ` ∈ Z, M(`) is the twisted modulegiven by M(`)d = M`+d. i.e., we are shifting the grading, but doing nothingelse.

Definition 3.16 (Homogeneous Homomorphism). A homomorphism ϕ : M →N of graded S-modules is homogeneous if ϕ(Md) ⊆Md for all d.

Definition 3.17 (Homogeneous Submodule). A submodule N ⊆ M is homo-geneous if N = ⊕d∈Z(N ∩Md).

This implies that N and M/N = ⊕Md/(N ∩Md) are graded and 0→ N →M →M/N → 0 is a short exact sequence of homogeneous homomorphisms.

Note: Let m ∈ M be homogeneous of degree `, that is, m ∈ M`, thenI = Ann(m) ⊆ S is a homogeneous ideal. Set N = S · m ⊆ M , then 0 →I → S → N → 0 is a short exact sequence, but it is not quite of homogeneousmaps, unless we change S, I to S(−`), I(−`). Thus, (S/I)(−`) is isomorphic toN = S ·m ⊆M .

Exercise: M f.g. module over a Notherian ring, then M is a Notherianmodule.

Lemma 3.27. M a f.g. graded module over a Notherian graded ring S, then∃ filtration 0 = M0 ⊆ . . . ⊆ Mr = M such that M i ⊆ M is a homogeneoussubmodule and M i/M i−1 ' S/Pi(`i) where Pi is a homogeneous prime idealand `i ∈ Z.

Proof. Let N ⊆ M be maximal homogeneous submodule such that the lemmais true for N .

Claim: N = M . Else M ′′ = M/N 6= 0, Take 0 6= m ∈ M ′′ homogeneoussuch that Ann(m) ⊆ S is as large as possible.

Claim: P = Ann(m) prime ideal. P 6= S. Let f, g ∈ S \ P . Enough toshow that fg /∈ P . Note: gm 6= 0 and Ann(gm) ) Ann(m), thus Ann(gm) =Ann(m). So we must have fg /∈ Ann(m), else f ∈ Ann(gm).

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Thus, Sm ' (S/P )(−`), ` = degm. So M → M/N = M ′′ ⊇ Sm, N ⊆ Mis the inverse image of Sm, and N ( N , and the lemma is true for N .

Definition 3.18 (Eventually Polynomial). f : Z→ Z is eventually polynomialif ∃P (z) ∈ Q[z] such that f(n) = P (n) for all n >> 0.

Set ∆f(n) = f(n+ 1)− f(n)

Lemma 3.28. f is eventually polynomial of degree r iff ∆f is eventually poly-nomial of degree r − 1.

Proof. ⇒: Obvious.

⇐: Assume ∆f(n) = Q(n) for all n >> 0 where Q(z) = c1 + c2

(z1

)+ . . .+

cr

(z

r − 1

). Set P (z) = c1

(z1

)+ . . .+ cr

(zr

).

Then ∆P = Q, so ∆(f − P )(n) = 0 for all n >> 0. Thus f(n)− P (n) = c0a constant for n >> 0.

Set S = k[x0, . . . , xn], M a f.g. graded S-module.Hilbert Function: HM (d) = dimk(Md), and Supp(M) = V+(Ann(M)) ⊆ Pn.Note: If 0 → M ′ → M → M ′′ → 0 is a short exact sequence of graded

S-modules, then Supp(M) = Supp(M ′) ∪ Supp(M ′′).⊇: Ann(M) ⊆ Ann(M ′) ∩Ann(M ′′).⊆: Let x ∈ Pn. If x /∈ Supp(M ′) ∪ Supp(M ′′), then ∃f ∈ Ann(M ′) such

that f(x) 6= 0 and ∃g ∈ Ann(M ′′) such that g(x) 6= 0, then fg ∈ Ann(M), butfg(x) 6= 0.

Ann(M ′) Ann(M ′′) ⊆ Ann(M) ⊆ Ann(M ′) ∩Ann(M ′′).

Theorem 3.29. M f.g. graded module over S = k[x0, . . . , xn] then HM (d) iseventually equal to a polynomial PM (d) of degree dim(Supp(M)).

Proof. ∃ filtration 0 = M0 ⊂ . . . ⊂Mr = M . such that M i/M i−1 ' (S/Pi)(`i)where Pi ⊆ S is a homogeneous prime.

Note: HM (d) =∑ri=1HMi/Mi−1(d) =

∑ri=1HS/Pi(d+ `i).

Supp(M) = ∪ri=1 Supp(M i/M i−1) = ∪ri=1V+(Pi). WLOG, M = S/P , P ahomogeneous prime. Induction of V+(P ):

If P = (x0, . . . , xn), then the theorem is true if we take dim ∅ = deg 0 = −1.Otherwise, some xi /∈ P . Set I = P + (xi). Then V+(I) ( V+(P ), so

dimV+(I) = dimV+(P )− 1 by the projective dimension theorem.By induction, HS/I(d) is eventually polynomial and deg(PS/I) = dimV+(P )−

10 → (S/P )(−1) → S/P → S/I → 0, so ∆HS/P (d − 1) = HS/P (d) −

HS/P (d− 1) = HS/I(d), so HS/P (d) is eventually polynomial of degree equal todimV+(P ).

Note: PM (z) = c0 + c1

(z1

)+ . . .+ cr

(zr

)∈ Q[z] is a numeric polynomial,

so ci ∈ Z.r = dim Supp(M), so r!(lead coef of PM (z)) ∈ Z ≥ 0

35

Definition 3.19 (Hilbert Polynomial of a Variety). Let X ⊆ Pn be a closedsubvariety of dimension r, then PX(z) = PS/I(X)(z) is the Hilbert Polynomialfor X.

We now define deg(X) = r!(lead coef of PX(z)).

Examples

1. deg V+(f) = deg f

2. V ⊆ Pn linear subspace. WLOG, V = V+S(xr+1, . . . , xn). S/I(V ) '

k[x0, . . . , xr], so PS/I(X)(d) =(r + dr

), so lead coef is 1/r!, so we get

degree 1.

Proposition 3.30. X1, X2 ⊆ Pn closed, dimX1 = dimX2 = r, no componentsin common, then deg(X1 ∪X2) = degX1 + degX2

Proof. I1 = I(X1), X2 = I(X2).I(X1 ∪X2) = I1 ∩ I2, so 0→ S/(I1 ∩ I2)→ S/I1 ⊕S/I2 → S/(I1 + I2)→ 0.The first takes f 7→ (f, f) and the second takes (f, g) 7→ f − g. They

are injective and surjective, and so we see this this is a short exact sequence.Thus PX1(d) + PX2(d) = PX1∪X2(d) + PS/(I1+I2)(d), so Supp(S/(I1 + I2)) =V+(I1 + I2) = X1 ∩X2. dim < r, so LC(PX1) + LC(PX2) = LC(PX1∪X2).

Corollary 3.31. If X ⊆ Pn has dim zero, then deg(X) =the number of pointsin X.

Definition 3.20 (Simple Module). If R is a ring and M an R-module, thenM is simple if M has no nontrivial submodules and M 6= 0. This is equivalentto M ' R/P where P ⊆ R is a maximal ideal.

Definition 3.21 (Decomposition Series). A decomposition series for M is afiltration 0 = M0 ( M1 ( . . . ( Mr = M such that Mi/Mi−1 is simple for all i.

Definition 3.22 (Artinian Module). If there exists a decomposition series, thenM is an Artinian module and we define length(M) = r.

Assume that R is Notherian and that M is a finitely generated R-module.Then there exists a filtration 0 = M0 ( M1 ( . . . ( Mr = M . such thatMi/Mi−1 is isomorphic to R/Pi where Pi ⊆ R is maximal.

Note: Ann(M) ⊆ Pi for all i.

Lemma 3.32. If P ⊆ R is a minimal prime over Ann(M) then MP is anArtinian RP -module which has lengthRP (MP ) = |i : Pi = P| in the filtrationof M .

Proof. 0 = (M0)P ⊆ (M1)P ⊆ . . . ⊆ (Mr)P = MP , and (Mi)P /(Mi−1)P =(Mi/Mi−1)P = (R/Pi)P .

If P = Pi, then we get RP /PRP , else we get 0.

36

X ⊆ Pn is a closed subvariety, S = k[x0, . . . , xn], then dimk(S/I(X))d =PX(d) for all d >> 0. dim(X) = deg(PX) and deg(X) = (dimX)! LC(PX)

Assume X ⊆ Pn has pure dimension r. That is, all components of X havedimension r.

Y = V+(f) ⊆ Pn a hypersurface such that no component of X is containedin Y . Also assume that f ∈ S is square-free.

Let Z ⊆ X ∩ Y be a component. Then dimZ = r − 1. Set M = S/(I(X) +(f)). Supp(M) = X∩Y , do if P = I(Z), then P is minimal prime over Ann(M).

Definition 3.23 (Intersection Multiplicity). I(X · Y ;Z) = lengthSP (MP )

Theorem 3.33. deg(X) deg(Y ) =∑Z⊆X∩Y I(X · Y ;Z) deg(Z).

Proof. Set d = deg(X), e = deg(Y ).

Then we have a short exact sequence 0→ S/I(X)f→ S/I(X)→M → 0.

This gives us 0 → (S/I(X))`−e → (S/(I(X)))` → M` → 0 will be a shortexact sequence for each `, where this is breaking it into homogeneous parts.

Thus PM (`) = PX(`)−PX(`−e). Thus LC(PM ) = r ·e ·LC(PX) = red/r! =de

(r−1)! .Take filtration 0 = M0 ( M1 ( . . . ( Mt = M , Mi ⊆ M is homogeneous

and Mi/Mi−1 is isomorphic to S/Pi where Pi ⊆ P is a homogeneous primeideal.

So PM (`) =∑ti=1 PMi/Mi−1(`) =

∑Q⊆S hom prime PS/Q(`)|i : Q = Pi|,

and from this we get∑Z⊆X∩Y ;component |i : Pi = I(Z)|PZ(`) + LOT . So we

have LC(PM ) = 1(r−1)!

∑Z⊆X∩Y I(X · Y ;Z) deg(Z).

Corollary 3.34 (Bezout’s Theorem). Let X,Y ⊆ P2 be curves of degree d ande such that X ∩ Y is a finite set, then

∑P∈X∩Y I(X · Y ;P ) = de.

Exercise: P ∈ X ∩ Y ⊆ P2 then I(X · Y ;P ) = 1 ⇐⇒ P is a nonsingularpoint of both X and Y and X and Y have different tangent directions at P .

Bezout’s Theorem for PnIdea: If X ⊆ Pn is closed and irreducible and Y ⊆ Pn is a hypersurface, then

[X] · [Y ] =∑Z⊆X∩Y I(X · Y ;Z) · [Z].

Suppose that Y1, Y2, . . . , Yn ⊆ Pn are hypersurfaces such that their inter-section is finite. Then (. . . ([Pn] · [Y1]) · [Y2] . . .) · [Yn] =

∑Ni=1 ci[Pi] where

Y1 ∩ . . . ∩ Yn = P1, . . . , PN. Then∑ci =

∏nj=1 deg(Yj).

In fact: (. . . ([Pn] · [Y1]) . . .) · [Ym] =∑Nmi=1 c

(m)i [Zi] where Y1 ∩ . . . ∩ Ym =

Z1 ∪ . . . ∪ ZNm components, then∏mj=1 deg(Yj) =

∑Nmi=1 c

(m)i deg(Zi).

Fact: ci, c(m)i do not depend on the order of multiplication.

Corollary 3.35. Y1 ∩ . . . ∩ Yn finite implies that |Y1 ∩ . . . ∩ Yn| ≤∏

deg(Yj).

Useful Fact: X1, . . . , Xm ⊆ Pn irreducible closed, d ∈ N then there existsirreducible hypersurface Y ⊆ Pn of degree d such that Xi 6⊂ Y for all i.

37

vd : Pn → PN , N =(n+ dn

)− 1 the veronese map, then Y = Pn ∩ H,

H ⊆ PN hyperplane.Intrinsic: hypersurfaces of degree d in Pn ↔ P(Sd)−nonsquarefree by

V+(f)↔ [f ]irreducible hypersurfaces ↔ P(Sd)\∪p+q=dϕp,q(P(Sp)×P(Sq)) where ϕp,q :

P(Sp)× P(Sq)→ P(Sp+q) : [f ]× [g] 7→ [fg].(n+ pn

)+(n+ qn

)≤(n+ p+ q

n

)so U ⊆ P(Sd) is a dense open subset.

Now, if X ⊆ Pn is closed, then X ⊆ V+(f) iff f ∈ I(X)d. Thushypersurfaces6⊇ X ↔ WX = P(Sd) \ P(I(X)d). Now, we can conclude

that irreducible hypersurfaces Y in Pn of degree d such that Xi 6⊆ Y for all icorrespond to U ∩WX1 ∩ . . . ∩WXm ⊆ P(Sd) is still a dense open subset.

4 Sheaves

Definition 4.1 (Presheaf). Let X be a topological space. A presheaf F ofabelian groups on X is an assignment U 7→ F (U) of an abelian group F (U) toeach open subseteq U of X plus group homomorphisms ρUV : F (U) → F (V )whenever V ⊆ U ⊆ X open with

1 ρUU = id

2 ρVW ρUV = ρUW when W ⊆ V ⊆ U

Notation: elements of F (U) are called sections of F over U , ρUV arecalled restriction maps and s ∈ F (U), V ⊆ U , then s|V = ρUV (s). Also,say Γ(U,F ) = F (U).

Definition 4.2 (Sheaf). A sheaf is a presheaf F such that

3 If s ∈ F (U) and U = ∪Vi an open cover, then s|Vi = 0 iff s = 0.

4 If U = ∪Vi an open cover and have sections si ∈ F (Vi) and si|Vi∩Vj =sj |Vi∩Vj for all i, j, then there exists s ∈ F (U) such that s|Vi = si.

Note: The section s of axiom 4 is unique by axiom 3.Remark: We can easily define sheaves of rings, sets, modules, etc.Examples:

1. X is a SWF, then we can define OX of k-algebras by OX(U) = k[U ]. Thissheaf is called the structure sheaf of X.

2. Let M be a manifold, OM (U) = C∞ f : U → R. Then there is π :TM → M the tangent bundle, π−1(x) = TxM . Then TM ↔ a sheaf Tof OM -modules. T (U) = s : U → TM such that π(s(x)) = x for allx ∈ U. T (U) is a OM (U)-module.

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Definition 4.3 (Morphism of Sheaves). A morphism ϕ : F → G of (pre)sheavesconsists of homomorphisms ϕU : F → G for all U ⊆ X open such that s ∈ F ,V ⊆ U open, then ϕU (s)|V = ϕV (s|V ) ∈ G (V ) and

F (V )

G (V )

F (U)

G (U).............................................................................................................................

ϕV

.............................................................................................................................

ϕU

................................................................................... ............

............................................................................................. ............

If ϕ : F → G and ψ : G →H are morphisms, we define ψ ϕ : F →H by(ψ ϕ)U = ψU ϕU . An isomorphism is a morphism with an inverse morphism.

Definition 4.4 (The stalk of F at x). If F is a presheaf on X, x ∈ X thenthe stalk of F at x is Fx = lim−→U3x F (U). This means

1. Fx is an abelian group (or whatever)

2. If x ∈ U we have homomorphism ρUx : F (U) → Fx, if x ∈ V ⊆ U thenρUx = ρV x ρUV .

3. If G is any abelian group with homomorphisms ΘU : F (U) → G suchthat ΘU = ΘV ρUV for all x ∈ V ⊆ U , then there exists a unique grouphomomorphism Θx : Fx → G such that ΘU = Θx ρUx

Example: X a variety, x ∈ X. Then OX,x = lim−→U3x OU =local ring of X atx.

Construction Fx =(⊕

U3x F (U))/〈(0, . . . , 0, s, 0, . . . , 0,−s|V , 0, . . . , 0) : ∀s ∈

F (U), x ∈ V ⊆ U〉.Notation: If s ∈ FU and x ∈ U write sx = ρUx(s) ∈ Fx.Exercise Let F be a presheaf.

1. All elements of Fx can be written as sx for some s ∈ F (U), x ∈ U .

2. s ∈ F (U), x ∈ U , sx = 0 ∈ Fx iff s|V = 0 ∈ F (V ) for x ∈ V ⊂ U .

Exercise: F is a sheaf. s ∈ F (U). s = 0 ⇐⇒ sx = 0 for all x ∈ U .Note: A morphism ϕ : F → G gives a homomorphism ϕx : Fx → Gx,

ϕx(sx) = ϕU (s)x, s ∈ F (U), x ∈ U .x ∈ U :

Fx

Gx

F (U)

G (U).............................................................................................................................

.............................................................................................................................

............. ............. ............. ....................... ............

....................................................................................................... ............

Proposition 4.1. ϕ : F → G is a morphism of sheaves. ϕ is an isomorphismiff ϕx : Fx → Gx are isomorphisms for all x ∈ X.

39

Proof. ⇒: Clear⇐: We must show that ϕ : F (U)→ G (U) are isomorphisms. ϕU is injective

as s ∈ F (U), if ϕU (s) = 0 ∈ G (U) then ϕx(sx) = ϕU (s)x = 0, but ϕx isinjective, and so sx = 0 for all x ∈ U , so s = 0.

To see that ϕU is surjective, take t ∈ G (U). ϕx surjective implies thattx = ϕx(s(x)x) ∈ Gx for some s(x) ∈ F (Vx) where Vx ⊆ U are open subsetscontaining x.

Now tx = ϕVx(s(x))x. We can make Vx smaller such that t|Vx = ϕVx(s(x)) ∈G (Vx). ϕ(s(x)|Vx∩Vy ) = t|Vx∩Vyϕ(s(y)|Vx∩Vy ). Thus s(x)|Vx∩Vy = s(y)|Vx∩Vy .

Patch: There exists a unique s ∈ F (U) such that s|Vx = s(x) ∈ F (Vx) for allx ∈ U . ϕU (s)|Vx = ϕVx(s|Vx) = ϕVx(s(x)) = t|Vx . Thus ϕU (s) = t ∈ G (U).

Remark: F is a sheaf on X, U ⊆ X is open. Define F |U to be the sheaf onU by Γ(V,F |U ) = Γ(V,F )

Example: Let S1 = (x, y) ∈ R2 : x2 + y2 = 1. Define sheaves F ,G byF (U) = f : U → R : f is locally constant and G (U) = g : U → R : ∂g∂θ = 1.If U ( S1 open then F |U ' GU by f(x, y) 7→ f(x, y) + Arg(x, y), so Fx ' Fx

for all x ∈ S1 but F and G are not isomorphic as F (S1) = R and G (S1) = ∅.SheafificationLet F be a presheaf on X, define a sheaf F+ as follows: U ⊆ X open set

F+(U) to be the set of all functions s : U →∐x∈U Fx such that

1. s(x) ∈ Fx for all x

2. ∀x ∈ U there exists an open set V with x ∈ V ⊆ U and t ∈ F (V ) suchthat s(y) = ty ∈ Fy forall y ∈ V .

Definition 4.5. We define a morphism Θ : F → F+ by t ∈ F (U), ΘU (t) =[x 7→ tx] ∈ F+(U).

Exercise: Θx : Fx → F+x .

Proposition 4.2. Let F be a presheaf and G be a sheaf. ϕ : F → G is anymorphism. Then there exists a unique ϕ+ : F+ → G such that ϕ = ϕ+ Θ.

Proof. Let s ∈ F+(U). i.e. s : U →∐x∈U Fx. For x ∈ U choose t(x) ∈ F (Vx)

s.t. x ∈ Vx ⊂ U open and s(y) = t(x)y for all y ∈ Vx.Set τ(x) = ϕVx = ϕVx(t(x)) ∈ G (Vx).If y ∈ Vx, then τ(x)y = ϕ(t(x))y = ϕy(t(x)y) = ϕy(s(y)) ∈ Gy. Thus,

τ(x1)|Vx1∩Vx2 = τ(x2)|Vx1∩Vx2 , which gives that τ(x) glue to τ ∈ G (U). SetϕU (s) = t. Exercise: Check the details!

Definition 4.6. ϕ : F ′ → F is a morphism of sheaves.We say ϕ is injective if ϕU is injective for all U ⊆ X. If ϕ is injective, then

we say F ′ ⊆ F is a subsheaf as F ′(U) ⊆ F (U).

Exercise: ϕ injective iff ϕx : F ′x → Fx is injective for all x ∈ X.Consequence: If ϕ : F → G is a morphism of presheaves such that ϕU :

F (U)→ G (U) injective for all U ⊆ X open, then ϕ+ : F+ → G + is injective.

40

Proof. Check: ϕp : Fp → Gp injective for all p ∈ X. If ϕp(sp) = 0 ∈ Gpfor s ∈ F (U), p ∈ U then ϕU (s)p = 0 so ϕU (s)|V = 0 for p ∈ V ⊆ U , soϕV (s|V ) = 0. Thus s|V = 0, so sP = 0.

In particular: If a presheaf F is a subpresheaf of a sheaf G then F+ is asubsheaf of G .

Definition 4.7 (Kernel and Image). Let ϕ : F → G be a morphism of sheavesof abelian groups.

Then kerϕ = the sheaf U 7→ ker(ϕU ) ⊆ F (U).Imϕ = the sheafification of the presheaf U 7→ Im(ϕU ) ⊂ G (U).

Note, kerϕ ⊂ F , Imϕ ⊂ G are subsheaves, and ϕ injective iff kerϕ = 0.

Definition 4.8 (Surjective). ϕ is surjective if Im(ϕ) = G .

WARNING: ϕ surjective DOES NOT IMPLY that ϕU is surjective.Exercise: ϕ surjective iff ϕp : Fp → Gp surjective for all p ∈ X.

Definition 4.9 (Quotient Sheaf). F ′ ⊂ F a subsheaf, then F/F ′ = thesheafification of [U 7→ F (U)/F ′(U)]

We have a surjective morphism F → F/F ′ which has kernel F ′.

Notation: A sequence of sheaves → F i φi→ F i+1 φi+1→ F i+2 → . . . is acomplex if φi+1 φi = 0 for all i and is exact if Imφi = kerφi+1 for all i.

Equivalently, complexes and exact sequences of the stalks in the category ofabelian groups.

Example, 0→ F ′ → F → F ′′ → 0 is exact iff F ′ ⊂ F and F ′′ ' F/F ′.

Definition 4.10. f : X → Y a continuous map, F a sheaf on X, then f∗F isa sheaf on Y defined by f∗F (V ) = F (f−1(V )).

Example: X a variety, Y ⊆ X closed, i : Y → X the inclusion, U ⊆ X open,then IY (U) = f ∈ OX(U)|f(y) = 0,∀y ∈ U ∩ Y , IY ⊆ OX is a subsheaf ofideals. Then OX(U)/IY (U) are the regular functions U ∩ Y → k which can beextended to all of U .

OX/IY ' i∗OY because we can extend locally. We have exact sequence0 → IY → OX → i∗OY → 0, which will often be written 0 → IY → OX →OY → 0.

Example, f : X → y a morphism of SWFs, we get morphism f∗ : OY →f∗OX by OY (V )→ f∗(OX(V )) = OX(f−1(V )), h 7→ h f = f∗h.

Exercise: Find a morphism f : X → Y of varieties such that f∗ : OY →f∗OX is an isomorphism, but f is NOT an isomorphism.

Definition 4.11 (Inverse Image Sheaf). Let f : X → Y continuous, G a sheafon Y . U ⊆ X open, define pre− f−1G (U) = lim−→V⊇f(U)

G (V ).

We have maps G (V ) → pre − f−1G (U), s 7→ f−1s, f(U) ⊆ V . pre −f−1G (U) = f−1s|s ∈ G (V ), V ⊇ f(U). f−1s = 0 ⇐⇒ s|W = 0 wheref(U) ⊆W ⊆ V . This is a presheaf on X. We define f−1G to be the sheafifica-tion of this presheaf.

41

Special Case: X ⊆ Y is a subset, i : X → Y the inclusion, G |X = i−1G .Example: X ⊆ Y open, pre − i−1G (U) = lim−→V⊇U G (V ) = U . It is already

a sheaf, and so the special case just mentioned above holds.Exercise: (f−1G )p = Gf(p).Adjoint Propertyf : X → Y continuous, F a sheaf on X, G a sheaf on Y , let ϕ : G → f∗F

be a morphism of sheaves on Y . U ⊆ X open, V ⊇ f(U), then G (V )ϕV→

F (f−1(V )) → F (U) and maps are compatible with restrictions of G , so thisinduces ψU : pre−f−1G (U)→ F (U), which gives a morphism ψ : pre−f−1G →F , we sheafify to get ψ+ : f−1G → F .

Exercise: hom(G , f∗F ) → hom(f−1G ,F ) : ϕ 7→ ψ+ is an isomorphism ofabelian groups.

Category Theory Interpretation: f−1 is a left adjoint functor to f∗ and f∗is a right adjoint functor to f−1.

Let X be a variety (ringed-space)

Definition 4.12 (OX -module). An OX-module is a sheaf F on X such thatF (U) is an OX(U)-module for all open U ⊆ X, such that if V ⊆ U openthen F (U) → F (V ) is OX(U)-homomorphism (ie, f ∈ OX(U), m ∈ F (U),(f ·m)|V = f |V ·m|V .)

An OX-homomorphism ϕ : F → G is a morphism of sheaves such thatϕU : F (U)→ G (U) is an OX(U)-homomorphism for all U ⊆ X open.

Note kerϕ ⊆ F and Imϕ ⊆ G are sub OX -modules, F is an OX -moduleimplies that FP is an OX,P -module.

Definition 4.13 (Tensor Product). F ,G , OX-modules, F ⊗OX G = [U 7→F (U)⊗OX(U) G (U)]+

Exercise: (F ⊗OX G )p = Fp ⊗OX,P Gp

Definition 4.14 (Locally Free). An OX-module F is locally free if ∃⋃α Uα =

X an open cover such that F |Uα ' O⊗rUα .

Example: π : An+1 \ 0 → Pn, let m ∈ Z.

Definition 4.15. A sheaf O(m) = OPn(m) of OPn-modules is Γ(U,O(m)) =h ∈ k[π−1(U)] : h(λx) = λmh(x)∀x ∈ π−1(U), λ ∈ k∗.

Note: f ∈ S = k[x0, . . . , xn] homogeneous of degree > 0. k[π−1(D+(f))] =k[D(f)] = Sf . So Γ(D+(f),O(m)) = (Sf )m.

O(m) is an invertible OPn-module (invertible means locally free of rank 1).On Ui = D+(xi) : OUi → O(m)|Ui , h 7→ xmi h. Note that OPn(0) = OPn .

Γ(Pn,O(m)) = Sm when m ≥ 0 and 0 else.

Lemma 4.3. O(m)⊗Pn O(p) ' O(m+ p).

Proof. U ⊆ Pn open. Γ(U,O(m))⊗ Γ(U,O(p))→ Γ(U,O(m+ p)), f ⊗ g 7→ fg.Sheafification gives a map O(m)⊗ O(p)→ O(m+ p), restricting to U we haveOUi ⊗ OUi → OUi .

42

Consequence: O(m) ' O(p) iff m = p. If m ≤ p and O(m) ' O(p) impliesthat O(m− p) ' O(p− p), on the right we have nonzero global sections, on theleft we only do if m− p nonnegative, so m = p.

Later: Any invertible sheaf on Pn is isomorphic to O(m), m ∈ Z.Coherent SheavesLet X be an affine variety and A = k[X], let M be an A-module.

Definition 4.16 (Quasi-Coherent Sheaf). M = [U 7→M ⊗AOX(U)]+ is calleda quasi-coherent OX-module.

Note: M ⊗A OX(D(f)) = M ⊗A Af = Mf .Examples A = OX , Y ⊆ X closed, I = I(Y ) ⊂ A. I = IY ⊂ OX .Exercise: A/I = i∗OY .Claim: Mp = MI(p) for p ∈ X, I(p) = I(p) ⊂ A.MI(p) → Mp by m/f 7→ (m ⊗ 1/f)p, this tensor product is in M(D(F )).

Surjectivity is clear.Injective: if m/f 7→ 0 ∈ Mp then (m⊗ 1/f)|D(h) = 0 for some h ∈ A \ I(p),

so m⊗1/f = 0 ∈M⊗AAh = Mh, thus hnm = 0 ∈M , and so m/f = 0 ∈MI(p).Consequence M(U) =set of functions s : U →

∐p∈U MI(p) such that ∀p ∈ U ,

there exists p ∈ V ⊂ U and m ∈ M,f ∈ A such tthat s(p) = m/f ∈ MI(q) forall q ∈ V .

Proposition 4.4. M(D(f)) 'Mf .

Corollary 4.5. 1. OX(D(f)) = Af

2. Γ(X, M) = M .

We now prove the proposition.

Proof. ψ : Mf → M(D(f)) the obvious mIt is injective, as if ψ(m/fn) = 0 then m/fn = 0 ∈ MI(p) for all p ∈ D(f).

Thus, for all p ∈ D(F ), there exists hp ∈ A \ I(p) with hpm = 0 ∈ M . D(f) ⊆⋃D(hp) ⇒ V (f) ⊇ V (hp), so f ∈ I(V (hp)) means fN =

∑p aphp for

ap ∈ A. fNm =∑aphpm = 0, thus m/fn = 0 ∈Mf .

Surjectivity: Let s ∈ M(D(f)), there exists a cover D(f) =⋃ri=1 Vi, mi ∈

M , hi ∈ A such that s = mi/hi on Vi. WLOG Vi = D(gi) and Vi = D(hi)(replace mi 7→ gimi by hi 7→ gihi).

On D(hihj) = D(hi)∩D(hj), s = mi/hi = mj/hj ∈ M(D(hihj)), injectivityfor D(hihj) : mi/hi = mj/hj ∈ Mhihj . So (hihj)N (hjmi − himj) = 0 ∈ M .Replace mi 7→ hNi mi, hi 7→ hN+1

i , hjmi = himj .D(f) ⊆

⋃D(hi) so fn =

∑aihi where ai ∈ A. Set m =

∑aimi ∈ M .

Claim: s = ψ(m/fn).For all j, hjm =

∑i hjaimi = (

∑i aihi)mj = fnmj .

So m/fn = mj/hj = s on D(hj).

43

Definition 4.17 (Quasi-Coherent and Coherent). Let X be any variety. AnOX-mdule F is quasi-coherent if there exists an open affine cover X =

⋃Ui

and k[Ui]-modules Mi such that F |Ui ' Mi as OUi-modules.F is coherent if Mi finitely generated k[Ui]-modules for all i.

Examples:

1. All locally free OX -modules are coherent, U = Spec−m(A), ˜A⊗r = O⊗rU .

2. Y ⊆ X closed, IY and i∗OY are coherent OX -modules.

Example: X = A1, F (U) =

OA1(U) 0 /∈ U0 0 ∈ U . This is called the exten-

sion of the OA1\0 by zeros. F is an OX -module, but is not quasi-coherent, asif U ⊆ X is open affine and 0 ∈ U , then Γ(U,F ) = 0 but F |U is not the zerosheaf.

Exercises:

1. F is quasi-coherent iff F |U ' ˜F (U) for all open affine U ⊆ X.

2. F is coherent implies F (U) is finitely generated k[U ]-module for all affineopen U .

3. f : X → Y a morphism of varieties.

(a) f affine implies that f∗OX quasi-coherent.

(b) f finite f∗OX coherent.

Example: M a finitely generated A-module, X = Spec−m(A). Then M islocally free OX -module of rank r iff M is a projective A-module of const. rankr.

(M loc free iff X =⋃D(fi), MD(fi) ' O⊗D(fi)

r iff Mfi = A⊗rfi iff M projec-tive.)

Recall: X is complete implies Γ(X,OX) = k. More general fact: F acoherent sheaf on a complete variety X then dimk Γ(X,F ) < ∞. We will usethis without proof. (Projective case in Hartshorne, General in EGA.)

Note that Γ(A1,OA1) = k[t] which has infinite dimension.Pushforward, PullbackLet f : X → Y be a morphism of varieties. F an OX -module, then f∗F is

a f∗OX -module, we have ring homomorphism f∗ : OY → f∗OX , and so f∗F isan OY -module.

Let G be a OY -module f−1G is an f−1OY -module. (f−1h ·f−1s = f−1(hs))OY → f∗OX is the same as f−1OY → OX by the adjoint property.

Definition 4.18 (Pullback). Define f∗G = f−1G ⊗f−1OY OX , we call it thepullback.

Examples

44

1. f∗OY = OX .

2. (f∗G )p = (f−1(G))p ⊗(f−1OY )p OX,p = Gf(p) ⊗OY,f(p) OX,p.

3. U ⊆ Y open, i : U → Y inclusion, i∗G = G |U ⊗OY |U OU = G |U .

Adjoint Propertyf : X → Y a morphism of SWFs. G an OY -module. We have f−1OY -

homomorphism f−1G → f∗G by s 7→ s⊗1. This gives an α : OY -homomorphismF → f∗f

∗G .

Lemma 4.6. G is an OY -module, F is an OX-module. Then homOX (f∗G ,F ) 'homOY (G , f∗F ).

Proof. Given ψ : f∗G → F we obtain φ : Gα→ f∗f

∗Gf∗ψ→ f∗F .

Given φ : G → f∗F , we obtain φ : f−1G → F which is an f−1OY -hom.Take ψ : f∗G = f−1F ⊗f−1OY OX → F by s⊗ h 7→ h · φ(s).

Functoriality Xf→ Y

g→ Z morphism of SWFs. If F is a sheaf on X,g∗(f∗F ) = (gf)∗F .

Proposition 4.7. 1. G on Z implies that (gf)−1G = f−1(g−1G )

2. G an OZ-module implies that (gf)∗G = f∗(g∗G ).

Proof. We will prove case 2.id : (gf)∗G → (gf)∗G gives G → (gf)∗(gf)∗G = g∗f∗(gf)∗G gives g∗G →

f∗(gf)∗G which gives f∗(g∗G )→ (gf)∗G .We have a global homomorphism, so enough to check stalks. f∗(g∗G )p =

(g∗G )f(p) ⊗OY,f(p) OX,p. This is (Gg(f(p)) ⊗OZ,gf(p) OY,f(p)) ⊗OY,f(p) OX,p =Ggf(p) ⊗OZ,gf(p) OX,p = ((gf)∗G )p

Let f : X → Y be a morphism of SWFs. G is an OY -module, and f−1G isan f−1OY -module.

Definition 4.19 (Pullback). f∗G = f−1G ⊗f−1OY OX is an OX-module.

f∗ : OY → f∗OX induces a map f−1OY → OX .Some sections: σ ∈ G (V ), we set f∗σ = f−1σ ⊗ 1 ∈ Γ(f−1(V ), f∗G ). The

stalks (f∗G )p = Gf(p) ⊗OY,f(p) OX,p.

If Zg→ X

f→ Y , then (fg)∗G = g∗(f∗G ).

Corollary 4.8. If G is a locally free OY -module, then f∗G is a locally freeOX-module of the same rank.

Proof. Let Y =⋃Vi be an open cover such that G |Vi ' O⊕rVi . Set Ui =

f−1(Vi) ⊂ X.

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Ui Vi

X Y

................................................................................................................. ............f ′

................................................................................................................. ............f

........

........

........

........

........

........

........

........

........

........

........

........

.................

............

p

........

........

........

........

........

........

........

........

........

........

........

........

.................

............

q

f∗G |Ui = p∗f∗G = f ′∗q∗G ' f ′∗q∗(O⊕rVi ) = O⊕rUi

Lemma 4.9. f : X → Y a morphism, 0→ G ′ → G → G ′′ → 0 is a short exactsequence of OY -modules. Then f∗G ′ → f∗G → f∗G ′′ → 0 is an exact sequenceof OX-modules. If G ′′ is locally free, then the first map is injective.

Proof. On stalks we start with 0 → G ′f(p) → Gf(p) → G ′′f(p) → 0 exact. Tensorproduce is right exact and gets us to f∗, ad so we have the first part of thetheorem immediately.

If G ′′ is locally free, then G ′′f(p) is a free OY,f(p)-module, and so the originalsequence is split-exact.

Definition 4.20 (Generated by Finitely Many Global Sections). The OX-module F is generated by finitely many global sections iff ∃ a surjective mapO⊕mX → F .

Equivalently, ∃s1, . . . , sm ∈ Γ(X,F ) such that Fp generated by (s1)p, . . . , (sm)pas an OX,p-module.

Example: Any quasi-coherent OX -module, if X is affine (this is just gener-ated by global sections, requires coherent to be generated by finitely many)

Example: OPn(1) is generated by x0, . . . , xn ∈ Γ(Pn,O(1)).Suppose that f : X → Pn is a morphism, then O⊕n+1

Pn → O(1) → 0 exactimplies that O⊕n+1

X → f∗O(1) → 0 is exact, so f∗O(1) is generated by globalsections f∗(x0), . . . , f∗(xn).

Proposition 4.10. X a variety, L invertible OX-module generated by globalsections s0, . . . , sn ∈ Γ(X,L ). Then ∃!f : X → Pn such that f∗O(1) ' L andf∗(xi)↔ si.

Proof. Set Ui = p ∈ X|(si)p /∈ mpLp.Ui open: If V ⊆ X open with L |V ' OV , then L |V is generated by

t ∈ Γ(V,L ) so we write si = hit on V with hi ∈ k[V ]. Then Ui ∩ V = p ∈V |hi(p) 6= 0 is open in V .

Note: L is generated by s0, . . . , sn implies that X =⋃ni=1 Ui, and OUi '

L |Ui implies that 1 7→ si|Ui .On Ui, we can write si = hijsj for some hij ∈ k[Ui]. We define a map

g : Ui → Pn by f(p) = (hi0(p) : . . . : hin(p)).The maps are compatible: On Ui ∩ Uj , h`js` = sj = hijsi = hijh`is`, so

h`j = h`ihij . The map on U` : p 7→ (h`0(p) : . . . : h`n(p)) = h`i(p)hi0(p) : . . . :h`i(p)hin(p). Thus, we have a morphism f : X → Pn.

Claim: ∃ isomorphism L → f∗O(1) by si 7→ f∗(xi). On Ui, we defineL |Ui → f∗O(1)|Ui by hsi 7→ hf∗(xi). This means that s` = hi`si 7→ hi`f

∗(xi),

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we must check that f∗(x`) = hi`f∗(xi). The definition of f implies that (x`/xi)

f = hi`hii

= hi`, so f∗(x`) = f∗(x`xi xi) = f∗(x`xi )f∗(xi) = hi`f

∗(xi).And now we prove uniqueness: If f : X → Pn is any morphism such

that L ' f∗O(1) and si ↔ f∗(xi) on Ui, hi`si = f∗(x`) = f∗(x`xi xi) =f∗(x`xi )f

∗(xi) = f∗(x`xi )si, so f∗(x`/xi) = hi` on Ui.

Definition 4.21 (Very Ample Sheaf). Let L be an invertible sheaf on X.L is very ample iff L is generated by (finitely many) global sections and themap f : X → Pn given by generators s0, . . . , sn ∈ Γ(X,L ) is an isomorphismf : X →W ⊆ Pn locally closed.

Exercise: OPn(m) is very ample iff m ≥ 1.

Definition 4.22 (PGL). PGL(n) = GL(n+ 1)/k∗.

Exercise: PGL(n) is an affine algebraic group.FACT: Every invertible sheaf on Pn is isomorphic to O(m) for some m.

Corollary 4.11. Aut(Pn) ' PGL(n).

Proof. PGL(n) ≤ Aut(Pn) is trivial.Let f : Pn → Pn be an automorphism. The fact implies that f∗O(1) '

O(m) for some m. In fact,(n+mm

)= dimk Γ(Pn,O(m)) = dim Γ(f∗O(1)) =

dim Γ(O(1)) = n+ 1, so m = 1.f∗x0, . . . , f

∗xn ∈ Γ(Pn,O(1)) form a basis. We write f∗(xi) =∑ni=0 aijxj

for aij ∈ k. Then A = (aij) ∈ GL(n+ 1).Define ϕ : Pn → Pn by ϕ(x0 : . . . : xn) = (

∑a0jxj : . . . : anjxj). ϕ∗(x`xi ) =P

a`jxjPaijxj

= f∗(x`xi ).Thus, f = ϕ ∈ PGL(n).

Corollary 4.12. Pn is not an algbriac group.

Normal Varieties

Definition 4.23 (Normal Variety). X is irreducible. Then X is normal iffOX,P is normal (integrally closed) for all p.

Example: Nonsingular varieties.Note: X affine, then X is normal iff k[X]m normal for all maximal m iff k[X]

is normal.Exercise: If A is a domain, S ⊆ A multiplicative, then S−1A = S−1A.

Definition 4.24 (Normalization). If f is an affine variety, k[X] ⊂ k(X),then k[X] ⊆ k(X) is the integral closure. The normalization of X is X =Spec−m(k[X]).

As we have the inclusion k[X] → k[X], we get a projection map X → Xwhich is finite.

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Lemma 4.13. ϕ : U → X morphism of affines, ϕ an open embedding iff∃f1, . . . f−n ∈ k[X] such that (ϕ∗f1, . . . , ϕ

∗fn) = (1) ⊂ k[U ] and ϕ∗ : k[X]fi →k[U ]ϕ∗fi is an isomorphism for all i.

Proof. ⇒: Take open cover U =⋃ri=1D(fi), fi ∈ k[X].

⇐: Set V =⋃ri=1D(fi) ⊆ X. (ϕ∗f1, . . . , ϕ

∗fr) = (1) ⊆ k[U ] implies thatϕ(U) ⊆ V . Thus, we have that ϕ : ϕ−1(D(fi)) → D(fi) is an isomorphism, soϕ : U → V isomorphism.

Lemma 4.14. Assume X affine, U ⊆ X is an open affine, then U ⊆ X openaffine.

Proof. k[X] ⊆ k[U ] ⊆ k(X). Thus, k[X] ⊆ k[U ], so we have a morphismϕ : U → X. Take f1, . . . , fn ∈ k[X] as in lemma wrt U ⊂ X.

(f1, . . . , fn) = (1) ⊆ k[U ], and k[U ]fi = k[U ]fi = k[X]fi = k[X]fi . And so,the lemma implies that ϕ is an open embedding.

Exercise: Given pre-varieties X1, . . . , Xn, open subsets Uij ⊆ Xi and iso-morphisms ϕij : Uij → Uji such that Uii = Xi, ϕii = id, for all i, j, k,ϕij(Uij ∩ Uik) = Uji ∩ Ujk and ϕik = ϕjk ϕij on Uij ∩ Uik, then ∃! preva-riety X with morphisms ψi : Xi → X such that ψi : Xi → open⊂ X is anisomorphism, X =

⋃ni=1 ψi(Xi), ψi(Uij) = ψi(Xi) ∩ ψj(Xj) and ψi = ψj ϕij

on Uij .

Definition 4.25 (Normal). A variety X is normal iff X is irreducible and OX,Pare normal for all p.

Construction: X an irreducible variety, X = X1∪ . . .∪Xn open affine cover,set Uij = Xi ∩ Xj . Then Uij is affine, have ψij : Uij → Uji nbe the identity.Now Uij ⊆ Xi and still have φij : Uij → Uji is the identity, so the satisfy thehypotheses of the exercise. Thus, there exists a prevariety X = X1 ∪ . . . ∪ Xn.We call this the normalization of X.

Note: k[Xi] is a finitely generated k[Xi]-module, so we have finite π : Xi →Xi, which we can glue to a morphism π : X → X.

Exercise: π : X → X is finite. (Check that π−1(Xi) = Xi)Exercise: ϕ : X → Y affine morphism of pre-varieities. Then Y is separated

implies that X is serpated, and so X is an irreducible normal separated variety.Example: X irred curve implies π : X → X resoluton of singularities.

Definition 4.26 (Local Ring along Subvariety). Let X be a variety, V ⊆ Xirreducible and closed. Then OX,V = lim−→U⊆X,U∩V 6= OX(U).

If X is irreducible, then OX,V = f ∈ k(X) : f is defined at one point ofV .

General Case: U ⊆ X open affine, U ∩ V 6= ∅, P = I(U ∩ V ) ⊆ k[U ],OX,V = k[U ]P .

Definition 4.27 (Regular along V ). X is regular along V if OX,V is a reg-ular local ring. i.e., the maximal ideal in OXV is generated by dim(OX,V ) =codim(V ;X) elements. This happens iff V 6⊆ Xsing.

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5 Divisors

Let X be a normal variety.

Definition 5.1 (Prime Divisor). A prime divisor on X is a closed irreduciblesubvariety of codimension 1.

Note: Y a prime divisor implies that OX,Y is a normal Notherian local ringof dimension 1. That is, OX,Y is a DVR.

Consequences

1. codim(Xsing;X) ≥ 2.

2. Have valuation map vY : k(X)∗ → Z for each prime divisor Y ⊆ X.

Lemma 5.1. Let f ∈ k(X)∗. Then vY (f) = 0 for all but finitely many Y .

Proof. Show vY (f) < 0 for finitely many Y . Set U ⊆ X open set where fdefined. Z = X \ U . vY (f) < 0 iff f /∈ OX,Y iff f is not defined at any point ofY iff Y ⊆ Z component.

Definition 5.2 (Divisor Group). Define Div(X) =free abelian group generatedby all prime divisors.

An element D =∑ni[Yi] is a finite sum and is called a Weil Divisor.

Definition 5.3. For f ∈ k(X)∗, set (f) =∑Y vY (f) · [Y ] ∈ Div(X).

Note (f−1) = −(f), (fg) = (f) + (g) Thus k(X)∗ → Div(X) by f 7→ (f) isa group homomorphism.

Definition 5.4 (Class Group). Define C`(X) = Div(X)/(f) : f ∈ k(X)∗.

Example: C`(An) = 0, as every hypersurface corresponds to a prime divisor.Remark: X complex, dim(X) = n, then C`(X)↔ H2n−2(X; Z).Remark: X irreducible but not normal, we can still define C`(X), use

vY (f/g) = lengthOX,Y (OX,Y /(f))− lengthOX,Y (OX,Y /(g)).Divisors on PnNote: All prime divisors are hypersurfaces Y = V+(h) where h ∈ S =

k[x0, . . . , xn] is an irreducible form.

Definition 5.5. Degree of a Divisor deg : Div(Pn) → Z by deg(∑mi[Yi]) =∑

mi deg Yi.

Let f ∈ k(Pn)∗, mi ∈ Z, g =∏ri=1 h

mii , hi ∈ S irreducible form. Then∑

mi deg(hi) = 0, so Yi = V+(hi) ⊆ Pn is a prime divisor, so vYi(hi) = 1,vYi(f) = mi.

(f) =∑ri=1mi[Yi] implies that deg(f) =

∑mi deg(hi) = 0, thus, deg :

C`(Pn)→ Z is well-defined.Claim: Isomorphism.

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Surjective: H ⊆ Pn hyperplane, deg(m[H]) = m. Injective: Let D =∑mi[Yi] ∈ Div(Pn), suppose deg(D) = 0, then Yi = V+(hi), hi ∈ S irred form,

so∑mi deg(hi) = deg(D) = 0, so f =

∏hmii ∈ k(Pn)∗ and D = (f).

Later: X nonsingular implies C`(X) ' Pic(X), thus Pic(Pn) = Z.X normal, Y ⊆ X prime divisor implies that OX,Y is a DVR.

Theorem 5.2. R normal Notherian domain implies R = ∩ht p=1Rp whereht p = dimRp, the max m such that ∃0 ( p1 ( . . . ( pm = p.

Corollary 5.3. If X is normal, f ∈ k(X)∗, then f ∈ k[X] iff vY (f) ≥ 0 for allY ⊆ X prime divisors.

Lemma 5.4. R Notherian, then R is a UFD iff all prime ideals of height oneare principal.

Proof. ⇒: Assume R a UFD. P ⊆ R prime of height 1, let x ∈ P be anirreducible element, 0 ( (x) ⊆ P , so P = (x).⇐: Exercise: R any Notherian domain then every element of R is a product

of irreducible elements.Unique Factorization: Show x ∈ R irred and x|fg implies that x|f or x|g,

ie, (x) ⊆ R is prime, let P ⊇ (x) min prime, PIT implies ht(P ) = 1 impliesP = (y), x = ay, a ∈ R a unit.

Proposition 5.5. X irreducible affine variety, k[X] a UFD iff X normal andC`(X) = 0.

Proof. ⇒: UFD implies normal. Let Y ⊆ X a prime divisor, P = I(Y ) ⊆ k[X]prime of height 1. P = (h) ⊆ k[X], h ∈ k[X]. So (h) = [Y ] implies [Y ] = 0 ∈C`(X).⇐: Let P ⊆ k[X] prime of height 1, Y = V (P ) ⊆ X a prime divisor,

[Y ] = 0 ∈ C`(X) so [Y ] = (h) ∈ Div(X), h ∈ k(X)∗. vZ(h) ≥ 0 for all Z ⊆ Xprime divisors implies h ∈ k[X]. Claim: P = (h) ⊆ k[X]. ⊇ is clear. Let g ∈ P ,then vY (g) ≥ 1, so vZ(g/f) ≥ 0 for all Z, so g/f ∈ k[X], and g = af ∈ (f).

Proposition 5.6. X normal, Z ⊆ X is a proper closed subset, U = X \ Z.Then

1. C`(X)→ C`(U) by [Y ] 7→ [Y ∩ U ] if Y ∩ U 6= ∅ and 0 else is surjective

2. If codim(Z;X) ≥ 2, then C`(X) = C`(U).

3. If Z prime divisor, then Z→ C`(X)→ C`(U)→ 0 is exact.

Proof. 1. Well defined Div(X)→ Div(U). f ∈ k(X)∗, (f) 7→ (f |U ) (becauseif Y ⊆ X is a prime divisor, Y ∩ U 6= ∅ then OX,Y = OU,U∩Y ).

Thus, C`(X) → C`(U) is well defined. Surjective: If V ⊆ U a primedivisor, V ⊆ X is a prime divisor [V ] 7→ [V ].

2. Div(X) = Div(U), so (f) = (f |U ).

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3. If D =∑nY [Y ] 7→ 0 ∈ C`(U), then ∃f ∈ k(X)∗: vY (f) = nY for all

Y 6= Z. D − (f) = m[Z]⇒ D = m[Z] ∈ C`(X).

Example: X = V (xy − z2) ⊂ A3.Exercise: X (above) is normal. L = V (y) ∩X = V (y, z) is a prime divisor

on X.Max ideal of OX,L is generated by z, y = z2

x ∈ OX,L.Set U = X \ L affine, k[U ] = (k[x, y, z]/(xy − z2))y = k[y, z]y is a UFD, so

C`(U) = 0. Z→ C`(X)→ C`(U) = 0. So C`(X) = m[L] : m ∈ Z, y ∈ k(X)∗.(y) = vL(y)[L] = vL(z2)[L] = 2[L]. Thus, C`(X) = Z/2Z or C`(X) = 0. Ask[x, y, z]/(xy − z2) is not a UFD, C`(X) = Z/2Z.

Picard GroupInvertible OX -module = line bundle.Let X be any variety, L1 and L2 are line bundles, then L1 ⊗OX L2 is

a line bundle. L is invertible implies that we can define L −1 = [U 7→homOU (L |U ,OU )].

Exercise: L −1 is an invertible OX -module and L ⊗L −1 ' OX .

Definition 5.6 (Picard Group). Pic(X) = isomorphism classes of invertiblesheaves on X.

This is a group under tensor product.Notation: X irreducible variety, L an invertible sheaf on X and s ∈ L (U),

t ∈ L (V ) are nonzero sections. Take W ⊂ U ∩ V open such that L |W ' OWgenerated by u ∈ L (W ). Then s|W = fu and t|W = gu, f, g ∈ k[W ]. Defines/t = f/g ∈ k(X)∗.

Example: s0, . . . , sn ∈ Γ(X,L ). Define f : X 99K An ⊂ Pn. f(x) =(s1/s0(x), . . . , sn/s0(x)) = (s1/s0(x) : . . . : sn/s0(x) : 1). If s0, . . . , sn generateL , then f extends to a morphism f : X → Pn.

X is a normal variety, s ∈ L (U), s 6= 0, Y ⊆ X is a prime divisor, takeV ⊆ X open such that L |V ' OV generated by t ∈ L (V ) and V ∩ Y 6= ∅.

Definition 5.7. VY (s) = VY (s/t).

Well defined, if V ′ ⊆ X, V ′ ∩ Y 6= ∅, L |V ′ generated by y′ ∈ L (V ′) impliesthat t/t′ nowhere vanishing function on V ∩ V ′, so t/t′ is a unit in OX,Y .

Thus, VY (s/t′) = VY (s/t · t/t′) = VY (s/t) + 0.

Definition 5.8. (s) =∑Y VY (s)[Y ] ∈ Div(X).

Note: If s′ ∈ L (U ′) then (s′) = (s) + (s′/s). Then (s′) = (s) ∈ C`(X), soall nonzero sections of a line bundle are equivalent in the class group.

Thus, we have a well defined map Pic(X)→ C`(X) by L 7→ (s).Check: s1 ∈ L1(U), s2 ∈ L2(U), then s1⊗s2 ∈ L1⊗L2(U) and (s1⊗s2) =

(s1) + (s2), so this map is a group homomorphism.Cartier DivisorsX normal.

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Definition 5.9 (Cartier Divisor). A Cartier is a Weil Divisor D =∑ni[Yi]

which is locally principal. I.E. there exists an open covering⋃ni=1 Ui = X such

that D|Ui = 0 ∈ C`(Ui) for all i.

Note: D|Uj = (fj) where fj ∈ k(Uj)∗. We can think about D as thecollection fj of these generators.

Definition 5.10 (Cartier Class Group). CaC`(X) = Cartier Divisors onX/(f) : f ∈ k(X)∗.

Recall: L is an invertible OX -module, s ∈ L (U) nonzero section, then(s) =

∑Y prime vY (s) · [Y ] ∈ Div(X) where vY (s) = vY (s/t) for t ∈ L (V )

generator of L |V ' OV , Y ∩ V 6= ∅.Note: (s) is Cartier.Thus, we have a group homomorphism Pic(X)→ CaC`(X) ⊂ C`(X).Line Bundles from DivisorsLet D =

∑nY [Y ] ∈ Div(X).

Definition 5.11. OX-module L (D) or OX(D) Γ(U,L (D)) = f ∈ k(X)∗|vY (f) ≥−nY for all prime divisors Y such that Y ∩ U 6= ∅ ∪ 0.

Example: L (0) = OX(0) = OX .Example: X = P1, Q = (a : b) ∈ X, D = n[Q], Q = V+(h), h = bx0 − ax1 ∈

k[x0, x1]. So OP1(n[Q]) ' OP1(n) by f 7→ hnf .Note: If h ∈ k(X)∗ then OX(D + (h)) → OX(D) by f 7→ hf is an isomor-

phism. vY (f) ≥ −nY − vY (h) ⇐⇒ vY (fh) ≥ −nY .Consequence: D is a Cartier Divisor implies that OX(D) is an invertible

OX -module.If D|U = (h) ∈ Div(U), then OX(D)|U = OU ((h)) ' OU . Note, this says we

have a map CaC`(X)→ Pic(X) by D 7→ OX(D).WARNING: If f ∈ Γ(U,OX(D)) then f a rational function. The notation

(f) means two things!

Proposition 5.7. Pic(X) ' CaC`(X) as abstract groups.

Proof. Will check that Pic(X) ←→ CaC`(X) are inverse maps.Let D =

∑nY [Y ] a Cartier Divisor. Set V = X \ (∪nY <0Y ) ⊆ X open.

Then 1 ∈ k(X)∗ is a section of OX(D) over V . (vY (1) ≥ −nY ⇐⇒ nY ≥ 0).Claim: (1) = D ∈ Div(X). If D|U = (g) ∈ Div(U), then OX(D)|U ' OU

gneerated by h−1 ∈ Γ(U,OX(D)), Y ∩U 6= ∅ implies that vY (1) = vY (1/h−1) =vY (h). Thus, (1)|U = (h)|U = D|U .

Thus, CaC`(X)→ Pic(X)→ CaC`(X) is the identity.Let L be a line bundle on X. t ∈ Γ(U,L ) a non-zero section.Note: If 0 6= s ∈ L (V ), then Y ∩ V 6= ∅ implies vY (s/t) = vY (s)− vY (t) ≥

−vY (t), and so s/t ∈ Γ(V,OX((t))).Claim: L ' OX((t)) by s 7→ s/t. If L |V ' OV gneerated by u ∈ L (V )

then (t)|V = (t/u) ∈ Div(V ) implies that OX((t))|V is generated by u/t asu 7→ u/t we get L |V ' OX((t))|V .

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Examples:

1. Pic(An) = CaC`(An) ⊂ C`(An) = 0, so all line bundles on An are trivial.

Fact: Any locally free OAn -module of finite rank is trivial.

2. Pn =⋃ni=0D+(xi), C`(D+(xi)) = 0, so all Weil divisors are Cartier, thus

Pic(Pn) = CaC`(Pn) = C`(Pn) = Z.

By the maps we have, any line bundle is isomorphic to OPn(m[H]), whereH ⊂ Pn is a hyperplane. I(H) = (h), OPn(m[H]) ' OPn(m) by f 7→ hmf .Thus, Pic(Pn) = O(m).

3. X = V (xy − z2) ⊂ A3. L = V (y) ∩ X, I(L) = (y, z) ⊂ k[A3]. Claim:[L] is not Cartier. Otherwise there exists open affine U ⊂ X such thatP = (0, 0, 0) ∈ U with [L ∩ U ] = (f |U ) ∈ Div(U). Thus f ∈ k[U ] andI(L ∩ U) = (f) ⊂ k[U ], so I(L) · OX,P = (y, z) ⊂ OX,P is principal.

But P ∈ X is a singular point, so dimk(mP /(m)2P ) = 3, so x, y, z is a

basis, and so dim((y, z) + m2P /m

2P ) = 2, so (y, z) ⊂ OX,P is not principal,

which is a contradiction. Thus CaC`(X) = Pic(X) = 0 6= C`(X).

Note: OX,P is not a UFD, as (y, z) ⊂ OX,P is height 1 prime but notprincipal.

Definition 5.12 (Locally Factorial). An irreducible variety X is locally facto-rial if OX,P is a UFD for all p ∈ X.

Example: Nonsingular implies locally factorial implies normal.

Proposition 5.8. X locally factorial implies Pic(X) = C`(X).

Proof. Show that any prime divisor [Y ] is Cartier. First: U = X \Y , [Y ]|U = 0.Let P ∈ Y , then I(Y ) · OX,P ⊂ OX,P is a height 1 prime, so I(Y ) · OX,P =(f) ⊂ OX,P , f ∈ OX,P ⊂ k(X).

Note: vY (f) = 1, if Z 6= Y prime divisor, p ∈ Z, then f ∈ OX,Z (defined atP ) and f /∈ I(Z) · OX,P . Thus, vZ(f) = 0, and we have (f) = [Y ] +

∑ni[Zi]

where p /∈ Zi for all i.Set U = X \ (∪Zi) open in X, p ∈ U . Then [Y ]|U = (f)|U ∈ Div(U)

principal, so [Y ] is Cartier.

Example: X = V (xy − z2) ⊂ A3, X0 = X \ 0, 0, 0, X0 is nonsingular, soPic(X0) = C`(X0) = C`(X) = Z/2Z, so there exists a unique nontrivial linebundle on X0 which is NOT equal to the restriction of a line bundle on X.

Definition 5.13 (Affine, Finite). Let f : X → Y be a morphism of varieties.f is affine if f−1(V ) ⊂ X is affine for all V ⊂ Y is open affine.

f is finite if it is affine and k[f−1(V )] is a finitely generated k[V ]-module.

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Exercise: Enough that this is true for an open affine cover of Y .Examples: X,Y affine, f : X → Y morphism is affine.X ⊂ Y closed, then the inclusion is finite.Divisors on Non-Singular CurvesRecall that X nonsing complete curve implies that X is projective. X ⊂ CK

open, K = k(X), X = CK .

Lemma 5.9. Let X be a complete, nonsingular curve, then any nonconstantmorphism f : X → Y is finite.

Proof. WLOG: Y is a curve. Thus, f∗ : k(Y ) ⊂ k(X) is a finite field extension.Take V ⊆ Y open affine, k[V ] ⊂ k(Y ). Set A = k[V ] ⊂ k(X), then A is afinitely generated k[V ]-module.

U = Spec−m(A) a nonsingular curve, k(U) = k(X)⇒ we have diagram

U V................................................................................................................. ............

X.............................................................................................................................

⊂

Y.............................................................................................................................

⊂

................................................................................................................. ............f

Claim: f−1(V ) = U . x ∈ f−1(V ) ⇒ k[V ] ⊆ OX,x, so A ⊂ OX,x, thusOX,x = AP for some P ⊂ A prime.

Thus, x = P ∈ U = Spec−m(A).

Definition 5.14 (Degree of f). Let f : X → Y be a finite, dominant morphism,then deg(f) = [k(X) : k(Y )].

Pullback of Divisors on Curvesf : X → Y a finite morphism of nonsingular curves. Q ∈ Y , mQ = (t) ⊆

OY,Q. If f(P ) = Q then f∗ : OY,Q → OX,P , f∗t ∈ mP .

Definition 5.15. f∗ : Div(Y )→ Div(X) : [Q]→∑P∈f−1(Q) vP (t)[P ].

Alternatively, if D ∈ Div(Y ), set V = Y−Supp(D), then s = 1 ∈ Γ(V,L (D)).Note: (s) = D. Then f∗s ∈ Γ(f−1(V ), f∗L (D)) is the pullback.

Exercise: f∗D = (f∗s) ∈ Div(X).

Definition 5.16 (Torsion Free). Let R be a domain, M an R-module. M istorsion free if ∀a ∈ R, x ∈M , then ax = 0 implies a = 0 or x = 0.

Fact: Any f.g. torsion-free module over a PID is free.

Definition 5.17 (Degree of a Divisor). X a nonsingular curve, D =∑ni[Pi] =∑

niPi ∈ Div(X). Set deg(D) =∑ni

Warning: If X is not complete, then deg is not defined on C`(X).

Proposition 5.10. f : X → Y is a finite morphism of nonsingular curves,D ∈ Div(Y ). Then deg(f∗D) = deg(f) deg(D).

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Proof. ETS if Q ∈ Y a point, then deg(f∗Q) = deg(f). V ⊆ Y open affine withQ ∈ V . Then f−1(V ) = Spec−m(A) ⊂ X, A = k[V ] ⊂ k(X).

Q ⊂ k[V ] a max ideal, set B = AQ = (k[V ] \ Q)−1A. A finitely generatedk[V ]-module implies B f.g. k[V ]Q = OY,Q-module. OY,Q a DVR, B torsion free,so B is free OY,Q-module.

rankOY,Q(B) = dimk(Y ) k(X) = deg(f). mQ = (t) ⊂ OY,Q. OY,Q/tOY,Q =k. Thus, dimk(B/tB) = deg(f).

Note: points in f−1(Q) correspond to max ideals in P ⊂ A such that P ∩k[V ] = Q, which correspond to max ideals in AQ = B.

Write f−1(Q) = P1, . . . , Ps, Pi ⊆ A max ideals, B = ∩si=1BPi ⇒ tB =∩si=1tBPi = ∩si=1(tBPi ∩B).

By the Chinese Remainder Theorem, B/tB ' ⊕si=1B/(tBPi ∩B).Injective: clearSurjective: t ∈ Pi for all i, BPi DVR, so tBPi = (PiBPi)

ni , so tBPi∩B ⊆ PiBand tBPi ∩B 6⊆ PjB for j 6= i.

Thus, this is an isomorphism after ⊗BPi (only the ith summand survives).Now: B/(tBPi ∩ B) = (B/(tBPi ∩ B))Pi = (B/tB)Pi = BPi/tBPi =

OX,Pi/(t). Thus dimk B/(tBPi ∩B) = vPi(t).Thus, deg(f∗Q) =

∑vPi(t) = dimk(B/tB) = deg(f).

Lemma 5.11. h ∈ k(Y )∗ implies f∗((h)) = (f∗h). f∗h = h f ∈ k(X).

Proof. Let P ∈ X, Q = f(P ) ∈ Y . mP = (s) ⊆ OX,P . mQ = (t) ⊂ OY,Q.h = utm, u ∈ OY,Q a unit. f∗t = vsn, v ∈ OX,P a unit.Coef of [P ] in f∗((h)) = vQ(h)vP (t) = mn.f∗h = f∗(utm) = (f∗u)(f∗t)m = (f∗u)vmsnm, so coef of [P ] in (f∗h) is

nm.

So, we have a group homomorphism f∗ : C`(Y )→ C`(X).

Corollary 5.12. X complete nonsing curve, f ∈ k(X)∗ implies deg((f)) = 0.

Proof. f is defined on open U ⊂ X. Then f : U → A1 ⊂ P1 is a regularfunction. As P1 is complete, f extends to f : X → P1. As X is complete, f isfinite.

k[A1] = k[t], f∗(t) = f ∈ k(X). (f) = (f∗t) = f∗((t)), so deg((f)) =deg(f∗((t))) = deg(f) deg(t).

(t) = [0]− [∞] ∈ Div(P1) so deg((t)) = 0.

So if X is a complete nonsingular curve, there exists deg : C`(X)→ Z.Notation: D,D′ ∈ Div(X), D ∼ D′ iff D = D′ ∈ C`(X).

Proposition 5.13. If X complete nonsingular curve, then X rational iff ∃P 6=Q ∈ X such that P ∼ Q.

Proof. ⇒: X = P1, then P ∼ Q for all P,Q ∈ P1.⇐: ∃f ∈ k(X)∗ such that (f) = [P ] − [Q] ∈ Div(X). f : X → P1 a

morphism. Then (f) = (f∗t) = f∗((t)) = f∗([0]− [∞]).This tells us that f∗([0]) = [P ] and f∗([∞]) = [Q]. So deg(f∗[0]) = 1 =

deg(f) · 1, so f is degree 1, so it is birational. Thus isomorphism.

55

Definition 5.18 (Elliptic Curve). An elliptic curve is a nonsingular closedplane curve E ⊂ P2 such that deg(E) = 3.

Example: V+(zy2 − x3 + z2x) ⊂ P2.Claim: No elliptic curve is rational.Exercise: Set OE(1) = OP2(1)|E . Then Γ(E,OE(1)) = (S/I(E))1, S =

k[x, y, z].Therefore, dimk Γ(E,OE(1)) = 3.Let L ⊂ P2 be a line. L.E =

∑P ′∈L∩E I(L·E;P ′)[P ′] = P+Q+R ∈ Div(E).

Exercise: OE(1) = OP2([L])|E = OE([L.E]), so ∃D(= L.E), D ∈ Div(E)such that deg(D) = 3 and dimk Γ(E,OE(D)) = 3.

Let D ∈ Div(P1) such that deg(D) = 3. Then OP1(D) = OP1(3), this givesus that Γ(P1,OP1(D)) = Γ(P1,OP1(3)) = k[x0, x1]3. The dimension of this is 4.

Conclude: E is not rational.Let X ⊂ P2 be any nonsingular curve. deg : C`(X)→ Z, [P ] 7→ 1.

Definition 5.19. C`0(C) = ker(deg). So we have a short exact sequence 0 →C`0(X)→ C`(X)→ Z→ 0 that splits, so C`(X) = C`0(X)⊕ Z.

Fact: C`0 corresponds to a nonginular complete abelian algebraic group, theJacobi Variety of X.

Let L = V+(f), M = V+(g) be lines in P2.X.L =

∑P I(X · L;P )P = P1 + . . . + Pn where n = deg(X). X.M =

Q1 + . . .+Qn.Exercise: f/g ∈ k(X)∗ and (f/g) = X.L−X.M ∈ Div(X).X.L−X.M = P1 + P2 + P3 −Q1 −Q2 −Q3 = 0 ∈ C`(X) for X = E.

Theorem 5.14. Let P0 ∈ E be any point, then E → C`0(E) by P 7→ P −P0 isbijective.

Proof. Injective: If P − P0 = Q − P0 ∈ C`0(E) then P ∼ Q so P = Q as E isnot rational.

Surjective: Let M ⊂ P2 be tangent line to E at P0. M.E = 2P0 +R, R ∈ E.Let D ∈ C`0(E). Write D =

∑ni(Qi − P0) for Qi ∈ E, ni ∈ Z.

Assume that ni < 0. Then L =line through Qi and R, L.E = Qi +R+Q′i,0 = L.E −M.E = Qi +R+Q′i − 2P0 −R so Qi − P0 = −(Q′i − P0) ∈ C`0(E).

Replace Qi by Q′i, ni 7→ −ni, WLOG, ni ≥ 0.Claim: D = P − P0 ∈ C`0(E), P ∈ E. Induction on

∑ni.

If∑ni ≥ 2, then Q1 − P0, Q2 − P0 have positive coefficients. L =line

through Q1, Q2, L.E = Q1 +Q2 +Q′ ∈ Div(E).Let L′ be the line through Q′ and P0. Then L′.E = Q′ + P0 + Q′′. L.E −

L′.E = Q1+Q2−P0−Q′′ = 0, so (Q1−P0)+(Q2−P0) = (Q′′−P0) ∈ C`0(E).

Example: char k 6= 2, λ ∈ k, λ 6= 0, 1. Eλ = V+(zy2−x(x−z)(x−λz)) ⊂ P2.Take P0 = (0 : 1 : 0) ∈ E Eλ corresponds to C`0(Eλ) by P 7→ P − P0. Let ⊕be a group op on W . Q1 ⊕Q2 = Q′′ (Picture omitted)

Fact: Any elliptic curve is isomorphic to Eλ by P0 ↔ (0 : 1 : 0).

56

Theorem 5.15. E is an algebraic group.

Proof. (char k 6= 2): WLOG, E = Eλ, P0 = (0 : 1 : 0). Define ϕ : E × E → Eby ϕ(P,Q) = R the unique point such that ∃ a line L with L.E = P +Q+ R.It is enough to show that ϕ : E × E → E is a morphism.

P ⊕Q = ϕ(P0, ϕ(P,Q)), −P = ϕ(P1, P0). Set U1 = D+(z) and U2 = D+(y)subsets of E. E = U1 ∪ U2.

Show that (Ui×Uj)∩ϕ−1(U`)→ U` by ϕ is a morphism for all i, j, ` ∈ 0, 1.U1 = V (y2−x(x−1)(x−λ)) ⊂ A2. (U1×U1)∩ϕ−1(U1)→ U1 → k is the regu-

lar function (x1, y1)×(x2, y2) 7→

(y2−y1)2

(x2−x1)2 − (x1 + x2) + 1 + λ x1 6= x2(x21+x1x2+x2

2+λ−(1+λ)(x1+x2)y1+y2

)2

+ 1 + λ− (x1 + x2) y1 + y2 6= 0

DifferentialsR is a ring, S is a commutative R-algebra, M an S-module.

Definition 5.20 (R-derivation). A function D : S → M is an R-derivation ifD(fg) = fD(g)+gD(f) for all f, g ∈ S, D(f+g) = D(f)+D(g), and D(f) = 0for all f ∈ R.

Remark: the third conditioon holds iff D is an homomorphism of R-modules.⇒: f ∈ R⇒ D(fg) = fD(g) and ⇐ is an exercise (use D(1) = 0).

Definition 5.21 (Module of Kahler differentials). F=free S-module with basisd(f)|f ∈ S = ⊕f∈SS · d(f). F ′ =submodule generated by d(f) for f ∈ R,d(fg)− fd(g)− gd(f), d(f + g)− d(f)− d(g).

We define ΩS/R = F/F ′ is the module of Kahler differentials of S over R

We define d = dS = dS/R : S → ΩS/R by f 7→ d(f) + F ′. This is theuniversal R-derivation of S.

It has the universal property that given any R-derivation D : S →M , thereexists a unique map S-homomorphism D : ΩS/R →M such that D = D dS .

Exercise: Let P (x1, . . . , xn) ∈ R[x1, . . . , xn] and f1, . . . , fn ∈ S, and D : S →M is an R-derivation. Then D(P (f1, . . . , fn)) =

∑ni=1

∂P∂xi

(f1, . . . , fn)D(fi).Consequence: If S generated by f1, . . . , fn as an R-algebra, then ΩS/R is

gererated by dS(f1), . . . , dS(fn) as an S-module.

Proposition 5.16. S = R[x1, . . . , xn]. Then ΩS/R is the free S-module ondx1, . . . , dxn.

Proof. Have a surjectve S-hom from Sn → ΩS/R which sends ei 7→ dxi. This

is surjective. We define D : S → Sn by P (x1, . . . , xn) 7→(∂P∂x1

, . . . , ∂P∂xn

). By

the universal property, there is a unique S-homomorphism D : ΩS/R → Sn, bydefinition, d(xi) 7→ D(xi) = ei, so this is an inverse.

Proposition 5.17. Given ring homomorphisms R → S → Y then we have anexact sequence of T -modules ΩS/R ⊗S T → ΩT/R → ΩT/S → 0.

57

Proof. S → TdT→ ΩT/R is an R-deriv of S. So we get S-hom ϕ : ΩS/R → ΩT/R

via ϕ(dS(f)) = dT (f). Thus, we have a T -hom ΩS/R ⊗ Tϕ→Ω T/R by ω ⊗ h 7→

hϕ(ω).Note: Image(ϕ) =submodule of ΩT/R generated by dT (f) for f ∈ S. Thus

ΩT/R/ Im(ϕ) = ΩT/S .

Note: I ⊂ S an ideal, T = S/I, then I/I2 is a T -module, T × I/I2 → I/I2

by (f + I) · (h+ I2) = fh+ I2 ∈ I/I2.

Proposition 5.18. T = S/I. We have an exact sequence of T -modules I/I2 →ΩS/R ⊗S T → ΩT/R → 0, where the first map is given by h+ I2 7→ ds(h)⊗ 1.

Proof. Set M equal to the image of I/I2 in ΩS/R ⊗S T . Then M is generatedby ds(h)⊗ 1|h ∈ I.

We define D : T → (ΩS/R ⊗ T )/M by D(f + I) = (dS(f)⊗ 1) +M . This isan R-derivation.

Thus, there is a unique T -hom D : ΩT/R → (ΩS/R⊗S T )/M by dT (f + I) 7→(ds(f)⊗ 1) +M

Example: S = R[x1, . . . , xn], I = (f1, . . . , fp) ⊂ S. T = S/I. ΩS/R ⊗S T =⊕ni=1Tdxi = T⊕n.

The image of fi under I/I2 → T⊕n: ds(fi) ⊗ 1 =(∂fi∂x1

, . . . , ∂fi∂xn

). Set

J =(∂fi∂xj

)∈Mat(p× n;T ) is the Jacobi matrix.

So the image of fi is eiJ , so ΩT/R = coker(I/I2 → ΩS/R⊗T ) = coker(T p J→Tn).

e.g. T = k[x, y]/(y2 − x3 + x), so J = [1− 3x2, 2y], so ΩT/k = T ⊕ T/〈(1−3x2)e1 + (2y)e2〉.

Proposition 5.19. S an R-algebra, U ⊆ S multiplicatively closed subset, thenΩU−1S/R = U−1ΩS/R

Proof. S → U−1S → ΩU−1S/R is an R-derivation. Thus, it induces an S-homomorphism ΩS/R → ΩU−1S/R, dS(f) 7→ d(f), where d is the universalderivation of U−1S.

This induces U−1S-hom U−1ΩS/R → ΩU−1S/R by ds(f)/u 7→ u−1d(f).We define D : U−1S → U−1ΩS/R by D(s/u) 7→ ud(s)−sd(u)

u2 . Exercise: D iswell defined R-derivation.

This induces D : ΩU−1S/R → U−1ΩS/R is the inverse map.

Let X be a topological space. R,S sheaves of rings on X, R → S a ringhom.

Definition 5.22. pre−ΩS /R(U) = ΩS (U)/T (U) for U ⊆ X open. For V ⊂ Uopen, S (U)→ S (V ) d→ pre−Ω(V ) is an R(U)-derivation. So, we get S (U)-hom pre− Ω(U)→ pre− Ω(V ).

We define ΩS /R = (pre− ΩS /R)+, the sheafification.

58

Let ϕ : X → Y morphism of varieties, then we have ring hom ϕ∗ : ϕ−1OY →OX .

Definition 5.23 (Relative cotangent sheaf). ΩX/Y = ΩOX/ϕ−1OY is called therelative cotangent sheaf

Special case: X → pt, ΩX = ΩX/k = ΩX/pt. This is called the cotangentsheaf.

Proposition 5.20. ϕ : X → Y a morphism of affine varieties, then ΩX/Y =˜Ωk[X]/k[Y ]

Proof next time.As a consequence, ΩX/Y is always coherent.

Lemma 5.21. If (A,m) is a local Notherian domain, N a finitely generatedA-module, then we set r = dimA/m(N/mN). If r ≤ dimA0(N0), then N is freeof rank r.

Proof. Nakayama’s Lemma implies that N can be generated by r elements.Thus, there exists an exact sequence 0 → K → Ar → N → 0, localizationis exact, so 0 → K0 → Ar0 → N0 → 0 is exact, so the last morphism is anisomorphism of vector spaces, so Ar0 ' N0, so K0 = 0. Thus, K = 0 as it istorsion free and localizes to zero, so Ar ' N .

Recall: Let X ⊂ An be a closed irreducible variety. Let I = I(X) =(f1, . . . , fs). Let P ∈ X. Set M = I(p) ⊂ k[An], M/M2 ' kn via h+M2 7→(∂h∂x1

(P ), . . . , ∂h∂xn (P ))

So we have an exact sequence (I + M2)/M2 → M/M2 → mP /m2P → 0.

mP ⊂ OX,P a max ideal, therefore ks → kn → mP /m2P → 0 is also exact where

the first map is J(P ), and we call the second φ.If h ∈M , then φ

(∂h∂x1

(P ), . . . , ∂h∂xn (P ))

= h+ m2P .

Note: rank(k(X)s J→ k(X)n) ≤ c = codim(X; An). (If h is any (c+ 1)× (c+1)-minor of J , then h ∈ k[X], and h(P ) = (c+ 1)× (c+ 1)-minor. J(P ) = 0 forall P ∈ X. So h = 0 ∈ k[X].)

Theorem 5.22. Assume X is an irreducible variety of dimension r, let P ∈X. Then P is a nonsingular point iff ΩX,P ' O⊕rX,P . If mP is generated byh1, . . . , hr ∈ mP , then dh1, . . . , dhr ∈ ΩX,P is a basis for ΩX,P .

Proof. WLOG: X ⊆ An affine. I(X) = (f1, . . . , fs), J =(∂fi∂xj

).

k[X]s J→ k[X]n → Ωk[X]/k → 0 yields OsX,P

J→ OnX,P → ΩX,P → 0, which

we will call (∗).We mod out by mP , and get ks

J(P )→ kn → ΩX,P /mPΩX,P → 0.Thus, mP /m

2P ' ΩX,P /mPΩX,P by h+ m2

P 7→∑nj=1

∂h∂xj

(P )dxj .Assume that ΩX,P is free of rank r, then dimk(mP /m

2P ) = r, thus P is a

nonsingular point.

59

Assume that dimk(mP /m2P ) = r. (∗) ⇒ k(X)s J→ k(X)n → (ΩX,P )0 → 0 is

exact.Note: r = dimk(ΩX,P /mPΩX,P ) ≤ dimk(X)((ΩX,P )0).The lemma implies that ΩX,P ' O⊕rX,P .

Lemma 5.23. ϕ : X → Y is a morphism of affine varieties. Then Γ(X, pre−ΩX/Y ) = Ωk[X]/k[Y ]

Proof. S = Γ(X,ϕ−1OY ), ring homomorphisms k[Y ] → S → k[X]. Thus,ΩS/k[Y ]⊗Sk[X]→ Ωk[X]/k[Y ] → Ωk[X]/S → 0 where the last is Γ(X, pre−ΩX/Y ),so enough to show that the first map is zero.

Let f ∈ Im(S → k[X]). We must show that df = 0 ∈ Ωk[X]/k[Y ]. Thereexists open cover X = ∪ni=1Ui such that f |Ui ∈ image of Γ(Ui, pre− ϕ−1OY ) =lim−→V⊃ϕ(Ui)

OY (V )WLOG, Ui = Xgi where gi ∈ k[X]. Enough to show that df = 0 in(

Ωk[X]/k[Y ]

)gi

for each i, since gNi df = 0 ∈ Ωk[X]/k[Y ] and (gNi , . . . , gNn ) =

(1) = k[X].But

(Ωk[X]/k[Y ]

)gi

= Ωk[Ui]/k[Y ]. So we replace X with Ui, we may assumethat f ∈ image of Γ(X, pre − ϕ−1OY ) = lim−→V⊃ϕ(X)

OY (V ). I.E. there existsV ⊂ Y open, f ′ ∈ OY (V ) such that ϕ(X) ⊂ V and f = ϕ∗(f ′) ∈ k[X]. NowV = ∪mi=1Yhi , hi ∈ k[Y ], f ′ ∈ k[Y ]hi , so hNi f

′ ∈ k[Y ] for all i, so hN+1i df =

d(hN+1i f) = 0 ∈ Ωk[X]/k[Y ].Now, X = ∪Xhi ⇒ (hN1 , . . . , h

Nm) = (1) ⊂ k[X], so df = 0.

Proposition 5.24. ϕ : X → Y morphism of affines. Then ΩX/Y = ˜Ωk[X]/k[Y ]

Proof. Set Ω = Ωk[X]/k[Y ]. We have Ω ' Γ(X, pre−ΩX/Y )→ Γ(X,ΩX/Y ), thisgives an OX -homomorphism Ω→ ΩX/Y .

Let f ∈ k[X]. Γ(Xf , Ω) = Ωf = Ωk[Xf ]/k[Y ] = Γ(Xf , pre− ΩX/Y ).Xf is a basis for the top, and so they have the same stalks.

Corollary 5.25. ϕ : X → Y any morphism of varieties. Then ΩX/Y is coher-ent.

Proof. Let Y = ∪Vi open affine cover. ϕ−1(Vi) = ∪Uij ⊆ X is an open affinecover of X. ΩX/Y |Uij = Ωk[Uij ]/k[Vi].

Corollary 5.26. If X irreducible, then X is nonsingular iff ΩX is a locally freeOX-module.

Example: X = P1, ΩP1 is a line bundle. The projective coordinate ring isk[x0, x1]. Set t = x1

x0∈ k(P1)∗. t ∈ OP1(D+(x0)), dt ∈ Γ(D+(x0),ΩP1). Find

(dt) ∈ Div(P1).Ui = D+(xi), U0 = A1 ⊂ P1. If p ∈ U0, then t− p generated mp, so ΩP1,p is

generated by d(t− p) = dt, so vp(dt) = 0 for all p ∈ U0.k[U1] = k[s], s = t−1, dt = d(s−1) = −s−2ds, v∞(dt) = v∞(s−2) = −2.

Thus (dt) = −2[∞] ∈ Div(P1), and so ΩP1 ' OP1(−2[pt]) = OP1(−2).

60

Example: E = Eλ ⊂ P2 an elliptic curve. Then ΩE ' OE .Linear SystemsLet X be a nonsingular projective variety. D =

∑nY [Y ] ∈ Div(X). We say

that D is effective if nY ≥ 0 for all Y . If D is effective, we write D ≥ 0.

Definition 5.24 (Complete Linear System of D). Given any D ∈ Div(X),define |D| = D′ ∈ Div(X) : D′ ∼ D and D′ ≥ 0.

Theorem 5.27. If X is projective and F is a coherent OX-module, thendimk Γ(X,F ) <∞.

Definition 5.25. Let `(D) = dimk Γ(X,OX(D)).

Theorem 5.28. P(Γ(X,OX(D)))→ |D| by s 7→ (s) is bijective.

Proof. Let s ∈ Γ(X,OX(D)), then (s) ≥ 0 and (s) ∼ D. So the map is welldefined.

Injective: Suppose s1, s2 ∈ Γ(X,OX(D)), assume that (s1) = (s2) ∈ Div(X).Then (s1/s2) = (s1)− (s2) = 0, so s1/s2 ∈ k[X] = k.

Surjective: Let D′ ∈ |D|. Then D′ ∼ D, so D′ = D + (f) where f ∈ k(X)∗.We define s to be the section given by f ∈ Γ(X,OX(D)). This is a global section,because vY (f) ≥ −nY for all Y , (D =

∑nY [Y ]). Set s0 = 1 ∈ Γ(U,OX(D)).

(s) = (f · s0) = (f) + (s0) = (f) +D = D′.

Lemma 5.29. X is a complete nonsingular curve, D ∈ Div(X), if `(D) 6= 0then deg(D) ≥ 0 and if `(D) 6= 0 and deg(D) = 0 then D ∼ 0.

Proof. `(D) 6= 0, then |D| 6= ∅, so D ∼ D′ ≥ 0. So deg(D) = deg(D′) ≥ 0.If deg(D) = 0, then deg(D′) = 0, but as D′ is effective, D′ = 0.

Riemann-Roch TheoremLet X be a complete nonsingular curve.

Definition 5.26 (Canonical Divisor). K ∈ Div(X) is a canonical divisor on Xif ΩX ' O(K).

Definition 5.27 (Genus). The genus g = `(K) = dimk Γ(X,ΩX)

Example: X = P1, ΩP1 = O(−2), so g = 0.Example: E = Eλ ⊂ P2 elliptic curve, ΩE ' OE . So g = 1.

Theorem 5.30 (Riemann-Roch). For any D ∈ Div(X) where X is a completenonsingular curve, we have `(D) + `(K −D) = deg(D) + 1− g.

Example: X = P1, K = −2P for some P ∈ P1. The RRT theorem says that`(nP ) + `(−2P − nP ) = n+ 1− 0, so if n ≥ 0, we have that `(nP ) = n+ 1. Ifn = −1, then 0 + 0 = −1 + 1 = 0. If n ≥ −2, we also see that it works.

Example: Set D = K, then `(K)− `(K −K) = deg(K) + 1− g, so g − 1 =deg(K) + 1− g, so deg(K) = 2g − 2.

Corollary 5.31. A nonsingular curve is either affine or projective.

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Proof. C nonsingular curve implies that C = CK \ P1, . . . , Pn where K =k(C), X = CK . Ifm >> 0, then `(mPi) = m+1−g ≥ 2. 1, fi ∈ Γ(X,OX(mPi)),fi /∈ k.

(fi) = −ri[Pi] effective divisor in Div(X). Set f =∑ni=1 fi ∈ k(X)∗. f is

defined exactly on C ⊆ X, so C ' Spec−m(k[f ]) is affine.

Corollary 5.32. X is rational iff g = 0.

Proof. ⇒: genus of P1 is 0.⇐: Let P 6= Q ∈ X. Set D = P − Q ∈ Div(X). By Riemann-Roch,

`(D) ≥ deg(D) + 1− g, so `(D) ≥ 1. Thus |D| 6= ∅, so there is D′ ≥ 0 such thatD ∼ D′, but deg(D′) = 0, so D′ = 0.

Corollary 5.33. X complete nonsingular curve of g = 1, P0 ∈ X, char(k) 6= 2.Then ∃ isomorphism X ' Eλ = V+(zy2 − x(x − z)(x − λz)) ⊂ P2 for some λnot 0 or 1, that sends P0 7→ (0 : 1 : 0).

Proof. deg(K) = 2g − 2 = 0, so Riemann-Roch implies `(nP0) = n+ 1− 1 = nfor all n ≥ 1, k = Γ(OX) = Γ(OX(P0)) ( Γ(OX(2P0)) ( . . . ( k(X).

Take x ∈ Γ(OX(2P0)) \ k. vP0(x) = −2 (x) = A+B − 2P0 for A,B ∈ X.x : X → P1 is a morphism, x∗([0]− [∞]) = A+B− 2P0, so x∗([0]) = A+B,

thus [k(X) : k(x)] = deg(x) = 2.Take y ∈ Γ(OX(3P0)) \ Γ(OX(2P0)). vP0(y) = −3, but as this is odd,

y /∈ k(x). Thus k(X) = k(x, y).1, x, y, x2, xy is a basis for Γ(OX(5P0)), 1, x, y, x2, xy, x3, y2 ∈ Γ(OX(6P0)),

dim = 6.So there exists a linear relations. Rescale x, y: y2 +b1xy+b0y = x3 +a2x

2 +a1x + a0. Replace y with y + 1

2 (b1x + b0), y2 = (x − a)(x − b)(x − c) wherea, b, c ∈ k.

Claim: a 6= b 6= c 6= a.ϕ : X \ P0 → C = V (y2 − (x− a)(x− b)(x− c)) ⊂ A2, P 7→ (x(P ), y(P )).

This is birational, as k(X) = k(x, y).Assume a = b = c. Then C is a curve with a cusp, and is rational, so X

would be rational, contradiction.Assume a = b 6= c, then C is the nodal curve, which is also rational.Replace x with x−a

b−a , rescale y, y2 − x(x− 1)(x− λ) where λ = c−ab−a 6= 0, 1.

X \ P0 → C ⊂ A2 extends to an isomorphism X → Eλ.

Lemma 5.34. X complete nonsingular curve of genus g. Let P0, Q0, . . . , Qg ∈X, then there exists P1, . . . , Pg ∈ X such that

∑gi=0 Pi ∼

∑gi=0Qi.

Proof. WLOG P0 6= Qi for all i.Set D =

∑Qi. `(D) ≥ deg(D) + 1− g = 2. Thus, ∃h ∈ Γ(X,OX(D)) such

that h /∈ k. Set f = h− h(P0) ∈ k(X)∗. (f) = −D + P0 + P1 + . . .+ Pg = 0 ∈C`(X).

Corollary 5.35. The Map Xg = X × . . . × X with g factors to C`0(X) by(P1, . . . , Pg) 7→

∑gi=1(Pi − P0) is surjective.

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Proof. Note: Let Q ∈ X, the lemma implies that there are P1, . . . , Pg ∈ X suchthat (g + 1)P0 ∼ Q+ P1 + . . .+ Pg, so −(Q− P0) =

∑gi=1(Pi − P0) ∈ C`0(X).

Let D =∑nQ(Q − P0) ∈ C`0(X). The note implies that we can assume

nQ ≥ 0, and the lemma implies that we may assume∑nQ ≤ g.

Blow-Up of VarietiesY affine variety, X ⊆ Y closed, I = I(X) = (f0, . . . , fn) ⊂ k[Y ]. Let

ϕ : Y \X → Pn by ϕ(y) = (f0(y) : . . . : fn(y)).

Definition 5.28 (Blowup of Y along X.). B`X(Y ) = (y, ϕ(y)) : y ∈ Y \X ⊂Y × Pn.

Note: If Y \X ⊆ Y is dense, then π : B`X(Y )→ Y is surjective because Pnis complete. And π : π−1(Y \X)→ Y \X is an isomorphism.

Point: If Y is singular along X, then usually B`X(Y ) is less singular.Example: Y = V (y2−x2(x+1)) ⊂ A2. I(X) = (x, y) ⊂ k[Y ]. ϕ : Y \0 →

P1, P 7→ line through O and P .B`X(Y ) = (P,ϕ(P )), (0, (1 : −1)), (0, (1 : 1)).Note: π−1(X) ⊂ B`X(Y ) is an effective Cartier divisor. i.e. codim = 1 and

the ideal of π−1(X) is locally generated by a single element.In fact: L = π∗Pn(OPn(1)). Define si = π∗Pn(zi) ∈ Γ(B`X(Y ),L ), zi ∈

Γ(Pn,OPn(1)) for 0 ≤ i ≤ n. We define s =∑ni=0 fisi ∈ Γ(B`X(Y ),L ). Then

π−1(X) = Z(s) ⊂ B`X(Y ).Note: B`X(Y ) is independent of the generators fi of I(X). B`X(Y ) ⊂

Y × Pn, set J = I(B`X(Y )) ⊂ k[Y ][z0, . . . , zn], a graded ideal.It is a fact that one can recover B`X(Y ) from k[Y ][z0, . . . , zn]/J .Claim: k[Y ][z0, . . . , zn]/J ' ⊕d≥0I

d and this is by definition ⊕Idtd ⊂ k[Y ][t]and it is the subring of k[Y ][t] generated by k[Y ] as well as tf0, . . . , tfn.

Morphism ψ : Y × A1 → Y × An+1 by ψ(y, t) 7→ (y, (tf0(y), . . . , tfn(y))).ψ(Y × A1) =cone over B`X(Y ), so J = I(ψ(Y × A1)) ⊂ k[Y ][z0, . . . , zn]. ψ∗ :k[Y ][z0, . . . , zn]→ k[Y ][t] with zi 7→ tfi.

J = ker(ψ∗), so k[Y ][z0, . . . , zn]/J ' Im(ψ∗) = k[Y ][tf0, . . . , tfn] = ⊕d≥0Id.

Now let Y be any variety, X ⊂ Y any closed subset. Take and openaffine cover Y = ∪Yi. B`Yi∩X(Yi) can be glued together to get B`X(Y ) =∪iB`Yi∩X(Yi).

IX ⊂ OY a sheaf of ideals. ⊕d≥0I dX is a sheaf of graded OY -algebras, which

can be turned into a variety, B`X(Y ) (see next semester).

6 Schemes

We will try to state definitions and theorems from commutative algebra, butwill not prove many of them, as our focus is geometry.

Let A be a ring

Definition 6.1 (SpecA). We define SpecA = prime ideals P ⊂ A.

If I ⊆ A, we set V (I) = P ∈ SpecA : I ⊆ P.

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Lemma 6.1. 1. V (IJ) = V (I ∩ J) = V (I) ∪ V (J)

2. V (∑Iα) = ∩V (Iα)

3. V (0) = SpecA, V (A) = ∅

4. V (I) ⊆ V (J) ⇐⇒√I ⊇√J

Proof. We begin with 4: This follows from the fact that V (I) = V (√I) and√

I = ∩P∈V (I)P .1: IJ ⊂ I ∩ J ⊆ I, so V (IJ) ⊇ V (I ∩ J) ⊇ V (I) ∪ V (J). Let P ∈ V (IJ).

Assume it is not in V (I). Then there exists a ∈ I with a /∈ P . Then for anyb ∈ J , ab ∈ IJ ⊆ P , so b ∈ P , thus J ⊂ P , so P ∈ V (J).

2 and 3 are straightforward exercises.

Topology: Let U ⊂ Spec(A) be open iff Spec(A)\U = V (I) for some I ⊂ A.We note that P ∈ Spec(A), then P = V (P ), so P is a closed point iff P is

maximal.Example: Take k = k an alg closed field, A = k[x, y]. Then Spec(A) =

(x−a, y− b) : a, b ∈ k∪(f) : f(x, y) irreducible∪0, and these correspondto (a, b) ∈ k × k ∪ irred curves ⊂ k × k ∪ k × k. The points (a, b) arethe closed points, the others are called generic points for curves or the plane,because their closure is either everything (generic point of the plane) or are allof the points that lie on the curve.

Structure Sheaf: For P ∈ Spec(A), set AP = a/f : a, f ∈ A, f /∈ P. ForU = Spec(A), open define, O(U) = s : U →

∐P∈U AP : s(P ) ∈ AP and s

is locally a quotient, that is, ∀P ∈ U , there exists an open neighborhood V ,P ∈ V ⊂ U and a, f ∈ A such that s(Q) = a/f ∈ AQ for all Q ∈ V .

Note: (1) O is a shead of rings on Spec(A)(2) The O(U) are not ”really” functions, so care must be taken.

Definition 6.2 (Spectrum of a Ring). The spectrum of A is (Spec(A),O).

Definition 6.3. For f ∈ A, set D(f) = P ∈ Spec(A) : f /∈ P = Spec(A) \V (f).

Note: D(f) is a basis for the topology on Spec(A). Let U ⊆ Spec(A) beopen, P ∈ U . Write Spec(A) \ U = V (I). Then P /∈ V (I), so I 6⊂ P , thus∃f ∈ I, f /∈ P , so P ∈ D(f) ⊂ U

Proposition 6.2. 1. OP = AP for all P ∈ Spec(A)

2. O(D(f)) = Af , for all f ∈ A.

3. O(Spec(A)) = A.

Proof. 1. For P ∈ U open, we have a ring homomorphism O → AP bys 7→ s(P ). This induces ϕ : OP → AP by ϕ(sP ) = s(P ). This is surjective,as we can let a/f ∈ AP . Then a/f ∈ O(D(f)) and ϕ(a/f) = a/f . It isinjective, as we can assume that ϕ(sP ) = 0 ∈ AP , where s ∈ O(U), U

64

open, P ∈ U . WLOG, s = a/f for all Q ∈ U , a/f = 0 ∈ AP , so f /∈ Pand there exists g ∈ A \ P such that ga = 0 ∈ A. s|U∩D(g) = ag

fg = 0 ∈O(U ∩D(g))⇒ sP = 0 ∈ OP .

2. Define ψ : Af → O(D(f)) by ψ(a/fn) = [P 7→ a/fn ∈ AP ]. For injec-tivity, assume that ψ(a/fn) = 0 ∈ O(D(f)). Set I = Ann(a) ⊂ A, forP ∈ D(f), we have a/fn = 0 ∈ AP , so there exists H ∈ A \ P such thatha = 0. Thus h ∈ I, h /∈ P , so P /∈ V (I). Therefore, D(f) ∩ V (I) = ∅, soV (I) ⊂ V (f), thus f ∈

√I, so fm ∈ Ann(a), so a/fn = 0 ∈ Af .

For surjectivity, let s ∈ O(D(f)). There exists an open cover D(f) = ∪Visuch that s = ai/gi on Vi for all i. WLOG, Vi = D(hi) for hi ∈ A. D(hi) ⊂D(gi) implies that V (hi) ⊇ V (gi), so hi ∈

√(gi). Thus hnii = cigi with

ci ∈ A. Note that ai/gi = ciai/cigi = ciai/hnii . So we replace ai with ciai

and hi with hnii , and WLOG, s = ai/hi pm D(hi) for all i. We claim thatD(f) =union of finitely many D(hi). D(f) ⊆ ∪D(hi) iff V (f) ⊇ ∩V (hi) =V (〈hi〉) iff f ∈

√〈hi〉 iff fm ∈ 〈hi〉 iff fm =

∑bihi for a finite sum with

bi ∈ A. Note that s = ai/hi = aj/hj on D(hi) ∩ D(hj) = D(hihj). ψinjective implies ai/hi = aj/hj ∈ Ahihj , so (hihj)n(hjai− hiaj) = 0 Thatis, hn+1

j (hni ai) = hn+1i (hnj aj) for all i, j. Replace ai with hni ai and hi with

hn+1i . WLOG, hjai = hiaj ∈ A.

fm =∑bihi, set a =

∑biai, hja =

∑i biaihj =

∑biajhi = fmaj , so s =

aj/hj = a/fm on D(hi), the D(hi) cover D(f), so s = a/fm ∈ O(D(f)).

3. Follows from part 2.

Definition 6.4 (Ringed Space). A Ringed Space is a pair (X,OX) where X isa topological space and OX is a sheaf of rings on X. (Usually just denoted byX).

A morphism of ringed spaces f : X → Y is a pair f = (f, f ]) where f : X →Y is continuous and f ] : OY → f∗OX a ring homomorphism.

Let f : X → Y be a morphism of ringed spaces. P ∈ X. For V ⊂ Y open,

f(P ) ∈ V , then we have OY (V )f]→ f∗OX(V ) = OX(f−1(V )) → OX,P induces

f ]P : OY,f(P ) → OX,P .

Definition 6.5 (Locally Ringed Space). X = (X,OX) is a locally ringed spaceif it is a ringed space such that OX,P is a local ring for all P ∈ X. We call themaximal ideal mP ⊆ OX,P .

f : X → Y is a morphism of locally ringed spaces iff it is a morphism ofringed spaces such that f ]P : OY,f(P ) → OX,P is a local homomorphism for allP ∈ X, that is, f ]P (mf(P )) ⊆ mP . (iff (f ]P )−1(mP ) = mf(P )).

Example: (SpecA,O) is a locally ringed space.Why do we want locally ringed spaces? Look at all the morphisms Spec(A)→

Spec(B), there are a lot of morphisms of ringed spaces, however, the morphismsof locally ringed spaces are in correspondence with ring homomorphisms B → A.

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Set X = Spec(A), Y = Spec(B). Let f : A → B be a ring homomorphism.Define ϕ : Y → X by ϕ(Q) = f−1(Q).

Continuous: I ⊂ A an ideal, ϕ−1(V (I)) = Q ∈ Y : f−1(Q) ⊃ I =V (f(I)B) ⊂ Y

For Q ∈ Y , let fQ : Af−1(Q) → BQ be a map on local rings. Let U ⊂ X open,OX(U) = s : U →

∐p∈U Ap|s is locally a quotient. Define ϕ] : OX(U) →

ϕ∗OY (U) = OY (ϕ−1(U)) by s 7→ [Q 7→ fQ(s(ϕ(Q))) ∈ BQ]Check: ϕ]Q = fQ : OX,ϕ(Q) → OY,Q.

Proposition 6.3. There is a bijective correspondence between ring homomor-phisms f : A→ B and morphisms of LRS ϕ : Y → X by f 7→ ϕ.

Proof. Note: f = ϕ] : Γ(X,OX) = A→ Γ(Y,OY ) = B.Must Show: Any morphism of LRS ϕ : Y → X is determined by ϕ] : A→ B.Let Q ∈ Y , set P = ϕ(Q) ∈ X. The following commutes:

AP = OX,P

A

OY,Q = BQ

B................................................................................................................. ............ϕ]

......................... ............ϕ]Q

.............................................................................................................................

.............................................................................................................................

ϕ]Q is a local ring homomorphism so mP = (ϕ]Q)−1(mQ), so P = (ϕ])−1(Q)

Remark: If X is an LRS, U ⊆ X open, then (U,OX |U ) is an LRS and U → Xthe inclusion is a morphism of LRS.

Definition 6.6 (Affine Scheme). An affine scheme is an LRS (X,OX) suchthat (X,OX) ' Spec(A) for some ring A.

Definition 6.7 (Scheme). A scheme is an LRS (X,OX) such that there is anopen cover X = ∪Uα such that (Uα,OX |Uα) is an affine scheme for all α.

Example: If k is a field, then Ank = Spec k[x1, . . . , xn], AnZ = Spec Z[x1, . . . , xn],and the dimension of AnZ is n+ 1.

Example: Assume that char(k) = p > 0, and A any k-algebra. Then thereis a ring homomorphism A → A : a 7→ ap. This gives the Absolute FrobeniusMorphism F : Spec(A) → Spec(A), F (P ) = a ∈ A : ap ∈ P = P , so it is theidentity on points, but is NOT the identity of schemes.

Example: Suppose k = k, X a (pre)variety over k. Define X = closedirreducible subvarieties in X. Then i : X → X is the inclusion. Open subsetsof X are U for U ⊆ X open. Define OX = i∗OX .

Exercise: (X,OX) is a scheme.Note: There exists a unique morphism of varieties X → pt = Spec(k),

and this gives a structure morphism from X → Spec(k). (X is a scheme overSpec(k))

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If ϕ : X → Y is a morphism of varieties, then

X Y

Spec(k)

....................................................................................................................................................................................................................................................................... ............ϕ

............................................................................................................................................................................................

......................

......................

......................

......................

......................

.............................................

commutes.Exercise: Morphisms of varietiesX → Y correspond to morphisms of schemes

as above.Example: The absolute Frobenius map F : X → X is NOT a morphism of

varieties.

Definition 6.8 (Reduced Scheme). A scheme X is reduced if OX(U) is a re-duced ring for all U ⊂ X open.

Definition 6.9 (Finite Type). A morphism ϕ : X → Y of schemes is offinite type if for all open affine V ⊂ Y there exists a finite open affine coverϕ−1(V ) = ∪Uα such that OX(Uα) is a finitely generated OY (V )-algebra for allα.

Exercise: The category of prevarieties over an algebraically closed field k isequivalent to the reduced schemes of finite type over Spec(k).

Projective SchemesS = ⊕d≥0Sd graded ring, Sd · Se ⊆ Sd+e. Exercise: 1 ∈ S0.I ⊆ S a homogeneous ideal, if I = ⊕(I∩Sd) iff I is generated by homogeneous

elements. EG S+ = ⊕d>0Sd.

Definition 6.10. Proj(S) = P ⊆ S homogeneous prime such that S+ 6⊂ P

For I a homogeneous ideal, set V (I) = P ∈ Proj(S)|P ⊇ I.Check V (I) ∪ V (J) = V (IJ) = V (I ∩ J). ∩V (Iα) = V (

∑Iα)

Topology: U ⊂ Proj(S) open iff Proj(S) \ U = V (I).Let P ∈ Proj(S), set T = homogeneous elements of S \ P, T−1S =

a/f : a ∈ S, f ∈ T = ⊕d∈Z(T−1S)d, deg(a/f) = deg a − deg f . DefineS(P ) = (T−1S)0

For U ⊂ Proj(S) open, define O(U) = s : U →∐p∈U S(p)|s(p) ∈ S(p) and

is locally a quotient .ie, for all P ∈ U , there is an open nbhd P ∈ V ⊂ U and homogeneous

elements a, f ∈ S of the same degree such that s(Q) = a/f ∈ S(Q) for allQ ∈ V .

Note: S(P ) is a local ring, S(P ) ⊂ T−1S → SP , max ideal S(P ) ∩ PSP .

Proposition 6.4. S graded ring, (Proj(S),O) as above, OP = S(P ) for allP ∈ Proj(S).

Proof. OP → S(P ), tP 7→ t(P ), and this is injective and surjective, following theproof for SpecA.

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For f ∈ S+ homogeneous, define D+(f) = P ∈ Proj(S) : f /∈ P =Proj(S) \ V (f)

Exercise: D+(f) is a basis for the topology on Proj(S).Define: S(f) = (Sf )0 = a/fn : deg(a) = ndeg(f).

Proposition 6.5. (D+(f),O|D+(f)) ' Spec(S(f)).

Proof. S → Sf ⊇ S(f). Define ϕ : D+(f)→ Spec(S(f)) by P 7→ PSf ∩ S(f).Exercise: For Q ∈ Spec(S(f)) we have

√〈Q〉 ⊂ Sf prime ideal.

Inverse map ϕ−1(Q) =√〈Q〉 ∩ S.

Homeomorphism: D+(h) ⊆ D+(f)ϕ→ Spec(S(f)) sendsD+(h) toD(hdeg(f)/fdeg(h)).

Note: (S(f))ϕ(P ) ' S(P ).For V ⊂ Spec(S(f)) open, then ϕ∗OProj(V ) = OProj(ϕ−1(V )) = s : ϕ−1(V )→∐

P∈ϕ−1(V ) S(P ) = s : V →∐Q∈V (S(f))Q = OSpecS(f)(V ).

Example: A a ring, PnA = ProjA[x0, . . . , xn], covered by

D+(xi) = SpecA[x0, . . . , xn](xi) = SpecA[x0/xi, . . . , xn/xi] = AnA

Definition 6.11 (Properties of Schemes). A scheme X is

1. connected if it is connected as a topological space

2. irreducible if it is irreducible as a topological space

3. reduced if OX(U) is a reduced ring for all open U ⊂ X iff OX,P reducedfor all P ∈ X.

4. integral if OX(U) a domain for all U ⊂ X open

5. noetherian if X = ∪Spec(Ai) where Ai Noetherian and the cover is finite

6. locally noetherian if X = Spec(Ai) for Ai noetherian, not necessarily afinite cover

Example: X = Spec(A) is irreducible iff√

0 ⊂ A is prime, it is reduced iff√0 = 0 and it is integral iff A is a domain.

Proposition 6.6. A scheme X is integral iff X is irreducible and reduced.

Note: An open subscheme of (locally) Noetherian scheme is (locally) Noethe-rian

Proposition 6.7. Spec(A) is locally noetherian iff notherian iff A is noetherian

Proof. Assume that Spec(A) is locally noetherian. Let U = Spec(B) ⊂ Spec(A)be open, B is noetherian. There exists f ∈ A such that D(f) ⊆ U , f |U ∈O(U) = B. Spec(Af ) = Spec(Bf |U ) = D(f). Af = Bf |U Noetherian.

We can write Spec(A) = ∪i Spec(Afi), Afi Noetherian, so ∅ = V (fi) ⇒(fi) = (1) = A⇒ 1 =

∑aifi with ai ∈ A a finite sum.

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Thus, there are f1, . . . , fr ∈ A such that Afi noetherian and (f1, . . . , fr) = A.Let I ⊆ A ideal, choose g1, . . . , gm ∈ I such that Ifi = IAfi = (g1, . . . , gm) ⊆

Afi for all i. Claim: I = (g1, . . . , gm). b ∈ I, fNi b ∈ (g1, . . . , gm) for all i. So(fN1 , . . . , f

Nr ) = (1) ⊆ A, so A is Noetherian.

Definition 6.12 (Properties of Morphisms). A morphism f : X → Y is

1. locally of finite type if for all open affine V ⊆ Y there exists open affinecover f−1(V ) = ∪Ui s.t. OX(Ui) is finitely generated OY (V )-algebra.

2. of finite type if ∃ a finite cover of f−1(V ) as above.

3. affine if for all open affine V ⊆ Y , f−1(V ) ⊆ X is open affine.

4. is finite if affine and OX(f−1(V )) is a finitely generated OY (V )-modulewhenever V is affine.

Exercise: In all cases, it is enough to know the property on a single openaffine cover.

Definition 6.13 (Open Immersion). f : X → Y is an open immersion if it canbe factored f : X '→ U ⊆ Y open.

Definition 6.14 (Closed Immersion/Closed Subscheme). f : X → Y is a closedimmersion if

1. f is a homeomorphism of X with closed subset of Y .

2. f ] : OY → f∗O∗ is surjective.

Example: Y = Spec(A), I ⊆ A ideal, A → A/I gives a closed immersionSpec(A/I)→ Spec(A) with image V (I).

Exercise: All closed immersions X → Spec(A) have this form.Exercise: Y scheme, V ⊆ Y a closed subset, then there exists a unique closed

immersion X → Y with image V such that X is reduced.

Definition 6.15 (Dimension). dim(X) =supremum of n such that ∃ ∅ 6= Z0 (Z1 ( . . . ( Zn ⊆ X with Zi closed irreducible subset.

Z ⊆ X closed irreducible. Then codim(Z;X) is the supremum of n suchthat ∃Z = Z0 ( Z1 ( . . . ( Zn ⊆ X with Zi closed irreducible.

Y ⊆ X is any closed subset, then codim(Y ;X) = infcodim(Z;X) : Z ⊆ Yclosed irreducible.

WARNING: Z ⊆ X closed and irreducible, dim(Z)+codim(Z;X) = dim(X)does not always hold!

Products

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Definition 6.16 (Product). Let X,Y, S be schemes with morphisms α : X → Sand β : Y → S. A product of X and Y over S is a scheme X ×S Y withmorphisms

S

X Y

X ×S Y...........................................................

............... p

.............................................................. ............q

.................................................................. ............α

.............................................................................. β

along with the universal property that for any scheme Z with morphisms f, gto X and Y s.t. αf = βg, there exists a unique morphism ϕ : Z → X ×S Ysuch that f = pϕ and g = qϕ.

Exercise: X ×S Y is unique up to unique isomorphism.Exercise: X any scheme, A is a commutative ring, then there is a correspon-

dence between morphisms X → Spec(A) and ring hom A→ OX(X).Consequence: X = Spec(A), Y = Spec(B) and S = Spec(R), then α, β

make A and B into R-algebras, and X ×S Y = Spec(A⊗R B).Observe that if S ⊆ T is an open subscheme, then X ×T Y = X ×S Y , as if

j : S → T is the inclusion, then αf = βg ⇐⇒ jαf = jβg. Also observe that ifU ⊆ X open, then U ×S Y = p−1(U) ⊆ X×S Y , and so it is an open subschemeof X ×S Y , because if h : Z → X is a morphism, so that h(Z) ⊆ U , then wecan factor h : Z → U ⊆ X.

Consequence: If X ′ ⊆ X and Y ′ ⊆ Y , S′ ⊆ S are all open such thatα(X ′), β(Y ′) ⊆ S′, then X ′ ×S′ Y ′ = X ′ ×S Y ′ and this is p−1(X ′)∩ q−1(Y ′) ⊆X ×S Y .

ConstructionAssume that S is affine. Take open affine covers X = ∪Xi, Y = ∪Yj . We

glue X×SY :=⋃i,j Xi×SYj by (Xi×SYj)∩(Xk×SY`) = (Xi∩Xk)×S (Yj∩Y`).

If S is any scheme, we take an open affine cover S = ∪Si. Xi = α−1(Si),Yi = β−1(Si) in X and Y are open sets. And here we glue X×SY := ∪iXi×SiYiby (Xi ×Si Yi) ∩ (Xj ×Sj Yj) = (Xi ∩Xj)×Si∩Sj (Yi ∩ Yj)

Examples: X is a scheme

1. U, V ⊆ X open. U ×X V = U ∩ V .

2. Y,Z ⊆ X closed subschemes. We define Y ∩ Z := Y ×X Z, the schemetheoretic intersection. This is still a closed subscheme of X.

X = A2k = Spec k[x, y], Y = V (y − x2) = Spec k[x, y]/(y − x2), Z = V (y) =

Spec k[x, y]/(y). We have a diagram of commutative rings

k[x, y]/(y − x2, y)

k[x, y]/(y − x2) k[x, y]/(y)

k[x, y]................................................

...............

................................................... ............

................................................... ............

...............................................................

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So Y ×XZ = Y ∩Z = Spec k[x, y]/(x2, y) = Spec k[x]/(x2). dimk k[x]/(x2) =2.

Let X be a scheme and p ∈ X a point.

Definition 6.17 (Residue Field at p). k(P ) = OX,P /mP is called the residuefield at P .

If P ∈ U = Spec(A) ⊆ X open, then k(P ) = AP /PAP , the field of fractionsof A/P .

Note: A→ k(P ) gives a morphism Spec(k(P ))→ Spec(A)→ X with imageP.

Examples: X an irreducible algebraic variety over k. If P ∈ X is a closedpoint, we get k(P ) = k. If P0 ∈ X is a generic point, we get k(P0) = k(X).

For p ∈ U ⊆ X, f ∈ OX(U), set f(P ) =the image of f under OX(U) →OX,P → k(P ).

Note: V (f) = P ∈ U : f(P ) = 0 ∈ k(P ) is relatively closed in U .Let ϕ : Y → X a morphism, ϕ(Q) = P , P ∈ U . Then ϕ]Q : OX,P → OY,Q

is a local ring homomrophism so it induces ϕ]Q : k(P )→ k(Q) a field extension.ϕ](f) ∈ OY (ϕ−1(U)), ϕ](f)(Q) = ϕ]Q(f(P )) = ϕ]Q(f(ϕ(Q))) ∈ k(Q).

E.G. X,Y varieties over k = k, Q ∈ Y a closed point implies that ϕ](f)(Q) =f(ϕ(Q)) ∈ k.

Exercise: A rational map of irreducible varieties f : X 99K Y is the same asa morphism of schemes over k f : Spec k(X)→ Y

P ∈ X a closed point, then rational mapsX 99K Y defined at P correspondto morphismsSpec OX,P → Y over k.

Note: Spec(k(X))→ Spec(OX,P )→ X.Examples of ProductsX,Y varieties over k, the product X × Y from last semester corresponds to

X ×k Y = X ×Spec(k) Y .Fibers: ϕ : X → Y morphism of schemes, Q ∈ Y , then XQ = ϕ−1(Q) :=

X ×Y Spec k(Q) with the following diagram:

XQ Spec(k(Q))

X Y

...................................................................... ............

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Example: ϕ : A1 → A1, ϕ(t) = t2. For a ∈ A1 a closed point, ϕ−1(a) =A1 ×A1 a = Spec(k[t] ⊗k[t2] k[t2]/(t2 − a)) = Spec(k[t]/(t2 − a)), if a 6= 0, weget ϕ−1(a) =

√a,−√a, if a = 0, ϕ−1(0) = Spec k[t]/(t2).

Example: P2k ×k A1

k = Proj k[x, y, z]× Spec k[t] = ProjA[x, y, z] where A =k[t].

E = V (zy2 − x(x− z)(x− tz)) ⊆ P2 × A1, ϕ : E → A1 projection, then forλ ∈ A1 a closed point, Eλ = V (zy2 − x(x− z)(x− λz)) ⊆ P2

k.Separated Morphisms

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Let f : X → Y a morphism, ∆ : X → X ×Y X the diagonal morphism (theunique map into this product which, when composed with either projection, isthe identity)

Definition 6.18 (Separated). f is separated if ∆ : X → X ×Y X is a closedimmersion.

X is separated if X → Spec(Z) is separated.

Note: A a ring, then there exists a unique Z→ A.Example: X = (A1 \ 0) ∪ 01, 02 the doubled line to Spec(k) is not

separated (See last semester).

Proposition 6.8. Any morphism f : X → Y of affine schemes is separated.

Proof. X = Spec(A), Y = Spec(B), f ] : B → A. ThenX×YX = Spec(A⊗BA).∆ : X → X ×Y X corresponds to ∆] : A⊗B A→ A, a1 ⊗ a2 7→ a1a2.

∆] is surjective, so A = A⊗B A/I, so X = V (I) ⊂ X ×Y X.

Corollary 6.9. f : X → Y separated iff ∆(X) ⊆ X ×Y X closed.

Proof. Assume that ∆(X) is closed. It is a homeomorphism as ∆ : X → ∆(X)has continuous inverse ∆(X)→ X ×X → X by p1.

We must now check that ∆] : OX×YX → ∆∗OX is surjective.If Q ∈ X ×Y X \∆(X), then OX×YX,Q → (∆∗OX)Q = 0 is surjective. Let

P ∈ X. Choose V ⊆ Y open affine such that f(P ) ∈ V . Choose U ⊆ f−1(U)open affine such that P ∈ U . Then ∆(P ) ∈ U ×V U ⊆ X ×Y X so ∆] issurjective in a nbhd of ∆(P ), because ∆ : U → U ×V U is separated.

Let (G,≤) be a totally ordered abelian group. IE, g1 ≤ g2 implies g1 + g3 ≤g2 + g3. K a field, and K× = K \ 0.Definition 6.19 (Valuation). A valuation of K with values in G is a mapv : K× → G s.t. v(xy) = v(x) + v(y), v(x+ y) ≥ minv(x), v(y).

e.g. v : k(t)→ Z by v(tmf(t)) = m if f is defined at 0 and f(0) 6= 0Note: x ∈ K× : v(x) ≥ 0 ∪ 0 ⊆ K is a subring.

Definition 6.20 (Valuation Ring). R is a valuation ring if R is a domain andthere exist a valuation v : K(R)× → G for some G such that R = x ∈ K(R)× :v(x) ≥ 0 ∪ 0.

R ⊆ K(R) gives us a morphism i : SpecK(R)→ SpecR.

Theorem 6.10. f : X → Y is a morphism, then f is separated iff the followingcondition holds: For any valuation ring R and morphisms α : SpecR→ Y andβ : SpecK(R)→ X such that αi = fβ

X Y

SpecK(R) Spec(R)

................................................................................................................. ............f

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β

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α

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Then there is at most one γ : SpecR→ X making this all commute, that is,fγ = α and γi = β.

Intuition: X,Y are prevarieties, C a curve and P ∈ C a nonsingular point.R = OC,P is a DVR. K = k(C). Then β : C 99K X is a rational map, andα : C 99K Y is also a rational map defined at P ∈ C. Then α = fβ is thecommutativity condition on the square. If f is separated, then there is at mostone possible value of β(P ).

e.g. If X is the doubled affine line, Y is Spec k then β : A1 99K X definedon A1 \ 0. There are two ways to define β(0), so this is not separated.

Corollary 6.11. 1. open and closed immersions are separated

2. compositions of separated morphisms are separated.

3. separated morphisms are stable under base extensions

X ′ = X ×Y Y ′ Y ′

X Y

.................................................. ............f ′

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Definition 6.21 (Base Extension). Then f ′ is the base extension of f by Y ′.

Proof. We will prove (b).

Let Xf→ Y

g→ Z are separated morphisms. Let R be a valuation ring, takeβ : SpecK(R)→ X and α : Spec(R)→ Z.

Assume that γ1, γ2 : SpecR → X are morphisms satisfying the diagram.Proof by diagram chasing.

Exercise: If g is separated, then gf is separated iff f is separated.Let X be a scheme, x1 ∈ X.

Definition 6.22 (Specialization). x0 ∈ X is a specialization of x1 if x0 ∈ x1.

This gives a ring homomorphism OX,x0 → OX,x1 which is NOT local.Note: Spec k(x0) → X has image x0, Spec k(x0) → Spec OX,x0 → X has

image x1 ∈ X : x0 is a specialization of x1.Assume R is a local domain, K = K(R) i : SpecK → SpecR, t0 = mR ∈

SpecR and t1 = 0 ∈ SpecR. Let β : SpecK → X a morphism, x1 = β(0),β] : k(x1)→ K

Lemma 6.12. There is a correspondence γ : SpecR → X|γi = β to x0 ∈

x1|image OX,x0 → OX,x1 → k(x1)β]→ K is a subset of R by γ 7→ γ(t0).

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Proof. Assume γi = β. t0 ∈ t1 implies that γ(t0) ∈ x1. mx0 7→ mR. We getthe following commutative square:

OX,x1 Ot1 = K

OX,x0 Ot0 = R

......................................................................... ............γ]t1

.............................................................................................................................

.............................................................................................................................

........................................................................... ............γ]t0

Assume that x0 ∈ x1. Then define γ : SpecR→ Spec OX,x0 → X.

Recall: A domain R is a valuation ring of K if K = K(R) and ∃ valuationv : K× → G such that R = a ∈ K|v(a) ≥ 0 ∪ 0.

Note: R is a local ring, mR = a ∈ K|v(a) > 0.Fact: Every local ring R′ ⊆ K is dominated by some valuation ring R of K

(that is, R′ ⊆ R with mR′ ⊆ mR). R′ ⊆ K a local subring is a valuation ring ofK iff R′ is not domainated by strictly larger R ⊆ K.

Theorem 6.13. X Notherian. f : X → Y a morphism, ∆ : X → X×Y X = P .Then ∆(X) ⊆ P is closed iff for all valuation rings R with morphisms α, β withαi = fβ there exists at most one γ sch that fγ = α and γi = β.

Proof. ⇒: Let γ1, γ2 : SpecR → X be given, with γji = β, fγj = α. Defineϕ : SpecR→ P by πjϕ = γj . Then γji = β imply that ϕ(t1) = ϕ(i(0)) ∈ ∆(X)and ϕ(t0) ∈ ϕ(t1) ⊆ ∆(X) = ∆(X). Thus, π1(ϕ(t0)) = γ1(t0) = γ2(t0) =π2(ϕ(t0)), so γ1 = γ2 by the lemma.⇐: Let z1 ∈ ∆(X) and z0 ∈ z1 ⊆ P . Claim: z0 ∈ ∆(X).Set K = k(z1), R′ = Im(OP,z0 → OP,z1 → K) ⊆ K. The fact implies that

there is a valuation ring R of K such that OP,z0 → R ⊆ K is a local ringhomomorphism. β : SpecK → P , πk : P → X.

Exercise: z1 ∈ ∆(X) implies that π1β = π2β : SpecK → X.Now the lemma implies that there is ϕ : SpecR → P such that β = ϕi,

ϕ(t0) = z0. Set γj = πjϕ : SpecR → X. Then fγ1 = fπ1ϕ = fπ2ϕ = fγ2. Sothe algebraic assumption says that γ1 = γ2, so ϕ = ∆π1ϕ, thus z0 = ϕ(t0) ∈∆(X), so the claim holds.

So now X Notherian implies that X = X1 ∪ . . .∪Xn where Xi is closed andirreducible.

So Xi = xi, xi ∈ X, we set zi = ∆(xi) ∈ P . So X ⊆ x1, . . . , xn, thus∆(X) ⊆ z1, . . . , zn = z1 ∪ . . . ∪ zn ⊆ ∆(X) by the claim, and so thetheorem holds.

Definition 6.23 (Properness). f : X → Y is proper if f is separated, of finitetype, and f is universally closed, which means that any base extension of f is aclosed map. (takes closed sets to closed sets.)

Definition 6.24 (Redefinition of a Variety). Let k be a field. A prevariety overk is a reduced scheme of finite type over k. A variety is a separated pre-varietyX, that is, the structure morphism X → Spec(k) is separated.

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Definition 6.25 (Complete). A variety is complete if X → Spec(k) is proper.ie, X ×k Y → Y is closed for all Y .

Theorem 6.14. X Notherian, f : X → Y of finite type. Then f is proper iff∀ valuation rings R with a commutative diagram

SpecK(R)

X

SpecR

Y

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∃!γ

there ∃!γ such that fγ = α, γi = β

Corollary 6.15. X a complete variety, C a curve, P ∈ C a nonsingular point,then f : C \ P → X is a morphism of varieties, then we can extend f toC → X.

This follows by setting Y = Spec(k), R = OC,P a DVR, and K(R) = K(C)the function field of C.

Corollary 6.16. 1. Closed immersions are proper.

2. Compositions of proper morphisms are proper.

3. Base extensions of proper maps are proper

4. Properness is determined locally. That is, f : X → Y a morphism it isproper iff for all V ⊆ Y open subscheme, f : f−1(Y )→ Y is proper.

Exercise: Assume that P is a property of morphisms such that closed em-beddings are P , compositions of P -morphisms are P and base extensions of Pmorphisms are P . Then products of P -morphisms are P , X

f→ Yg→ Z, if gf is

P and g is separated then f is P , and if f : X → Y is P then fred : Xred → Yredis P .

Projective MorphismsPnA = ProjA[x0, . . . , xn] = PnZ ×Spec Z SpecA.

Definition 6.26 (Projective Space over a Scheme). For any scheme Y , setPnY = PnZ × Y

Definition 6.27 (Projective Morphism). f : X → Y is projective if there existsa closed immersion i : X → PnY s.t. f = πY i.

f is quasi-projective if there is an open immersion j : X ⊂ X ′ and a projec-tive morphism f ′ : X ′ → Y s.y. f = f ′ j.

Example: A variety X is quasi-projective iff X → Spec k is quasiprojective,and likewise projective.

Example: S is a graded ring, gneerated bny finitely many elements of S1 asan S0-algebra. Then Proj(S)→ Spec(S0) is projective.

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Theorem 6.17. A projective morphism of Notherian schemes is proper.

Proof. Enough to show that PnZ → Spec Z is proper.Separated: Show that ∆ : PnZ → Pn×Pn is a closed immbedding. D+(xixj)→

D+(xi)×D+(xj) gives Z[x0, . . . , xn](xixj) ← Z[x0, . . . , xn](xi)⊗ZZ[x0, . . . , xn](xj).Properness; R a valuation ring, K = K(R). We want γ such that πγ = α

and γi = β. Let t0 = mR, t1 = 0 ∈ Spec(R). Set p1 = β(t1) ∈ Pn. If p1 ∈ V (xi),then induction imples that there is a γ : SpecR→ V (xi) ' Pn−1 ⊆ Pn.

WLOG: p1 ∈ D+(x0x1 . . . xn)⇒ xi/xj ∈ OP,p1 for all i, j.β] : k(p1)→ K, set fij = β](xi/xj) ∈ K.A valuation v : K → G, R = v ≥ 0. Choose m such that v(fm0) is

minimal. v(fim) = v(fi0/fm0) = v(fi0) − v(fm0) ≥ 0, so fim ∈ R for all i. Sowe have a ring homomorphism Z[x0/xm, . . . , xn/xm]→ R by xi/xm 7→ fim.

Thus, SpecRγ→ D+(xm) ⊆ Pn is as desired.

Exercise:1) There exists a closed Segre Embedding PnZ × PmZ → Pnm+n+m

Z .2) Composition of projective morphisms is projective.

Example: Xaffine→ Y → Spec k, X a prevariety, Y a variety, then X is a

variety.OX modulesX is a scheme, F ,G are sheaves of OX -modules.homOX (F ,G ) = homomorphisms of OX -modules F → G .We define a sheaf of OX -modules H omOX (F ,G ) by U 7→ homOX |U (F |U ,G |U ).F ⊗OX G by the sheafification of U 7→ F (U)⊗OX(U) G (U).Let f : X → Y be morphisms of schemes, H an OY -module. f∗F is an

f∗OX -module, f ] : OY → f∗OY a ring homomorphism implies that f∗F is anOY -module.

The adjoint property: homOY (H , f∗F ) ⇐⇒ homf−1OY (f−1H ,F ).

Definition 6.28 (Pullback). f∗H = f−1H ⊗f−1OY OX , this is an OX-module.

Continued...see Sheaves from last semester

Proposition 6.18. MP = MP , M(D(a)) = Ma.

Corollary 6.19. Γ(X, M) = M and M 7→ M is an exact, fully faithful functorfrom A-mod to OX-mod with inverse the global section functor, ˜⊕iMi = ⊕iMi

and ˜M ⊗A N = M ⊗OX N

We have done this several times before, and so will not prove them again.

Proposition 6.20. f : SpecA = X → SpecB = Y a morphism, M an A-module and N a B-module. Then f∗M = MB, f∗N = ˜N ⊗B A where MB isM as a B-module using f ] : B → A.

Proof. Let b ∈ B, Γ(D(b), f∗M) = Γ(f−1(D(b)), M) = Γ(D(f ](b)), M) =Mf](b) = (MB)b.

Let P ∈ X. Q = (f ])−1(P ) = f(P ) ∈ Y . (f∗N)P = Nf(P ) ⊗OY,f(P ) OX,P =NQ ⊗BQ AP = (N ⊗B A)P .

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Definition 6.29 (Quasicoherent and Coherent Sheaves). Let X be a scheme.An OX-module F is quasi-coherent if there exists an open cover X = ∪Ui,Ui = SpecAi and Ai modules Mi such that F |Ui ' Mi.

F is coherent if the Mi are ginitely generated.

Example: i : Y → X a closed subscheme implies that i∗OY is a coherent OX -module. U = SpecA ⊆ X open, i−1(U) = Spec(A/I), i∗OY = i∗(A/I) = A/I.

Lemma 6.21. X = SpecA, f ∈ A. F a quasi-coherent OX-module.

1. If s ∈ Γ(X,F ) and s|D(f) = 0 ∈ Γ(D(f),F ), then ∃n > 0 such thatfns = 0 ∈ Γ(X,F )

2. If t ∈ Γ(D(f),F ) then ∃s ∈ Γ(X,F ) and m > 0 so that s|D(f) = fnt.

Proof. P ∈ X, there exists an open nbhd P ∈ U = Spec(B) ⊆ X and B-moduleM such that F |U ' M . So there exists g ∈ A such that P ∈ D(g) ⊆ U .F |D(g) = M |D(g) = ˜M ⊗B Ag. Thus, we can write X = D(g1) ∪ . . . ∪ D(gm)such that F |D(gi) = Mi, with Mi a Agi-module.

1. Let s ∈ Γ(X,F ), with s|D(f) = 0. Set mi = s|D(gi) ∈ Mi. Thenmi|D(fgi) = 0 ∈ (Mi)f , so fnmi = 0 ∈Mi for all i, so fns = 0 ∈ Γ(X,F ).

2. X = D(g1) ∪ . . . ∪ D(gr). F |D(gi) = Mi, where Mi is an Agi-module.t|D(fgi) ∈ (Mi)f . ∃ti ∈ Mi such that ti|D(fgi) = fnt|D(fgi). So then(ti − tj)|D(fgigj) = (fnt − fnt)|D(fgigj) = 0. By part 1, fm(ti − tj) =0 ∈ Γ(D(gigj),F ). That is, fmti = fmtj on D(gi) ∩ D(gj). We flues ∈ Γ(X,F ) s.t. s|D(gi) = fmti and s|D(f) = fn+mt, as s|D(fgi) = fmti =fnfmt|D(fgi).

Corollary 6.22. Γ(D(f),F ) = Γ(X,F )f .

Proposition 6.23. X a scheme, F an OX-module.

1. F is quasi-coherent iff for all open SpecA = U ⊆ X, there exists anA-module M such that F |U = M

2. X Notherian: F coherent iff the same is true but M finitely generated.

Proof. 1. Assume that F is quasicoherent, U = SpecA ⊆ X is open. ThenF |U is a quasicoherent OX -module. Write U = D(gi) ∪ . . . ∪ D(gr),gi ∈ A. F |D(gi) ' Mi. Set M = F (U). M 7→ Γ(U,F ) gives an OU -homomorphism α : M → F . The lemma implies that Mi = F (D(gi)) =Mgi . Therefore, α is an isomorphism over D(gi) implies that it is one overU .

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2. Assume X Notherian, F is coherent. Then A is Notherian, so Mi = Mgi

is a finitely generated module over Agi for all i. We must show that Mis a finitely generated A-module: take m1, . . . ,mN ∈M such that Mgi isgneerated by mj/1 for all i. Then M is generated by m1, . . . ,mN asan A-module.

Corollary 6.24. X = SpecA. Then there is an equivalence M 7→ M andF → Γ(X,F ) between A-modules and quasi-coherent OX-modules.

Recall: 0 → F ′ → F → F ′′ → 0 is exact iff 0 → F ′p → Fp → F ′′p → 0 isexact for all p ∈ X. We only get automatically that 0→ Γ(U,F ′)→ Γ(U,F )→Γ(U,F ′′) is exact.

Proposition 6.25. X is affine scheme, 0 → F ′ → F → F ′′ → 0 exactsequence, if F ′ is quasi-coherent, then Γ(U,F ) → Γ(U,F ′′) is surjective, soΓ(U,−) becomes an exact functor.

Proposition 6.26. Any kernel, cokernel or image of an OX-homomorphismof quasicoherent OX-modules is quasicoherent. An extension of quasi-coherentOX-modules is quasi-coherent.

If X is Notherian, then the same holds for coherent modules.

Proof. WLOG, X = SpecA. ϕ : M → N . This gives 0 → K → M → N →C → 0 an exact sequence, as − is exact, 0→ K → M → N → C → 0 is exact.So the kernel and cokernel must be quasicoherent. The image is coker(K → M),and so is also quasicoherent.

We now assume that 0 → M → F → N → 0 is a short exact sequencewith M, N quasicoherent and F an OX -module. Define F = Γ(X,F ). Theproposition states that 0 → M → F → N → 0 is a short exact sequence ofA-modules. Then we obtain0 M F N 0................................................................................................................. ............ ................................................................................................................. ............................................................................................................................. ............ ................................................................................................................. ............

0 M F N 0................................................................................................................. ............ ................................................................................................................. ............................................................................................................................. ............ ................................................................................................................. ............

..................................................'

..................................................'

..................................................

So by the five lemma, F ' F .

Definition 6.30 (Ideal Sheaf). Let X be a scheme, i : Y → X a closed immer-

sion. Then the ideal sheaf of Y is IY = ker(OXi]→ i∗OY ) ⊆ OX .

Proposition 6.27. closed subschemes of X correspond to quasicoherentsheaves of ideals ⊆ OX.

Proof. Assume that i : Y → X is a closed immersion, then we have 0IY →OX

i]→ i∗OY → 0. i∗OY is quasicoherent (we will prove this next, it is truebecause i is separated and quasicompact). Then IY is quasicoherent as well.

Assume that I ⊆ OX is quasicoherent. For p ∈ X, Ip ⊆ OX,p. DefineY = P ∈ X : IP ⊆ mP ( OX,P = Supp(OX/I ) = P ∈ X|(OX/I )P 6= 0.Let i : Y → X be the inclusion.

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Define OY = i−1(OX/I ). Then (Y,OY ) is a locally ringed space. OX,P =

(OX/I )P = OX,P /IP . Then i−1(OX/I ) id→ OY , which corresponds to a mapOX/I

ϕ→ i∗OY by the adjoint property.Then we define i] : OX → OX/I

ϕ→ i∗OY , so (i, i]) : Y → X is a morphismof locally ringed spaces. Claim: This is a closed immersion.

WLOG X = SpecA. Then I = I ⊆ OX , with I ⊆ A an ideal. ThenY = V (I) ⊆ X is closed, and i : Y → X is a homeomorphism onto its image.We must check that Y = SpecA/I.

OY,P = (OX/I )P = AP /IP = (A/I)P = OSpecA/I,P .

Corollary 6.28. All closed subschemes of SpecA are in correspondence withideals of A.

All closed subschemes of an affine scheme are affine.

Corollary 6.29. If U1, U2 ⊆ X open affine subschemes, and if X is separated,then U1 ∩ U2 is affine.

X X ×X............................................................................................. ............∆

U1 × U2U1 ∩ U2

........

........

........

........

........

........

........

........

........

........

........

........

.................

............

⊆

........

........

........

........

........

........

........

........

........

........

........

........

.................

............

................................................................. ............

Proof. As ∆ is a closed embedding, the bottom map must be as well. Theproduct of affines over an affine is affine, and so then U1 ∩ U2 is affine, as it isa subscheme of U1 × U2.

Proposition 6.30. Let f : X → Y be a morphism.

1. G a quasi-coherent OY -module, then f∗G is a quasi-coherent OX-module.

2. X,Y both Notherian, same for coherent.

3. Assume X is Notherian or that f is separated and quasicompact, then ifF is quasicoherent OX-module, then f∗F is a quasicoherent OY -module

Proof. 1. Let P ∈ X, Let SpecB = V ⊆ Y open affine such that f(P ) ∈ V .Take U = SpecA ⊆ X open such that P ∈ U ⊆ f−1(V ). G |V = M withM a B-module. Then (f∗G )|U = f∗(G |V ) = f∗(M) = ˜M ⊗B A.

2. Similar, but noting finite generation everywhere.

3. WLOG, Y is affine. Last time, we showed that (c) is true when X isaffine. Assumptions imply that there exists a finite open affine coverX = ∪Ui such that Ui ∩ Uj has a finite open affine cover Ui ∩ Uj = ∪Uijkfinite. As F is a sheaf, we obtain an exact sequence of OY -modules0 → f∗F → ⊕if∗F |Ui → ⊕i,j,kf∗F |Uijk . So f∗F is the kernel of a mapof quasicoherent OY -modules.

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Previously, we’ve shown that the following properties are all equivalent

1. Closed subschemes of affine schemes are affine

2. If X is separated and U1, U2 ⊆ X open affine, then U1 ∩ U2 affine.

3. Pushforward of a quasicoherent sheaf is quasicoherent

4. The ideal sheaf of a closed subscheme is quasi-coherent.

We now prove number 1:

Proof. If i : Y → X is a homeomorphism onto i(Y ), and i(Y ) is closed, then(i∗OY )P is OY,P if P ∈ Y and 0 else.

For f ∈ Γ(X,OX) write Xf = D(f) = P ∈ X : f /∈ mP ⊆ OX,P .Assume that i : Y → X = SpecA is a closed subscheme. Let P ∈ Y ,

then there exists open affine V ⊆ Y . Note: Yf = i−1(Xf ) is a basis for thetopology on Y .

Thus, there exists f ∈ A such that P ∈ Yf ⊆ V , Yf = Vf is affine. So we canwrite X = Xf1 ∪ . . . ∪Xfr such that Yfi affine for all i. (f1, . . . , fr) = (1) ⊆ A,thus (i](f1), i](f2), . . . , i](fr)) = (1) ⊆ Γ(Y,OY ).

By exercise 2.17b in Hartshorne, Y is affine iff ∃f1, . . . , fr ∈ Γ(Y,OY ) suchthat Yfi is affine for all i and (f1, . . . , fr) = 1 ⊆ Γ(Y,OY ).

S = ⊕d≥0Sd a graded ring, X = Proj(S), and M = ⊕d∈ZMd a graded S-module. Sd ·Me ⊆Me+d for P ∈ X defin an S(P )-module M(P ) = m/f : m ∈M,f ∈ S homogeneous of the same degree.

For U ⊆ X open, define OX -module M by Γ(U, M) = s : U →∐P∈U M(P )|s(P ) ∈

M(P )∀P ∈ U and s is locally a quotient.

Proposition 6.31. 1. MP = M(P )

2. M |D+(f) = M(f) for f ∈ S+, deg(f) > 0.

Corollary 6.32. M is quasicoherent. If S is Notherian and M finitely gener-ated S-module, then M is coherent.

Definition 6.31 (Twisted Modules). Define M(n) = M as an S-module, butwith grading M(n)e = Me+n.

If X = Proj(S), define OX(n) = ˜S(n). For any OX-module, F , defineF (n) = F ⊗OX OX(n).

Example:

1. OX(0) = S = OX .

2. S = A[x0, . . . , xm], X = Proj(S) = PmA . Then Γ(D+(xi),OX(n)) =S(n)xi = f/xri |f ∈ Sr+n = elements of degree n ⊆ Sxi

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Claim: Γ(X,OX(n)) = Sn. We have a map Sn = S(n)0, Let t ∈ Γ(X,OX(n)),set ti = t|D+(xi) ∈ Sxi element of degree n, if ti = tj on D+(xixj), thenti = tj ∈ Sxixj , not that Sxi ⊆ Sxixj , Sxi ∩ Sxj = S ⊆ Sxixj , thus ti = tj ∈ Snfor all i, j.

Proposition 6.33. Let X = Proj(S), S generated by S1 as an S0-algebra.

1. OX(n) is an invertible OX-module.

2. M ⊗OX N = ˜M ⊗S N .

Proof. 1. Let f ∈ S1. Γ(D+(f),OX(U)) = S(n)(f)·f−n→ S(f), which says that

OX(n)|D+(f) = ˜S(n)(f) ' S(f) = OD+(f).

2. M ⊗S N is a graded S-module, by (M ⊗S N)e =submodule generated bym⊗ n where m ∈ Mr, n ∈ Nt with r + t = e. Let P ∈ X. Then we havea map M(P ) ⊗S(P ) N(P ) → (M ⊗S N)(P ) by m/f ⊗ n/g 7→ m ⊗ n/fg. Sis generated by S1, so this is an isomorphism. Why? Because P 6⊇ S+ ⇒P 6⊇ S1 ⇒ ∃h ∈ S1 such that h /∈ P . Let m ⊗ n/f ∈ (M ⊗ N)(P ), go tom/hr ⊗ hrn/f ∈M(P ) ⊗N(P ) where m ∈Mr, n ∈ Nt and s ∈ Sr+t.

Construct an isomorphism ϕ : M⊗OX N → ˜M ⊗N by s ∈ Γ(U, M⊗N) 7→[P 7→ sp ∈ MP ⊗OX,P NP = (M ⊗N)(P )], which is an isomorphism.

Corollary 6.34. M(n) = M ⊗OX OX(n) = M ⊗ ˜S(n) = (M ⊗S S(n)) = ˜M(n).

This says that OX(m)⊗ OX(n) = OX(m+ n).

Definition 6.32. Let X = Proj(S), F an OX-module. Define Γ∗(F ) =⊕n∈ZΓ(X,F (n)). This is a graded S-module.

Let s ∈ Sd, t ∈ Γ(X,F (n)) gives s ∈ Γ(X,OX(d)), so t⊗ s ∈ Γ(X,F (n))⊗OX(d) = Γ(X,F (n+ d))

Note: X = ProjS where S = A[x0, . . . , xn]. Γ∗(OX) = S. This is NOTalways true!

(Side Note: Look at the operation Γ∗(ProjS). Is it a functor? What prop-erties does it have? Filling in lower degrees?)

Let X be a scheme, I an invertible LX -module, and f ∈ Γ(X,L ).

Definition 6.33. Xf = P ∈ X|fP /∈ mPLX. Xf ⊆ X is open.

Lemma 6.35. Let X be a quasi-compact scheme, L an invertible OX-module,f ∈ Γ(X,L ) and F a quasi-coherent OX-module.

1. Let s ∈ Γ(X,F ), s|Xf = 0, then s⊗ fn = 0 ∈ Γ(X,F ⊗L ⊗n).

Proof. 1. X = U1 ∪ . . . ∪ Ur where Ui = SpecAi and L |Ui ' OUi . F |Ui =Mi with Mi an Ai-module. si|Ui ∈ Mi. fi = f |Ui ∈ Γ(Ui,L ) ' Ai,F ⊗L ⊗n|Ui ' Mi ⊗ O⊗nUi = Mi and s ⊗ fn 7→ fni si = 0 for n >> 0, sos⊗ fn = 0 ∈ Γ(X,F ⊗L ⊗n).

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X a schemem L an invertible OX -module.

Definition 6.34. L −1 = H om(L ,OX).

If L |U ' OU then L −1|U 'H om(OU ,OU ) ' OU .Note: Global OX -hom: L ⊗L −1 → OX by s⊗α 7→ α(s) is an isomorphism

on U .Let f ∈ Γ(X,L ), Xf = P ∈ X : fP /∈ mPLP ⊆ X open. L −1|Xf →

OXf , α 7→ α(f).Define f−1 ∈ Γ(Xf ,L −1) to be the inverse image of 1 by this isomorphism.So f ⊗ f−1 = 1, adn L ⊗L −1 ' OX .

Lemma 6.36. Let X be a quasi-compact scheme, L an invertible OX-module,f ∈ Γ(X,L ) and F a quasicoherent OX-module.

1. s ∈ Γ(X,F ), s|Xf = 0⇒ s⊗ fn = 0 ∈ Γ(X,F ⊗L ⊗n)

2. Assume X = U1 ∪ . . . ∪ Ur with Ui open affine and L |Ui ' OUi for all i,and Ui ∩ Uj quasicompact. t ∈ Γ(Xf ,F ) ⇒ ∃s ∈ Γ(X,F ⊗L ⊗n) suchthat s|Xf = t⊗ fn.

Corollary 6.37. S = ⊕d≥0Γ(X,L ⊗d) is a graded ring, f ∈ S1. Γ∗F =⊕e∈ZΓ(X,F ⊗L ⊗e) is a graded Smodule. Then (Γ∗F )(f) = Γ(Xf ,F )

Proof. (Γ∗F )(f) = s/fd|s ∈ Γ(X,F ⊗L ⊗d)f−d ∈ Γ(Xf ,L ⊗−d). (Γ∗F )(f) → Γ(Xf ,F ) by s/fd 7→ s ⊗ f−d. This is

injective and surfective.

Let S = ⊕d≥0Sd be a graded ring, X = ProjS.Recall, OX(n) = ˜S(n). f ∈ Sd implies f ∈ S(d)0 ⇒ f ∈ Γ(X,OX(d)). F

an OX -module, Γ∗F = ⊕e∈ZΓ(X,F (e)). Construct an OX -hom β : ˜Γ∗F → Fby β|Xf : ˜Γ∗F(f) → F |Xf , which corresponds to a module homomorphism(Γ∗F )(f) → Γ(Xf ,F ) by s/fm 7→ s⊗ f−m.

Proposition 6.38. If F is quasi-coherent, and S is finitely generated by S1 asan S0-algebra, then β : ˜Γ∗F → F is an isomorphism.

Proof. Assume S0[f1, . . . , fr] → S is surjective, fi ∈ S. Ui = Xfi affine. X =U1 ∪ . . . ∪ Ur. O(1)|Ui = ˜S(1)(xi) = ˜S(xi) = OUi , Ui ∩ Uj = Xfifj affine, so thelemma implies that β|Xfi is an isomorphism for all i.

Let I ⊆ S be a homogeneous ideal. Natural inclusion i : Y = Proj(S/I) →X = Proj(S) i(Y ) = V (I), i∗OY = ˜S/I (exercise)

So OX → S → ˜S/I = i∗OY , so Y ⊆ X is a closed subscheme. Note thatIY = I ⊆ OX .

Corollary 6.39. 1. Assume Y ⊆ PrA a closed subscheme with PrA = ProjS,S = A[x0, . . . , xr]. Then there exists a homogeneous ideal I ⊆ S such thatY = Proj(S/I).

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2. A morphism ϕ : Y → SpecA is projective iff Y = Proj(S), S0 = A and Sis finitely generated by S1.

Proof. 1. IY ⊆ OPrA a quasicoherent subsheaf, IY ⊗ O(d) ⊆ O(d) impliesthat I = Γ∗IY ⊆ Γ∗OPrA = S a homogeneous ideal.

Proj(S/I) ⊆ PrA has ideal sheaf I = ˜Γ∗IY = IY ⊆ OPrA . Thus, Y =Proj(S/I).

2. ϕ is projective iff it factors through PrA as a closed immersion for some riff Y = ProjS, with S = A[x0, . . . , xr]/I.

Definition 6.35 (Twisting Sheaf). Let Y be any scheme. The twisting sheafof PrY = PrZ × Y

π→ PrZ is O(1) = π∗OPr (1).

Definition 6.36 (Immersion). A morphism i : X → Z is an immersion if wecan factor it as i : X → Z1 → Z with X → Z1 an open immersion and Z1 → Za closed immersion.

Exercise: A composition of immersions is an immersion.

Definition 6.37 (Very Ample). Let X be a scheme over Y and L an invertibleOX-module. L is very ample relative to Y if ∃ an immersion i : X → PrY suchthat L = i∗O(1).

Note: If ϕ : X → Y is projective, then ϕ is proper, and X has a very ampleinvertible sheaf relative to Y .

Suppose that ϕ : X → Y is proper and X has a very ample invertible Lrelative to Y . We have an immersion i : X → PrY , ϕ proper implies that i isproper, so i(X) ⊆ PrY is closed, and so i is a closed immersion.

Definition 6.38 (Generated by global sections). Let X be a scheme, F anOX-module, F is generated by globa section if ∃ ⊕i∈IOX

α→ F a surjectiveOX-hom.

IE, ∃si ∈ Γ(X,F ) such that FP is generayed by (si)P as an OX,P -modulefor all P ∈ X.

Examples:

1. X = SpecA, F = M .

2. X = ProjS, S generated by S1 as an S0-module. Then OX(1) is generatedby global sections.

Definition 6.39 (Ample Sheaf). An invertible OX-module L on a Notherianscheme X is ample if ∀ coherent OX-modules F , ∃n0 > 0 such that F ⊗L ⊗n

generatd by global sections for all n ≥ n0.

Later: If X is of finite type over SpecA, A Notherian, then L is ample iffL ⊗m is very ample relative to A for some m > 0.

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Theorem 6.40. Let X be projective over SpecA, A Notherian. If O(1) is veryample relative to A, then O(1) is ample.

Proof. i : X → PrA is a closed immersion, O(1) = i∗OPr (1). F a coherentOX -module. Then i∗F is coherent on PrA. For P ∈ X, (i∗F )P = FP .

Exercise: Γ(X,F ⊗ O(1)⊗m) = Γ(PrA, (i∗F )(m)).WLOG: X = PrA = ProjA[x0, . . . , xr].F coherent implies that F |D+(xi) ' Mi, with Mi a finitely generated Bi

module, where Bi = A[x0/xi, . . . , xr/xi]. Let si1, . . . , siN generate Mi as a Bi-module. Then the lemma implies that there exists tij ∈ Γ(X,F ⊗OX(n)) suchthat tij |D+(xi) = sij ⊗ xni .

Take n large enough for all i, j. Claim: F (n) = F ⊗OX(n) is generated bytij ⊂ Γ(X,F (n)).

Why? Because F |D+(xi)'→ F (n)|D+(xi) by s 7→ s⊗ xni .

Corollary 6.41. X projective over Notherian A, F a coherent OX-module.Then there exists ⊕OX(ni)→ F surjective with a finite sum.

Proof. The theorem says that F⊗O(n) = F (n) is generated by global sections.Thus, O⊕NX → F (U)→ 0 exact, tensor with O(−n), and we get the result.

Definition 6.40 (*-scheme). X is a *-scheme if it is Notherian, separated,integral and OX,P is a DVR whenever dim OX,P = 1.

Examples: nonsing alg varietyX normal, Notherian, separated and integral.

Definition 6.41 (Prime Divisor). Assume X is a *-scheme, a prime divisor isa closed integral subscheme of codimension 1.

Note that OX,Y is a DVR, we have a valuation vY : k(X)∗ → Z by OX,Y =f ∈ k(X)|vY (f) ≥ 0. We call vY (f) the order of vanishing of f along Y .v(fg) = vY (f) + vY (g) and vY (f + g) ≥ min(vY (f), vY (g)).

We set Div(X) to be the free abelian group on the prime divisors. A principaldivisor is one of the form (f) =

∑Y vY (f)[Y ] with f ∈ k(X)∗. For D,D′ ∈

Div(X) write D ∼ D′ ⇐⇒ D −D′ is principle.Define the divisor class group C`(X) = Div(X)/principal divisors.Example: X = Pnk , a prime divisor is Y = V (f) and deg(Y ) = deg(f).

There is an ismorphism C`(Pn)→ Z by∑ni[Yi] 7→

∑ni deg(Yi).

1. U ⊆ X open implies there exists a surjective group hom C`(X)→ C`(U)by [Y ] 7→ [Y ∩ U ] if Y ∩ U 6= ∅ and 0 otherwise. So if Z = X \ U andcodim(Z;X) ≥ 2, then C`(X) ' C`(U). If Z ⊆ X is a prime divisor, thenhave Z→ C`(X)→ C`(U)→ 0.

2. A is a Notherian domain. Then A is a UFD iff U = SpecA is normal andC`(U) = 0.

3. π : X × Am → X gives an isomorphism π∗ : C`(X) → C`(X × Am) by[Y ] 7→ [Y × Am].

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Example: X = Pn × Pm, p : X → Pn and q : X → Pm, then p∗ : C`(Pn) →C`(X) is injective, by [Y ] 7→ [Y × Pm]. Let H ⊆ Pm be a hyperplane, U =X \ Pn × H = Pn × Am. Then we have an isomorphism C`(Pn) → C`(X) →C`(Pn ×An). Similarly q∗ : C`(Pm)→ C`(X) is injective. In fact, the image ofq∗ is Z[Pn ×H].

So we get 0→ C`(Pm) = Z→ C`(X)→ C`(Pn×Am) = Z→ 0, but the lastmap has a section, so we get C`(X) = Z[Pn ×H]⊕ Z[H ′ × Pm].

Carier DivisorsLet X be any scheme. For U ⊆ X open, let S(U) ⊆ OX(U) be the set of

non-zerodivisors. We define K to be the sheafification of pre −K , which haspre−K (U) = S(U)−1OX(U). This is a sheaf of rings, OX -module, OX → K .

Example: if X is an integral scheme, S(U) = OX(U)\0. Then for U affine,Γ(U, pre −K ) = K(OX(U)) = k(U) and Γ(U,K ) = k(X) for any nonemptyU ⊆ X open.

Note that OX ⊆ pre−K is a sub-presheaf, so pre−K is a decent presheaf.Recall that a decent presheaf satisfies the first sheaf axiom, and OX ⊂ pre−K ⊂K .

Define K ∗(U) to be the set of invertible elements of K (U). This is a sheafof abelian groups. So O∗X ⊆ K ∗.

Definition 6.42 (Cartier Divisor). A Cartier Divisor is an element D ∈ Γ(X,K ∗/O∗).It is principleif it is in the image of Γ(X,K ∗)→ Γ(X,K ∗/O∗).

NOTE: This is not surjective! There is a sheafification involved.

Convention: We use additive notation.Let D be a Cartier Divisor. Then there exists an open cover X = ∪Ui

and fi ∈ Γ(Ui,K ∗) such that D|Ui is the image of fi and on Ui ∩ Uj , fi/fj ∈Γ(Ui ∩ Uj ,O∗X).

Whenever fi satisfy this condition, they define a Cartier divisor D.

Definition 6.43. We define CaC`(X) = cartier divisors/principal cartierdivisors. Write D ∼ D′ iff D −D′ = 0 ∈ CaC`(X).

Assume that X is a *-scheme. Then D is a Cartier, Y ⊆ X a prime divisor,we write vY (D) = vY (fi) where Y ∩ Ui 6= ∅ and D|Ui =image of di. This givesus a group homomorphism Γ(X,K ∗/O∗X)→ Div(X) by D 7→

∑Y vY (D)[Y ].

This induces a map CaC`(X)→ C`(X), and is injective when X is normal.

Definition 6.44 (Locally Factorial). X is locally factorial if OX,P is a UFDfor all P ∈ X.

Proposition 6.42. X is locally factorial *-scheme, then CaC`(X) ' C`(X).

Definition 6.45 (Picard Group). The set of all isomorphism classes of invert-ible OX-modules under tensor product.

Let D ∈ Γ(X,K ∗/O∗X) cartier. Then D|Ui is the image of fi ∈ Γ(U,K ∗).

Definition 6.46. L (D) ⊆ K is the sub OX-module such that L (D)|Ui isgenerated by f−1

i .

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L (D) is an invertible OX -mod as OUi → L (D)|Ui given by multiplicationby f−1

i .

Proposition 6.43. 1. Γ(X,K ∗/O∗X)↔ invertible subsheafs of L .

2. L (D1)⊗L (D2)−1 ' L (D1 −D2).

3. D1 ∼ D2 iff L (D1) ' L (D2).

Proof. 1. Let L ⊆ K be any invertible OX -module. There exists an opencover X = ∪Ui such that L |Ui ' OUi , so Γ(Ui,L ) ' Γ(Ui,OUi). De-fine fi to be the section corresponding to 1 ∈ Γ(Ui,OUi) by the chosenisomorphism. fi/fj ∈ Γ(Ui∩Uj ,O∗X)⇒ f−1

i define a Cartier divisor D.

2. L (D1)⊗L (D2)−1 → L (D1 −D2) by h1 ⊗ h2 7→ h1h2.

3. Assume L (D1) ' L (D2) as OX -modules. By the last part, K ⊂ L (D1−D2) ' OX . Let 1 ∈ Γ(X,OX) which corresponds to f ∈ Γ(X,K ). f ∈Γ(X,K ∗), so D1 −D2 =image of f−1.

Corollary 6.44. If X is any scheme, then the map CaC`(X) → Pic(X) isinjective.

Proposition 6.45. If X is integral, then CaC`(X) ' Pic(X).

Proof. Must show that any invertible L is a submodule of K .Let L |U ' OU . Then 1 ∈ Γ(U,OU ) corresponds to f ∈ Γ(U,L ). Assume

that ∅ 6= V ⊆ X open, h ∈ L (V ), then U ∩ V 6= ∅. h ⊗ f−1 ∈ Γ(U ∩ V,L ⊗L −1) = Γ(U ∩ V,OX) ⊆ Γ(U ∩ V,K ) = Γ(V,K ).

Define an OX -hom L → K by h 7→ h ⊗ f−1. This is injective, as fpgenerator for Lp for all p ∈ U .

Corollary 6.46. X Notherian, Integral, Separated, Locally Factorial impliesC`(X) = CaC`(X) = Pic(X)

Corollary 6.47. Pic(Pnk ) = O(m)|m ∈ Z.

Proof. X = Pnk . We know that C`(X) = Z[H] ' Pic(X).Must show that [H] ↔ O(1). H = V (h) where h ∈ k[x0, . . . , xn] is a linear

form. Set fi = h/xi ∈ OX(D+(xi)). fi/fj = xj/xi is a unit on D+(xixj), sofi defines a Cartier Divisor. D ∈ CaCl(X) thar corresponds to [H] sincevY (D) = 0 if Y 6= H and 1 if Y = H. So we get OX(1) → L (D) an isomor-phism, by s 7→ s/h.

Definition 6.47 (Effective Divisor). A Cartier Divisor D is effective if ∃ anopen cover X = ∪Ui and nonzerodivisors fi ∈ OX(Ui) such that D|Ui is theimage of fi.

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D is effective gives us a closed subscheme of X by ID generated by fi onU . This gives us a correspondence between effective Cartier Divisors and closedsubschemes that are locally generated by one nonzerodivisor.

Note: If X is a *-scheme and D an effectve Cartier Divisor, then vY (D) =vY (fi) ≥ 0 so

∑Y vY (D)[Y ] is effective. If X is a locally factorial *-scheme,

then effective Cartier divisors correspond to effective Weil divisors.Commutative Algebra Fact: If A is a normal Notherian Domain, then A =

∩ht(P )=1AP ⊆ K(A). So fi ∈ k(Ui) gives fi ∈ OUi(Ui) ⇐⇒ (fi) ∈ Div(Ui)effective.

Proposition 6.48. D ⊆ X an effective Cartier divisor implies that ID =L (−D) ⊆ K .

This is because OD ⊆ OX ⊆ K is locally generated by fi.Let ϕ : X → Y , f ∈ OY (Y ) then ϕ−1(Yf ) = Xϕ]f .In fact, we can do this for any line bundle L . Let L be an OY -module. Then

ϕ∗ = ϕ−1L ⊗ϕ−1OY OX . For s ∈ Γ(Y,L ) define ϕ∗s = ϕ−1s⊗ 1 ∈ Γ(X,ϕ∗L ).Xϕ∗s = P ∈ X|(ϕ∗s)P /∈ mP (ϕ∗L )P = ϕ−1(Ys).Morphisms to Pn.If ϕ : X → PnA is a morphism, then

1. X is a scheme over A

2. L = ϕ∗O(1) is an invertible OX -module.

3. Generated by si = ϕ∗xi ∈ Γ(X,L ) for 0 ≤ i ≤ n where O(1) is generatedby x0, . . . , xn ∈ Γ(Pn,O(1)).

Claim: 1+2+3 determines a unique ϕ : X → PnA.Note: s−1

i ∈ Γ(Xsi ,L−1). sj/si = sj ⊗ s−1

i ∈ Γ(Xsi ,L ⊗ L −1) =Γ(Xsi ,OX).

We must have Xsi = ϕ−1(D+(xi)). This defines ϕ : Xsi → D+(xi) by A-algebra homomorphism A[x0/xi, . . . , xn/xi]→ Γ(Xsi ,OX) by xj/xi 7→ sj/si.

Theorem 6.49. Let X be a scheme over A. Then there is a correspondenceϕ : X → PnA over A to (L , s0, . . . , sn)|L generated by s0, . . . , sn/ '.

Example: if k is a field, then Pnk = Ank ∪ Pn−1k , An = D+(x0) and P ∈ Pn.

Then P = 0 ∈ An. So π : An \ 0 → Pn−1, id : Pn−1 → Pn−1 defines theprojection from a point Pn \ P → Pn−1.

Let x1, . . . , xn ∈ Γ(Pn,O(1)). These generate O(1)|Pn\P. This defines theprojection from P .

Note: Assume ϕ : X → PnA is given by si = ϕ∗xi ∈ L = ϕ∗O(1). ϕ is aclosed immersion iff ϕ : Xsi → D+(xi) is a closed immersion for all i iff Xsi

affine and A[x0/xi, . . . , xn/xi]→ OX(Xsi) is surjective.Recall the definition of an ample line bundle.Example: If X is projective over a Notherian ring, L = O(1) is very ample.

If X is affine, and L is any invertible sheaf.

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Proposition 6.50. X a Notherian scheme, L an invertible OX-module. ThenTFAE

1. L ample

2. L ⊗m ample for all m ≥ 1

3. L ⊗m ample for some m ≥ 1.

Proof. 1⇒ 2⇒ 3: trivialAssume that L ⊗m is ample, F a coherent OX -module. F ⊗L ⊗k is coher-

ent. For 0 ≤ k ≤ m − 1 choose nk > 0 such that (F ⊗L ⊗k) ⊗ (L ⊗m)⊗n isgeneratedby global sections for all n ≥ nk.

N = maxk + nmk implies that F ⊗L ⊗n is generated by global sectionsfor all n ≥ N .

Theorem 6.51. X is of finite type over Notherian A. Then L is ample iffL ⊗m isvery amples relative to SpecA for some m > 0.

Pnk : O(m) is ample iff m ≥ 1. If m < 0 then Γ(Pn,O(m)) = 0. O(0) gen byglobal sections.

Remark: ϕ : X → Y a morphism, (X Notherian OR ϕ separated andquasi-compact) Let Z = ϕ(X) ⊆ Y . IZ = ker(ϕ] : OY → ϕ∗OX). IZ isquasicoherent implies that (Z,OX/IZ) ⊆ Y is a closed subscheme called thescheme-theoretic image of X.

Exercise: X is reduced implies that Z = ϕ(X) is reduced.Example: Y = Spec k[x, y]/(xy, y2), X = D(x) = Spec k[x, x−1] ⊆ Y . j :

X ⊆ Y . So j(X) = Yred = A1.

Application: Xf→ Y

g→ Z an immersion. Then X ⊆ gf(X) → Z is animmersion.

Exercise: Check This! Hint: Assume f is closed and g is open.

Lemma 6.52. X Notherian Scheme, U ⊆ X open, F a coherent OU -module.Then there exists a coherent OX-module F ′ such that F ′|U ' F .

Note: i : U → X, so i∗F is quasicoherent has (i∗F )|U ' F .

Proof. If not, let U ⊆ X be a maximal open set such that the lemma is false,and WLOG, F is a counterexample.

Let Y ⊆ X be open affine, Y 6⊆ U . Then j : U ∩ Y ⊆ Y . G = j∗(F |U∩V )is a quasicoherent OY -module, and G |U∩Y = F |U∩Y . Let M = Γ(Y,G ) =Γ(U ∩ Y,F ) be a module over A = Γ(Y,OX).

G = M . Now, U ∩Y = Yf1 ∪ . . .∪Yfn , fi ∈ A. Then Mfi = G |Yfi = F |Yfi iscoherent. Thus Mfi is a finitely generated Afi-module. Choose m1, . . . ,mN ∈M generated Mfi for all i. Set M ′ = (m1, . . . ,mN ) ⊆M . G ′ = M ′ is a coherentOY -module. G ′ ⊆ G and G ′|U∩Y = G |U∩Y = F |U∩Y .

Set X ′ = U ∪ Y . Define an OX′ -module F ′ by F ′(V ) = (a, b)|a ∈ F (U ∩V ), b ∈ G ′(Y ∩ V ) with a|U∩Y ∩V = b|U∩Y ∩V ∈ F (U ∩ Y ∩ V ).

F ′|U = F , and F ′|Y = G ′, so F ′ is a coherent OX′ -module. The contra-dicts that F is a maximal counterexample.

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Theorem 6.53. X is a scheme of finite type over a Notherian ring A. Lan invertible OX-module. Then L is ample iff L ⊗m is very ample relative toSpecA for some m > 0.

Proof. Assume that L ⊗m is very ample over A. Then there exists an immersionX ⊆ X → PnA where the first is open and the second is closed, such thatL ⊗m ' OX(1).

Let F be coherent OX -module. We know that OX(1) is ample, so thelemma implies that there exists a coherent OX -module F such that F |X = F .F ⊗ OX(N) generated by global sections for all N >> 0, so F ⊗ OX(N) isgenerated by global sections for N >> 0. Thus OX(1) = L ⊗m is ample, so Lis ample.

Now we assume that L is ample. Let P ∈ X. There exists an open affineneighborhood P ∈ U ⊆ X such that L |U ' OU . Y = (X \ U)red → X is aclosed subscheme, IY ⊆ OX is coherent implies htat IY ⊗L ⊗n is gneeratedby global sections for some n. So there exists s ∈ Γ(X,IY ⊗L ⊗n) such thatsP /∈ mP (IY ⊗L ⊗n)P .

So IY ⊆ OX , so I ⊗L ⊗n ⊆ L ⊗n, thus s ∈ Γ(X,L ⊗n).P ∈ Xs ⊆ U , so (IY )P = OX,P , (IY )Q ⊆ mQ for all Q ∈ Y , so L |Y ' OU .

s|U corresponds to f ∈ Γ(U,OX).Xs = Us = Uf is affine. Therefore for all P ∈ X, there exists n > 0

and s ∈ Γ(X,L ⊗n) such that P ∈ Xs and Xs affine. As X is Notherian,X = Xs1∪. . .∪Xsk , where s1, . . . , sk ∈ Γ(X,L ⊗n) and Xsi are affine. n =

∏ni,

so we can replace si with sn/nii ∈ Γ(L ⊗n).

WLOG, ni = n for all i. We replace L with L ⊗n. So WLOG, there exists1, . . . , sk ∈ Γ(X,L ) such that Xsi is affine and X = Xs1 ∪ . . . ∪Xsk .

Set Bi = Γ(Xsi ,OX). As X is of finite type over SpecA, we know that Bi isa finitely gneerated A-algebra generated by bi1, . . . , biN ∈ Γ(Xsi ,OX). So thereexists n > 0 and cij ∈ Γ(X < OX ⊗L ⊗n) such that cij |Xsi = bijs

ni .

Now, L ⊗n is an invertible OX -module generated by the global setionssni , cij.

We define a morphism over A, ϕ : X → Pk(N+1)−1A = ProjA[xij : 1 ≤ i ≤

n, 0 ≤ j ≤ N ]. Then ϕ∗O(1) = L ⊗n, ϕ∗(xij) = cij and ϕ∗(xi0) = sni . Note thatXsi = ϕ−1(D+(xi0))→ D+(xi0) is a closed immersion, so Bi ← O(D+(xi0)) issurjecitve, mapping xij/xi0 to cij/sni = bij .

Thus, ϕ : X → ∪ki=1D+(xi0) ⊆ Pk(N+1)−1A is an immersion, so L ⊗n '

ϕ∗O(1) is very ample.

Remark: Y a scheme, X ⊆ Y an integral closed subscheme. D ⊆ Y is aneffective Cartier divisor. Assume that X 6⊆ D, then D|X = X ∩ D ⊆ X is aneffective Cartier divisor, and L (D)|X = i∗L (D) = L (D ∩ X). So ID|X =ID∩X .

Example: P2k = Proj k[x, y, z]. X = V (zy2 − x3 + xz2) ⊆ P2, and P0 = (0 :

1 : 0) corresponds to (x, z) ⊆ k[x, y, z]. L (P0) is ample, but not very ample.

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Claim: OX(1) ' L (3P0). Let L = V (z) ⊆ P2, this is just the line at infinity.C`(P2) = Z[L] and OP2(1) = L ([L]). So OX(1) = OP2(1)|X = L ([L])|X =L (L ∩X).

Show that L ∩X = 3P0 ∈ Div(X). Notice that X = Xy ∪Xz IL∩X ⊆ OXis generated by z/y on Xy and 1 on Xz.

OX,P0 = k[x/y, z/y](x/y,z/y)/(z/y− (x/y)3 + (x/y)(z/y)3) and mP0 = (x/y).Then vP0(L ∩ X) = vP0(z/y) = 3 as z/y = (x/y)3 times a unit, thereforeL ∩X = 3P0 ∈ Div(X) so L (P0) is ample.

Claim: L (P0) is not very ample. Otherise there would exist s ∈ Γ(X,L (P0))such that sP0 /∈ mP0L (P0)P0 . (s)0 ⊆ X is an effective cartier divisor. IfL (P0)|U = OU then s|U corresponds to f ∈ Γ(U,OU ), and (s)0∩U = V (f) ⊆ Uis a closed subscheme. (s)0 ∼ P0 so deg((s)0) = degP0 = 1, so (s0) = Q ∈Div(X), Q 6= P0. So P0 ∼ Q and P0 6= Q on a nonsingular projective curve X,so X is rational and X ' P1, contradiction.

Let f : Y → X be an affine morphism, A = f∗OY is a quasi-coherent sheafof OX algebras. f ] : OX → f∗OY = A .

If U ⊆ X open affine, then V = f−1(U) ⊆ Y is open affine, and A (U) =OY (f−1(U)) = OY (V ), thus, f : V → U given by OX(U)→ A (U).

Let X be any scheme, A a quasi-coherent OX -algebra. Want: affine f :Y → X such that A = f∗OY .

Functor of PointsLet X be a scheme.

Definition 6.48 (Functor of Points). We define the functor of points to bea contravariant functor FX :schemes to sets with FX(Y ) = homsch(Y,X). Ifh : Y → Y ′ is a morphism, then F (h) = h∗ : FX(Y ′)→ FX(Y ) by g 7→ gh.

Example: X a variety over k = k. Then FX/k(Y ) = homk(Y,X), thenFX/k(Spec k) = the set of points of X.

Proposition 6.54. The set of morphisms ϕ : X → X ′ are in correspondencewith the natural transformations α : FX → FX′ .

Proof. Given ϕ : X → X ′, then αY : FX(Y ) → FX′(Y ) can be defined byg 7→ ϕg.

Given α : FX → FX′ , αX : FX(X)→ FX′(X) set ϕ = αX(id) : X → X ′.Let Y be a scheme, and g ∈ FX(Y ). As we have a natural transformation,

we know that αY g∗ = g∗αX . Thus, if we take id ∈ FX(X), it is mapped tog ∈ FX(Y ) then to αY (g) in FX(Y ). But also it is mapped to ϕ in FX′(X) andthen to ϕg in FX(Y ), so they must be equal, and so αY (g) = ϕg.

Corollary 6.55. X ' X ′ ⇐⇒ FX ' FX′ .

Example: Let X×SY be the fibered product over f : X → S and g : Y → S.Then X ×S Y is uniquely determined by FX×SY (Z) = (p, q)|p : Z → X, q :Z → Y, fp = gq = FX(Z)×FS(Z) FY (Z).

X is a scheme, A a quasi-coherent OX -algebra, γ : OX → A .

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We define FA :schemes→sets by FA (Y ) = (f, f)|f : Y → X, F : A →

f∗OY an OX -alg hom, and f ] : OXγ→ A

f→ f∗OY on objects and if h : Y → Y ′

is a morphism, then we define h∗ : FA (Y ′)→ FA (Y ) by (f, f) 7→ (fg, f∗(h])f).

Definition 6.49 (SpecA ). SpecA is the unique scheme represented by FA ifit exists. FSpecA = FA .

Note: FSpecA (SpecA ) = FA (SpecA ) so id maps to (π, π) the natural pro-jection π : SpecA → X, π] : OX

γ→ Aπ→ π∗O with O = OSpecA .

Example: f : Y → X an affine morphism, A = f∗OY , γ = f ] : OX → A ,then SpecA = Y

Let Z be any scheme, g : Z → Y a morphism, g] : OY → g∗OZ , so f∗(g]) :f∗OY = A → (gf)∗OZ define FY (Z)→ FA (Z) by g 7→ (gf : Z → X, f∗(g])).

Assume (ϕ, ϕ) ∈ FA (Z). ϕ : Z → X and ϕ] : OXf]→ A

ϕ→ ϕ∗OZ .Let U ⊆ X be open affine, V = f−1(U) ⊆ Y affine.ϕ : A (U) = OY (V ) → ϕ∗OZ(U) = IZ(ϕ−1(U)) defines a morphism

ϕ−1(U)→ V .

Proposition 6.56. Let X be a scheme, A be a quasi-coherent OX-algebra.Then SpecA exists, π : SpecA → X is affine, and π : A → π∗O is anisomorphism.

Proof. If U ⊆ X open affine, then SpecA |U = Spec(A (U)), because π :Spec A (U)→ X has π∗OSpecA = π∗ ˜A (U) = ˜A (U) = A |U .

Assume U ′ ⊆ U open subset, U affine, then SpecA |U ′ = π−1U (U ′) We

have Spec A |U ′ = Spec A |U ×U ′ U ⊆ Spec A |UπU→ U ⊇ U ′, so we glue

Spec A (U)|U ⊆ X affine together to Spec A . U1, U2 ⊆ X open affine, andSpec A (U1) ⊇ π−1

U1(U1∩U2) = Spec A |U1∩U2 = π−1

U2(U1∩U2) ⊆ Spec A (U2).

Let X be a Notherian Scheme, S = ⊕d≥0Sd a graded quasi-coherent OY -algebra. For U ⊆ X open affine, πU : Proj S (U) → U . If U ′ ⊆ U a smalleropen affine, then S (U)→ S (U ′) define Proj S (U ′)→ Proj S (U). So we havea fiber square

Proj S (U ′) U ′

Proj S (U) U

.................................................................... ............πU ′

......................................................................... ............πU

........

........

........

........

........

........

........

........

........

........

........

........

.................

............

inc

........

........

........

........

........

........

........

........

........

........

........

........

.................

............

inc

S is quasi coherent implies that S (U ′) = S (U)⊗OX(U) OX(U ′).Define FS (Y ) = homsch(Y,Proj S ) = (f, gU )|f : Y → X, gU : f−1(U) →

Proj S (U) compatible.

Proposition 6.57. There exists a unique scheme ProjS representing FS.

Proof. If U ⊆ X open affine, then Proj(S|U ) = Proj(S(U)) πU→ U . If U ′ ⊆ Uany open subset, then Proj(S|U ′) = π−1

U (U ′), now glue.

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Example: If X is a scheme, and S = OX [τ0, . . . , τn] then Proj(S) = PnX =Pn ×X.

Example: P1 = Proj(k[x, y]), S = ⊕d≥0Sd. Let a ∈ Z, then Sd = OP1 ⊕O(a) ⊕ O(2a) ⊕ . . . ⊕ O(da). So if f ∈ Sd, f

′ ∈ Sd′ , then f ∈ O(ia) andf ′ ∈ O(ja), so ff ′ ∈ O((i+ j)a).

So we get π : ProjS → P1 with π−1(D+(x)) = ProjS(D+(x)) = Proj k[y/x][s, xat] =A1 × P1, and π−1(D+(y)) = Proj(k[x/y][s, yat]) = A1 × P1. Set λ = y/x, thenthis is k[λ−1][s, xaλat], so we glue along the (A1 \ 0) × P1’s vis (λ, (s : t)) 7→(λ−1, (s : λat)).

Then F−a = Proj(S) π→ P1 is the Hirzebruch Surface.X is a scheme, S = ⊕d≥0Sd graded OX -algebra. Then ProjS is the unique

scheme such that hom(Y,ProjS ) = FS (Y ).Note: Have OU (1) on π−1(U), compatible: U ′ ⊆ U a smaller open affine,

then Proj S (U ′) ⊆ Proj S (U), so OU ′(1) =pullback of OU (1). Glue to getO(1) on ProjS : Γ(V,O(1)) = (σU )|σU ∈ Γ(V ∩ π−1(U)),OU (1) which arecompatible.

Remark: S = ⊕d≥0Sd is a graded ring, u ∈ S0 a unit. Then define θd : Sd →Sd by θd(s) = uds. This gives an isomorphism of graded S0-algebras θ : S → S.

Note: h ∈ S+ homogeneous implies that θ = id : S(h) → S(h), so it inducesid : ProjS → ProjS.

Definition 6.50. Let S = ⊕d≥0Sd graded OX-algebra. L an invertible OX-module. Then S ∗L = S ′ = ⊕d≥0S ′d where S ′d = Sd ⊗L ⊗d. Let π : P =ProjS → X and π′ : P ′ = ProjS ′ → X.

Lemma 6.58. We have an natural isomorphism ϕ : P ′ → P over X andOP ′(1) = ϕ∗OP (1)⊗ (π′)∗L .

Proof. Let U ⊂ X open affine, with OU → L |U an isomorphism with 1 corre-sponding to f ∈ γ(U,L ). This defines an isomorphism S (U) → S ′(U), withS (U)d → S ′(U)d = S (U)d ⊗ Γ(U,L ⊗d) by s 7→ s⊗ fd.

This defines an isomorphism ϕ : Proj S ′(U)→ Proj S (U).Remark implies that ϕ is independent of f ∈ Γ(U,L )∗. So we can glue to

an isomorphism ϕ : P ′ → P .The sections of OP ′(1) correspond to elements in Γ(U,S ′1) = Γ(U,S1) ⊗

Γ(U,L ) correspond to sections of OP (1)⊗ π∗L .

Definition 6.51. Assume X is Notherian. S = ⊕d≥0Sd satisfies (+) if

1. S0 = OX

2. S1 is a coherent OX-module

3. S is generated (locally) by S1.

Note: If U ⊆ X is a small open affine, then ∃S ⊗d1 |U → Sd|U → 0 impliesthat Sd is coherent.

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Lemma 6.59. Assume that S satisfies (+). Let π : P = Proj S → X. Thenπ is prokper and if there exists an ample invertible OX-module L on X, thenπ is projective, and OP (1)⊗L ⊗n is very ample relative to X for some n > 0.

Proof. Let U ⊂ X open affine such that S (U) generated by S1(U) as A-algebra,with A = OX(U). Then S1-coherent implies that S1(U) is a finitely gneeratedA-module, so ∃ a graded A-algebra homomorphism A[x0, . . . , xN ] → S (U)surjective. Thus, Proj S (U) close→ PNU

π→ U . Thus π|U is projective, so it isproper. Thus, π : Proj S → X is proper.

If L is ample, then S1 ⊗ L ⊗n is generated by global sections. As X isNotherian, S1⊗L ⊗n is coherent, so it is generated by finitely many global sec-tions. Thus, there exists a surjection of graded OX -algberas OX [T0, . . . , TN ]→S ∗ (L ⊗n), and so P ′ = Proj(S ∗L ⊗n)→ Proj OX [Ti] = PnX is closed, and soOP ′(1) = ϕ6 ∗ (OP (1)⊗ π∗(L ⊗n)) is very ample.

Definition 6.52 (Tensor Algebra). Let A be a ring and M be an A-module.Then T dM = M ⊗ . . . ⊗M , d-times. T (M) = ⊕d≥0T

dM is called the tensoralgebra. S(M) = ⊕d≥0S

d(M) = T (M)/(x⊗y−y⊗x) is the symmetric algbera.If M = A⊕r, then S(M) = A[T1, . . . , Tr].

If X is a Notherian scheme, E a locally free OX -module of rank r andE ∨ = H om(E ,OX) the dual sheaf is also locally free of rank r. Then S(E ∨) =[U 7→ S(Γ(U,E ∨))+ (sheafification) is a graded OX -algebra.

S(E ∨)0 = OX , and S(E ∨)1 = E ∨, ... this satisfies (+). Set π : Y =SpecS(E ∨)→ X.

If U = SpecA ⊆ X open, E |U ' O⊕rU , then S(E ∨)|U ' OU [T1, . . . , Tr], soπ−1(U) = Spec OU [T1, . . . , Tr] = U × Ar. Thus π : Y → X affine bundle (infact, a vector bundle), so assume f : U → π−1(U) ⊆ Y is a section, (πf = idU ),then

Γ(U,E ∨)→ Γ(U, S(E ∨))f]→ Γ(U,OX)

gives an OX -homomorphism f : E ∨ → OX over U , ie, a section f ∈ Γ(U,E ).THus E is the sheaf of sections of π : SpecS(E ∨)→ X.

Definition 6.53. P(E ) = ProjS(E ∨).π : P(E )→ X, OE (1).

If E |U ' O⊗rU , then π−1(U) = U × Pr−1 = Pr−1U .

Example: L an invertible OX -module, π : P(L ) → X an isomorphism,OL (1) = π∗(L −1).

Proposition 6.60. 1. If rank(E ) ≥ 2, then π∗OE = Sm(E ∨) for m ≥ 0 and0 for m < 0.

2. Have surjection π∗E ∨ → OE (1).

Proof. Have global Sm(E ∨) → π∗OE (m), and in U ⊆ X open affine, thenf ∈ Γ(U, Sm(E ∨)) gives a section of OE (m) over π−1(U) = Proj Γ(U, S(E ∨))

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isomorphism locally when E |U ' O⊗rU , π−1(U) = Pr−1U , already computed

Γ(Pr−1U ,O(m)).For the secton part, we have an OX -homomorphism E ∨ → π∗OE (1) which

gives an OP(E )-homomorphism π∗E ∨ → OE (1), and can check that it is sur-jecitve locally.

Universal PropertyFP(E )(Y ) = (f,L , θ)|f : Y → X,L an invertible OY -module, and θ :

f∗E ∨ → L surjective. Left to reader.Exercise: P2

k, k = k. Then P2 = L ⊂ k3|dimL = 3 = E2 ⊆ k3|dimE2 =2.

0 → E → O⊕3P2 → OP2(1) → 0 with E locally free of rank 2, then P(E ) =

F`(k3) = (E1, E2)|E1 ⊂ E2 ⊂ k3,dim(Ei) = i.π : P(E )→ P2. Flag of locally free OX -module, OE (−1) ⊆ π∗E ⊆ O⊕3

P(E ) cor-respond to flags of vector bundles B1 ⊆ B2 ⊆ P(E )×k3, Bi = ((E1, E2), ~v)|~v ∈Ei.

7 Schemes II

DifferentialsLet R be a ring, S a commutative R-algebra and M an S-module.

Definition 7.1 (R-derivation). D : S → M is an R-derivation if D(fg) =fD(g)+gD(f), D(f +g) = D(f)+D(g) for all f, g ∈ S and D(f) = 0 if f ∈ R(iff D is R-linear)

Universal Derivation: Let F be the free S-module with basis df |f ∈ S,and F ′ ⊂ F the submodule generated by d(fg)−fdg−gdf, d(f+g)−df−dg anddf for all f ∈ R. Then defime ΩS/R = F/F ′ and d = dS = dS/R : S → ΩS/R byd(F ) = df + F ′. This satisfies the universal property that if D : S →M is anyR-derivation, then ∃!S-homomorphism D : ΩS/R →M such that D = D dS/R.

Note: P (x1, . . . , xn) ∈ R[x1, . . . , xn] and f1, . . . , fn ∈ S. ThenD(P (f1, . . . , fn)) =∑ni=1

∂P∂xi

(f1, . . . , fn)D(fi).

Proposition 7.1. S = R[x1, . . . , xn], then ΩS/R ' S⊕n = ⊕ni=1S · dxi andd : S → S⊕n is given by df = (df/dx1, . . . , df/dxn).

Proposition 7.2. R → S → T ring homomorphisms, then ΩS/R ⊗S T →ΩT/R → ΩT/S → 0 is exact as T -modules.

Proposition 7.3. If S is an R-algebra and T = S/I, then I/I2 δ→ ΩS/R⊗ST →ΩT/R → 0 wotj δ taking f + I2 to df ⊗ 1.

Proposition 7.4. Let R′ and S be R-algebras and S′ = S⊗RR′. Then ΩS′/R′ =ΩS/R ⊗S S′

Proposition 7.5. U ⊆ S a multiplicative subset, then U−1ΩS/R = ΩU−1S/R.

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Corollary 7.6. If S is a localization of a finitely generated R-algebra, thenΩS/R is a finitely generated S-module.

Proof. S = U−1S′, S′ = R[s1, . . . , sn]. Then ΩS/R is generated by ds1, . . . , dsnas an S′-module, so ΩS/R = U−1ΩS′/R is generated by ds1, . . . , dsn as an S-module.

Sheaves of DifferentialsLet X be a topological space, R,S sheaves of rings, and R → S a sheaf

homomorphism.Define pre− Ω = pre− ΩS /R = [U 7→ ΩS(U)/R(U)]. If V ⊆ U is open, then

we get a diagram S (U)→ S (V )→ pre−Ω(V ), but also S (U)→ pre−Ω(U),which goes to pre− Ω(V ) by restriction, and this is an R(U)-derivation. Thenpre− Ω is a presheaf.

Define ΩS /R = (pre− Ω)+.Let ϕ : X → Y be a morphism of schemes, ϕ] : ϕ−1OY → OX .Define ΩX/Y = ΩOX/ϕ−1OY , and if we have X → Spec(k) is a scheme over

k, then ΩX = ΩX/k = ΩX/ Spec(k). We call this the relative cotangent sheaf andthe cotangent sheaf.

Proposition 7.7. If ϕ : X → Y is a morphism of affine schemes, X = Spec(S)and Y = Spec(R), then ΩX/Y ' ˜ΩS/R.

Corollary 7.8. ΩX/Y is always quasi-coherent. If ϕ : X → Y is locally offinite type, then ΩX/Y is coherent.

Proof. P ∈ X, take open affine neighborhoods ϕ(P ) ∈ V ⊆ Y adn P ∈ U ⊆ϕ−1(V ) ⊆ X. ΩX/Y |U = ΩU/V is quasicoherent. If ϕ is lcoally of finite type,then we can take U, V such that OX(U) finitely generated OY (V )-algebra.

Proposition 7.9. Take a fiber square:

X ′

X

Y ′

Y

........

........

........

........

........

........

........

........

........

........

........

........

.................

............

g′

........

........

........

........

........

........

........

........

........

........

........

........

.................

............

g

................................................................................................................. ............f

................................................................................................................. ............

Then g′∗ΩX/Y = ΩX′/Y ′

Proposition 7.10. If Xf→ Y

g→ Z morphisms then f∗ΩY/Z → ΩX/Z →ΩX/Y → 0 is exact.

Proposition 7.11. If g : Y → Z, X ⊆ Y a closed subscheme, then IX/I2X →

ΩY/Z ⊗ OX → ΩX/Z → 0 is exact.

Theorem 7.12. Let Y be a scheme. X = PnY = PnZ × Y . Then ∃ an exactsequence 0→ ΩX/Y → OX(−1)⊕n+1 → OX → 0.

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Y

X

Spec Z

PnZ.............................................................................................................................

................................................................................................................. ............

............................................................................................... ............

.............................................................................................................................

Proof. WLOG, Y = Spec(Z) and X = Proj(S) for S = Z[x0, . . . , xn]. SetE = S(−1)⊕n+1. We have a map E → S by ei 7→ xi, it has kernel M , so weget an exact sequence 0 → M → E → S. This gives us an exact sequence0→ M → O(−1)⊕n+1 → OX → 0

Notice: f ∈ Q(x0, . . . , xn) homogeneous of degree d, then∑ni=0

∂f∂xi

xi = df .Define d : OX → E by d(f) =

∑ni=0

∂f∂xi

ei. Note that d(OX) ⊆ M , ie d : OX →M is a derivation. This induces d : ΩX/Z → M , we will check that this is anisomorphism locally on D+(xi) = Spec Z[x0/xi, . . . , xn/xi].

Enough to check that Γ(D+(xi), M) is a free S(xi)-module with basis d(xj/xi)|j 6=i.

0→M(xi) → E(xi) → S(xi) → 0 which takes ej → xj but there’s a map fromS(xi) → E(xi) taking 1→ ei/xi. And therefore, d(xj/xi) = 1

xiej − xj

x2iei.

Singular VarietiesLet k = k

Definition 7.2 (Nonsingular Variety). A variety X over k is nonsingular atP ∈ X if OX,P is a regular local ring. X is nonsingular if all points are non-singular.

Theorem 7.13. X is an irreducible separated scheme of finite type over k = k.Set n = dim(X). Then ΩX/k is locally free of rank n iff X is a nonsingularvariety.

Recall that F ⊂ K is a field extension, and each a ∈ K has a minimalpolynomial pa(T ) ∈ F [T ] such that pa(a) = 0 ∈ K. K is separable over F ifpa(T ) does not have multiple roots for all a ∈ K.

Exercise: If K/F is separable, then ΩK/L = 0. (0 = dK(pa(a)) = p′a(a)dK(a)with p′a(a) 6= 0)

Corollary 7.14. X is a variety over k implies that a dense open subset of Xis nonsingular.

Proof. K = k(X).FACT: k is a perfect field implies that any finitely generated extenseion k ⊂

K is separaly generated. IE, there exists a transcendence basis x1, . . . , xn ∈ Ksuchj that k(x1, . . . , xn) ⊆ K is separable.

K is separable over F = k(x1, . . . , xn) for n = dim(X). Let S = k[x1, . . . , xn],F = S0 ⇒ ΩF/k = (ΩS/k)0 = F⊕n.

So ΩF/k ⊗F Kα→ ΩK/k → ΩK/F = 0.

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As any k-derivation D : F → K can be extended to D : K → K we canconclude that α is injective.

Thus, ΩK/k ' Kn. Spec(R) ⊆ X open. K = R0 ⇒ (ΩR/k)0 = ΩK/k = Kn.Thus ∃0 6= f ∈ R : (ΩR/k)f = ΩRf/k = (Rf )n.

Lemma 7.15. If (R,m) is a local ring, k = R/m, and k ⊆ R. Then δ : m/m2 '→ΩR/k ⊗R k is an isomorphism.

Theorem 7.16. X nonsing variety over k = k and Y ⊆ X irreducible closedsubscheme, then Y is nonsingular iff ΩY/k is locally free and 0 → IY /I 2

Y →ΩX ⊗ OY → ΩY → 0 is exact.

Proof. Assume the latter conditions, set q = rank(ΩY ), n = rank(ΩX) =dim(X). It is enough to show that q = dim(Y ). The second condition causesIY /I 2

Y to be locally free of rank n − q, and by Nakayama, IY is locally gen-erated by n− q elements.

The Principle Ideal Theorem says that dim(Y ) ≥ q. Let P ∈ Y , mP ⊆OY,P , then the lemma says that mP /m

2P = ΩY,P ⊗ OY,P /mP = ΩY,P ⊗ k. So

dimk(mP /m2P ) = q. Thus dim(Y ) = dim OY,P ≤ q, so dim(Y ) = q, so Y is

nonsingular.Assume that Y is nonsingular. ΩY/k is then locally free of rank q = dim(Y ).

We know that IY /I 2Y → ΩX ⊗ OY

ϕ→ ΩY → 0 is exact, so all that remains isshowing that δ : IY /I 2

Y → ΩX ⊗ OY is injective.Let P ∈ Y be any closed point. Localize at P . Then kerϕP is a free OY,P -

module of rank r = n−q, so there exist x1, . . . , xr ∈ IY,P such that dx1, . . . , dxrform a basis for kerϕP . Let I ′ ⊆ IY be a subideal generated by x1, . . . , xr andY ′ = Z(I ′) ⊆ X.

Thus, dx1, . . . , dxr must generate a free subsheaf of rank r in ΩX ⊗OY ′ . Sowe get that I ′/I ′2 δ→ ΩX ⊗ OY ′ → ΩY ′ → 0 and δ′ must be injective.

So the first part tells us that Y ′ is a nonsingular variety and dim(Y ′) =rank ΩY ′ = n − r = q, and Y ⊆ Y ′ ⊆ X closed subschemes, Y, Y ′ both havedimension q, so Y ′ = Y and δ′ = δ.

Exercise: (R,m) is a regular local ring, f ∈ m. Then R/(f) is regular localiff f /∈ m2. Use: Minimum number of gens of m = dimk m/m2, k = R/m.

Theorem 7.17 (Bertini’s Theorem). Let X ⊆ Pnk a closed irreducible nonsin-gular subvariety, k algebraically closed. Then there exists a hyperplane H ⊆ Pnsuch that X 6⊂ H and X ∩H is nonsingular.

Proof. V = Γ(Pn,O(1)) = kx0 ⊕ . . .⊕ kxn. Then points of P(V ) correspond tohyperplanes by f 7→ Z(f). Let P ∈ X be a closed point, BP = f ∈ P(V )|X ⊆Z(f) or P ∈ X ∩ Z(f) is a singular point.

Will show: ∪P∈XBP ( P(V ) is proper closed.Set r = dim(X). Claim: BP ⊆ P(V ) is a linear subspace of dimension

n − r − 1. Check that f0 ∈ V such that P /∈ Z(f0) ⊆ Pn. For f ∈ V , we have

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f/f0 ∈ OX,P . Then OX∩Z(f),P = OX,P /(f/f0). f/f0 ∈ mP ⇐⇒ P ∈ Z(f),f/f0 = 0 ∈ OX,P ⇐⇒ X ⊆ Z(f), f/f0 ∈ mP /m

2P ⇐⇒ f ∈ BP .

Note: ϕP : V → OX,P /m2P by f 7→ f/f0 is surjective because k = k so mP

is generated by linear forms.dimk V = n + 1, dimk(OX,P /m2

P ) = r + 1. So dim kerϕP = n − r, soBP = P(kerϕP ) so dimBP = n− r − 1.

Now we define B = (P, f) ∈ X × P(V )|f ∈ BP Check: This is actually a closed subvariety of X × P(V ). π1 : B → X is a

surjective morphism and π−11 (P ) = BP has dimBP = n−r−1 and dim(X) = r

and so dimB = dim(X) + dimBP = n − 1 and ∪P∈XBP = π2(B) ⊆ P(V ).Then π2 : X × P(V ) → P(V ) is a proper map, so π2(B) is closed of dimensionat most n− 1.

8 Cohomology

Let R be a ring.

Definition 8.1 (Injective). An R-modules I is injective if for every R-moduleM and submodule M ′ and R-homomorphism ϕ : M ′ → I, there exists an exten-sion to M .

Definition 8.2 (Divisible Abelian Group). A Z-module T is divisible if for alln ∈ Z the map T → T by multiplication by n is surjective.

Examples: Q,Q/Z.

Lemma 8.1. Any divisible Z-module is injective.

Proof. M ′ ⊂ M , ϕ′ : M ′ → T . Consider (N,ψ)|M ′ ⊂ N ⊂ M,ψ : N → Textends ϕ′.

Zorn’s Lemma says that there is a maximal element (N,ψ).Claim: N = M . Else take x ∈ M \ N . 0 → N → (N, x) → Z/mZ → 0.

Either m = 0 in which case (N, x) = N ⊕ Zx and we can extend ϕ(x) = 0 orm 6= 0 then mn ∈ N . So there exists y ∈ T with ϕ′(mx) = my, so ψ(x) = y.

Definition 8.3. If M is a Z-module, Mˆ = homZ(M,Q/Z).

Exercise: M →Mˆˆ, x 7→ [f 7→ f(x)] is injective.

Corollary 8.2. Every Z-module M is contained in a divisible module.

Proof. Take F free, F → Mˆ → 0. So M → Mˆˆ → F ˆ =direct product ofQ/Zs, with all the maps inclusions.

Note: T a Z-module, M an R-module, then homZ(R, T ) is an R-module.homZ(M,T ) ' homR(M, homZ(R, T )) by f 7→ [x 7→ [r 7→ f(rx)]].

Lemma 8.3. T a divisible Z-module implies that I = homZ(R, T ) is an injectiveR-module.

98

Proof. M ′ ⊆M are R-modules. Want:

homZ(M,T )

homR(M, I)

homZ(M ′, T )

homR(M ′, I)

........

........

........

........

........

........

........

........

........

........

........

........

.................

............

'

........

........

........

........

........

........

........

........

........

........

........

........

.................

............

'

........................................................................................ ............surj

........................................................................................ ............surj

Which is true.

Theorem 8.4. Every R-module M is a submodule of an injective R-module.

Proof. Lemma implies that there exists f : M → T injective Z-hom when T isdivisible. We define M → I = homZ(R, T ) by x 7→ fx with fx(r) = f(xr)

Note that M ⊆ I: this is because f was injective to begin with.

Definition 8.4 (Abelian Category). A category C is abelian if hom(A,B) isan abelian group for all A,B ∈ obj(C ), hom(B,C) × hom(A,B) → hom(A,C)is a group homomorphism, finite direct products exist, kernels exist, cokernelsexist, etc. See Weibel.

For any I an object of C , we have a contravariant functor hom(−, I) : C →Ab. It is left exact, ie 0→ A′ → A→ A′′ → 0 exact implies 0→ hom(A′′, I)→hom(A, I)→ hom(A′, I) exact.

Definition 8.5 (Enough Injectives). I is injective if hom(−, I) is exact.A category C has enough injectives if every object is a subobject of an injec-

tive object.

Corollary 8.5. If (X,OX) is a ringed space, then the category of OX-moduleshas enough injectives.

Proof. Let F be an OX -module. For p ∈ X, there exists an inclusion FP → IPan injective OX,P -module. Set I(U) =

∏p∈U IP .

Claim: I is an injective OX -module. For any OX -module, G , we havehomOX (G , I) =

∏p∈X homOX homOX,P (GP , IP ).

So if G ′ ⊆ G submodule, then hom(G , I)→ hom(G ′,I ) is surjective.

Corollary 8.6. Let X be a topological space, then the category Ab(X) of sheavesof abelian groups on X has enough injectives.

Proof. Set OX to be the constant sheaf Z, the sheaf of locally constant functionsto Z.

Injective ResolutionLet M be an object of C , then if C has enough injectives, there exists an

exacct sequence 0 → M → I0 → I1 → . . . with Ij injective, that is, M has aninjective resolution.

Let F : C → D be a left exact functor. Then we have a complex F (I ·) =0→ F (I0)→ F (I1)→ . . ..

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Definition 8.6. RjF (M) = Hj(F (I ·)).

Example: Left Exact implies that 0 → F (M) → F (I0) → . . .. We throwaway M so that H0(I ·) = R0F (M) = F (M).

Definition 8.7. Let X be a topological space, F a sheaf of abelian groups, andΓ : Ab(X) → Ab by Γ(F ) = Γ(X,F ) is left exact. Then set Hj(X,F ) =RjΓ(F ).

Let A·, B· be complexes in C , a morphism f : A· → B· is a collection offn : An → Bn such that the appropriate squares all commute.

This induces a map H(f) : Hn(A·)→ Hn(B·).f, g : A· → B· are homotopic (f ∼ g) iff there exists morphisms hn : An →

Bn−1 with fn − gn = dn−1B hn + hn+1d

nA. This implies that H(f) = H(g) :

Hn(A·)→ Hn(B·).

Lemma 8.7. Consider two chain complexes A· and I ·, 0 → M → A· and0 → M ′ → I · an injective resolution, with ϕ : M → M ′. Then there exists amorphism f : (M → A·)→ (M ′ → I ·) and any other is homotopic to f .

Proof. See Weibel.

Note: (f1 − g1 − d0Bh1)d0

A = d0B(f0 − g0 − h1d

0A) = d0

Bd−1B h0 = 0.

Corollary 8.8. RjF (M) is independent of the choice of injective resolution.

Corollary 8.9. A morphism ϕ : M → M ′ induces a unique ϕ∗ : RjF (M) →RjF (M ′)

Lemma 8.10 (Horseshoe Lemma). Let 0 → M ′ → M → M ′′ → 0 be a shortexact sequence in a category with enough injectives. Then there exists...

Theorem 8.11. Let C be abelian with enough injectives and F : C → C ′ a leftexact functor.

1. RnF : C → C ′ is additive, ie, RnF (M ⊕M ′) = RnF (M)⊕RnF (M ′)

2. R0F (M) = F (M) and FnF (I) = 0 if I is injective and n ≥ 1.

3. 0 → M ′ → M → M ′′ → 0 short exact gives a long exact sequence 0 →F (M ′)→ F (M)→ F (M ′′)→ R1F (M ′)→ . . .

4. The δ morphisms from RnF (M ′′)→ Rn+1F (M ′) are natural.

Definition 8.8 (F-acyclic). J ∈ obC is F -acyclic if RnF (J) = 0 for all n ≥ 1.

Example: J injective implies that J is F -acyclic for all left exact functorsF .

Lemma 8.12. 0 → Y 0 → Y 1 → . . . exact with Y i F -acyclic implies that0→ F (Y 0)→ F (Y 1)→ . . . is exact.

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Theorem 8.13. 0 → M → J0 → . . . a resolution by F -acyclic objects Jn.Then RnF (M) = Hn(F (J ·))

Definition 8.9 (δ-functor). If C and C ′ are abelian categories, then a δ-functorfrom C → C ′ is a sequence of functors T = (Tn : C → C ′)n≥0 togetherwith a morphism δn : Tn(M ′′) → Tn+1(M ′) for each short exactr sequence0→M ′ →M →M ′′ → 0 such that

1. Long exact sequence

2. Naturality.

Example: Tn = RnF .

Definition 8.10 (Universal δ-functor). A δ-functor T is universal if, given anyδ functor U and natural transformation f0 : T 0 → U0, then there are uniquefi : T i → U i such that the appropriate diagram commutes.

Definition 8.11 (Erasable). An additive functor F : C → C ′ is erasable if forall M ∈ ob(C ) there exists a monomorphism P : M → J such that F (P ) = 0 :F (M)→ F (J).

Example: RnF is erasable for n ≥ 1, M → I a monomorphism and Iinjective, then RnF (M)→ RnF (I) = 0.

Theorem 8.14. Let T = (Tn) be a δ-functor. If every Tn is erasable for alln ≥ 1 then T is a universal δ-functor.

Proof. Assume that U = (Un) is a δ-functor with f0 : T0 → U0 a naturaltransformation. Let M ∈ ob(C ).

0→Mp→ J

q→ X → 0 such that T 1(p) = 0.Then we take the long exact sequences, horizontally, and f0 vertically. We

use that ker δ0T = Im(q∗) ⊆ ker δ0

U f0. Then ∃!f1(M) : T 1(M)→ U1(M). Mustcheck that f1 is a natural transformation.

Check: f1 commutes with δ.To get f2, we use 0 → M → J → X → 0 such that T 2(p) = 0. Proceed by

induction using the long exact sequence.

Definition 8.12. Let X be a topological space, and F a sheaf of abelian groups(hereafter referred to as an abelian sheaf). Let Γ = Γ(X,−). Then Hn(X,F )is defined to be RnΓ(F ).

Definition 8.13 (Flasque). A sheaf F is flasque if all the restriction maps aresurjective.

Lemma 8.15. (X,OX) a ringed space. F injective as an OX-module. ThenF is flasque.

Proof. For U ⊆ X open, Set OU = j!(OX |U ) by j : U → X the inclusion andfor W ⊆ X open, OU (W ) = OX(W ) if W ⊆ U and 0 else.

Note, hom(OU ,F ) = F (U). If V ⊆ U , then 0 → OV → OU implies thathom(OU ,F )→ hom(OV ,F ) is surjective.

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Lemma 8.16. Assume that 0→ F ′ → F → F ′′ → 0 exact.

1. F ′ is flaque implies that 0→ F ′(X)→ F (X)→ F ′′(X)→ 0 is exact

2. F ′ and F flasque implies that F ′′ is flasque.

Proof. Let γ ∈ Γ(X,F ′). Consider (U, β)|U ⊆ X open and β ∈ Γ(U,F ) withβ 7→ γ|U ∈ F ′′(U).

Then (U ′, β′) ≤ (U, β) iff U ′ ⊆ U and β′ = β|U ′ . Zorn’s lemma gives us amaximal element (U, β).

Claim: U = X. Assume not, then there exists (V, σ) such that V 6⊆ U andσ ∈ Γ(V,F ), σ 7→ γ|V . Then β|U∩V − σ|U∩V ∈ F ′(U ∩ V ). Then flasqueimplies that ∃α ∈ F ′(V ) with α|U∩V = β − σ. Set β1 = σ + α ∈ F (V ). Thenβ|U∩V = β1|U∩V , so glue β, β1 to β ∈ Γ(U ∪ V,F ), then (U ∪ V, β) ≥ (U, β),contradiction.

Proposition 8.17. If F is flasque, then Hn(X,F ) = 0 for all n ≥ 1.

Proof. 0 → F → I → G → 0 is a short exact sequence, with I injective andG flasque. So we get a long exact sequence 0 → F (X) → I(X) → G (X) →H1(X,F ) → . . .. But Hi(X, I) = 0, so we get H1(X,F ) = 0, H2(X,F ) =H1(X,G ) = 0, etc.

Corollary 8.18. If 0→M → F 0 → . . . is a flasque resolution, then Hn(X,M ) =Hn(Γ(X,F ∗)).

Proposition 8.19. (X,OX) a ringed space. The right derived functos fo Γ :Mod(X)→Mod(OX(X)) are given as M 7→ Hn(X,M ).

Proof. Injective resolution in Mod(X) 0→M → I0 → . . ., we know that In isflasque for all n, so the corollary implies this.

Definition 8.14 (Direct Limit Sheaf). lim−→Fα = [U 7→ lim−→Fα(U)]+.

Lemma 8.20. X Notherian implies that Γ(U, lim−→Fα) = lim−→Γ(U,Fα).

Proof. Check sheaf axioms using finite open covers.

Lemma 8.21. X Notherian and Fα flasque, then lim−→Fα is flasque.

Proof. V ⊆ U ⊆ X open, Fα(U)→ Fα(V ) surjective implies that lim−→Fα(U)→lim−→Fα(V ) is surjective.

Proposition 8.22. X Notherian, Fα directed system, then Hn(X, lim−→Fα) =lim−→Hn(X,Fα).

Proof. Let I = indA(Ab(X)) be the category of all directed systems of abeliansheaves.

Tn : I → Ab by Tn(Fα) = lim−→Hn(X,Fα) and Un : I → Ab byUn(Fα) = Hn(X, lim−→Fα).

Note that lim−→ is exact, so Tn and Un form δ-functors.

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And T 0(Fα) = lim−→Γ(X,Fα) = Γ(X, lim−→Fα) = U0(Fα).By theorem from before, ETS that Tn and Un are universal δ-functors,

because they agree on the 0th term.Given F ∈ Ab(X), set F (U) =

∏p∈U Fp, the sheaf of discontinuous sec-

tions.0→ F → F exact and F is flasque.Fα ∈ ob(I), and Fα ∈ ob(I), and we have 0→ Fα → Fα.So now sett that Tn(Fα) = lim−→Hn(X, Fα) = 0 and simialry for Un.

Lemma 8.23. Y ⊆ X a closed subspace, j : Y → X the inclusion, and F anabelian sheaf on Y . Then Hn(Y,F ) = Hn(X, j∗F ).

Proof. 0 → F → I0 → I1 → . . . an injective resolution on Y . In is flasque, soj∗(In) is flasque. So 0 → j∗F → j∗I

0 → . . . is a flasque resolution (exactnessfollows from Y being closed).

So Hn(Y,F ) = Hn(Γ(Y, I∗)) = Hn(Γ(X, j∗I∗)) = Hn(X, j∗F )

Remark: Y ⊆ X closed, U = X \Y . Then F an abelian sheaf on X, we havej : Y → X and i : U → X inclusions. Then we can set FU = i!(F |U ) = 0 unlessV ⊆ U , and if V ⊆ U , then F (V ) and also FY = j∗(F |Y ) where FY (U) = 0if V ⊆ U and equals j−1F (Y ∩ V ) if V 6⊆ U . Then FY,P = FP if P ∈ Y andzero else, as FU = Fp iff p ∈ U and is 0 otherwise.

So we have 0→ FU → F → FY → 0 is exact.

Theorem 8.24. Let X be a Notherian topological space, and F an abeliansheaf on X. Then Hn(X,F ) = 0 for all n > dim(X).

Proof. Step 1: Reduce to X irreducible: If X is reducible, let Y be a maximalcomponent in X. Take U = X \Y , then we have the above short exact sequence.This gives us a long exact sequence on cohomology, Hn(X,FU )→ Hn(X,F )→Hn(X,FY ) = Hn(Y,F |Y ). If the theorem is true for irreducible spaces, thenthe last one is zero, and so we have isomorphisms Hn(X,FU )→ Hn(X,F ), soby induction, it is enough to show that the theorem holds when X is irreducible.

Step 2: dimX = 0, X irreducible implies that X has open sets ∅, X andnothing else. Then Ab(X) = Ab, so Γ : Ab(X)→ Ab is the identity, and so it isexact, so Hn(X,F ) = 0 for n ≥ 1.

Step 3: Assume that X is irreducible of dimension > 0. If σ ∈ F (U), thenget ZU → F by 1 7→ σ. Let α = σ1, . . . , σm with σi ∈ F (Ui). DefineFα = Im(⊕ZUi → F ) the subsheaf generated by α. Set B =

∐U⊆X F (U).

A = α ⊆ B|α finite. A is a directed poset, so Fα is a directed system.lim−→Fα = F . And ten, as Hn(X,F ) = Hn(X, lim−→Fα) = lim−→Hn(X,Fα), itsuffices to prove this for F = Fα finitely generated.

If α′ ⊆ α, then we have 0→ Fα′ → Fα → G → 0, with G generated by thenumber of sections in α \ α′. So WLOG, F is generated by one section. So wehave 0 → K → ZU → F → 0 exact. Thus, it is enough to show vanishing forF ⊆ ZU a subsheaf.

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For p ∈ U , Fp ⊆ ZU,p = Z. Define d = mine ∈ Z+|Fp = eZ for some p.Choose p such that Fp = dZ, d = σp, σ ∈ F (V ) ⊆ ZU (V ), p ∈ V . Must haveF |V = dZU |V = dZV .

So we have 0 → ZV·d→ F → F/ZV → 0. Supp(F/ZV ) ⊆ U \ V , so

dimU \ V < dim(X). So induction on dimension gives us that Hn(X,F/ZV ) =0 for n > dim(X).

Set Y = X \ U , 0 → ZU → ZX → ZY → 0 is ses, dimY < dimX, soHn(X,ZY ) = 0 for n ≥ dim(X). Hn−1(ZY ) → Hn(ZU ) → Hn(ZX). SoWLOG, F = ZX .

If X is irreducible, then ZX is flasque. So Hn(X,ZX) = 0 for all n ≥ 1.

NOTE: Above, FU is not a sheaf, to correct things, sheafify

Lemma 8.25. Z ⊆ X closed, G a sheaf on Z, then Hn(Z,G ) = Hn(X, j∗G ).

Note: Supp(FU ) = U ⊆ U , then FU = j∗(FU |U ), so Hn(X,FU ) =Hn(U ,FU |U ).

Next: IfX = Spec(A) and F is a quasi-coherent OX -module, thenHn(X,F ) =0 for n ≥ 1.

Exercise: If M is an A-module, assume homA(A,M) → homA(I,M) issurjective for all ideals I ⊆ A. Then M is injective. Hint: N ′ ⊆ N , ϕ : N ′ →M .

Let M ′ ⊆M an A-module, I ⊆ A an ideal, Ip(M ′∩InM) ⊆M ′∩In+pM forall p, n ≥ 0. The Artin-Rees lemma says that if A is Notherian and M finiteygenerated, then for some n > 0 we have equality for all p ≥ 0.

Definition 8.15. ΓI(M) = m ∈M |Inm = 0 for some n > 0.

Lemma 8.26. A Notherian, M injective implies that ΓI(M) is injective.

Proof. Let J ⊆ A be an ideal. ϕ : J → ΓI(M) an A-homomorphism. If a ∈ J ,then Inϕ(a) = 0.

J finitely generated implies that ∃p > 0 such that ϕ(IpJ) = 0. By Artin-Rees, there exists n > 0 such that Ip(J ∩ InA) = J ∩ Ip+nA.

So we have ϕ(J ∩ In+p) = 0. So the map from J → A gives a map fromJ/J ∩ In+p → A/In+p. As we have J/J ∩ In+p → ΓI(M) ⊆ M , and Minjective, we get a map A/In+p → M . The image is contained in ΓI(M), asIn+pψ(A/In+p) = 0, and so the inclusion really extends to a map to ΓI(M).

Lemma 8.27. M an injective A-module, A Notherian, f ∈ A. Then θ : M →Mf is surjective.

Proof. Ann(f) ⊆ Ann(f2) ⊆ . . . ⊆ Ann(fr) = Ann(fr+1) = . . .. Let x ∈ Mf ,then x = θ(y)/fn for some y ∈ M . So we have a ses 0 → Ann(fr) → A →(fn+r)→ 0. Then A→M by fry and (fn+r) ⊆ A, so we get a map (fn+r)→M by fn+r 7→ fry. This map can be extended to A, call the image of 1 in thismap z. Then fn+rz = fry, then θ(z) = θ(y)/fn = x.

F a sheaf on X, σ ∈ F (U). Supp(σ) = p ∈ U |σp 6= 0 ∈ Fp ⊆ U relativelyclosed. For Z ⊆ X closed, ΓZ(U,F ) = σ ∈ F (U)|Supp(σ) ⊆ Z

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Definition 8.16. Sheaf H 0Z (F ) by Γ(U,H 0

Z (F )) = ΓZ(U,F ).

Let j : X \ Z → X, then 0 → H 0Z (F ) → F → j∗(F |X\Z), which becomes

short exact if F is flasque.Assume X = Spec(A) and F = M . For m ∈ M = Γ(X, M), then

Supp(m) = V (Ann(m)), p ∈ Supp(m) ⇐⇒ m/1 6= 0 ∈ Mp ⇐⇒ Ann(m) ⊆P .

Lemma 8.28. X = Spec(A), Notherian, Z = V (I) ⊆ X. Then H 0Z (M) =

ΓI(M)

Proof. 0→H 0Z (F )→ F → j∗(F |U\Z) tells us that H 0

Z is the kernel of a mapof quasi-coherent sheaves.

Enough to show that Γ(X,H 0Z (M)) = ΓI(M), m ∈ ΓZ(X, M) ⇐⇒ Supp(m) ⊆

Z iff V (Ann(m)) ⊆ V (I), iff√

Ann(m) ⊇√I iff In ⊆ Ann(m) (by Notherian

property), and this is iff Inm = 0, iff m ∈ ΓI(M).

Proposition 8.29. A a Notherian ring, X = Spec(A). Then M an injectiveA-module implies that M is a flasque OX-module.

Proof. Notherian induction on Y = Supp M . On Y = point, clear.Let U ⊆ X open. Show Γ(X, M)→ Γ(U, M) surjective. WLOG, Y ∩U 6= ∅.

Choose f ∈ A such that Xf ⊆ U and Xf ∩Y 6= ∅. Set Z = V (f) = X \Xf . SetI = (f) ⊆ A.

We know that ΓI(M) is injective, and Supp( ˜ΓI(M)) ⊆ Y ∩Z. (fnm = 0 forall m ∈ ΓI(M), so f /∈ P implies ΓI(M)P = 0)

Induction implies that ΓI(M) is flasque, and M = Γ(X, M) → Γ(U, M) →Γ(Xf , M) = Mf gives us Γ(X, ˜ΓI(M)) = ΓZ(X, M)→ ΓZ(U, M) = Γ(U, ˜ΓI(M)).Recall that M →Mf is surjective.

Let σ ∈ Γ(U, M). Then ∃τ ′ ∈ Γ(X, M) with the same image in Γ(Xf , M).Set τ = τ ′|U . Then (σ − τ)|Xf = 0, so σ − τ ∈ ΓZ(U, M).

So ∃α′ ∈ ΓZ(X, M) with α′ 7→ σ− τ . Therefore, α′+ τ ′ ∈ Γ(X, M) maps toσ ∈ Γ(U, M).

Theorem 8.30. X affine Notherian scheme, F a quasi-coherent OX-module,then Hn(X,F ) = 0 for all n > 0.

Proof. X = Spec(A), F = M . A resolution of M by injective A modules givesus a resolution of F by flasque OX -modules.

The global section functor gives back the original sequence, so Hn(X, M) =Hn(I ·) which is M if n = 0 and 0 otherwise.

Corollary 8.31. X Notherian scheme, F a quasi-coherent OX-module, thenthere exists a monomorphism 0 → F → G where G is a flasque quasicoherentOX-module.

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Proof. X = U1 ∪ U2 ∪ . . . ∪ Un where Ui = SpecAi where Ai is Notherian.F |Ui = Mi, Mi an Ai-module. There exists Mi ⊆ Ii for some injective Ii.

FUi → Ii injective, so F → j∗(Ii) is injective over Ui. As Ii is injective, Iiis flasque, and j∗(Ii) is flasque.

0→ F → ⊕ni=1j∗(Ii) is flasque and quasicoherent.

Exercise: X be a scheme, f1, . . . , fr ∈ Γ(X,OX), then if Xfi is affine for alli, and (f1, . . . , fr) = (1) ⊆ Γ(X,OX), then X is affine.

The ideas is that A = Γ(X,OX), and we have ϕ : X → Spec(A). Then showthat Γ(Xfi ,OX) = Afi , and so we get isomorphisms ϕ : Xfi → Spec(Afi). SoX = ∪Xfi and so ϕ is a global isomorphism.

Theorem 8.32 (Serre’s Criterion). Let X be a Notherian scheme. TFAE

1. X is affine

2. Hn(X,F ) = 0 for all quasicoherent OX-modules F and n > 0.

3. H1(X,I ) = 0 for all coherent sheaves of ideals I ⊆ OX .

Proof. 1 implies 2 implies 3 are done already.Assume 3. Let P ∈ X be a closed point. Let P ∈ U be an open affine

neighborhood of P . Let Y = X \ U . We get 0 → IY ∪P → IY → k(P ) → 0,Γ(X,IY )→ Γ(X, k(P ))→ H1(X,IY ∪P ) = 0.∃f ∈ Γ(X,IY ) ⊆ A = Γ(X,OX) such that Y ⊆ V (f) and f 7→ 1 ∈ k(P ),

so P ∈ Xf ⊆ U , Xf = Uf is affine. Choose f1, . . . , fr ∈ A such that X =Xf1 ∪ . . . ∪Xfr and Xfi is affine. It remains to show that (f1, . . . , fr) = (1).

Claim: F ⊆ OrX is a coherent subsheaf implies that H1(X,F ) = 0. For

r = 1 this is three. For r > 1, we have 0 → Or−1X → Or

X → OX → 0 givingus 0 → F ∩ Or−1

X → F → I → 0, and so H1(I ) = 0 and by induction,H1(X,F ∩ Or−1

X ) = 0, so the claim is proved.Take Or

X → OX → 0 by (g1, . . . , gr) 7→∑gifi. This map is surjective, and

so we take the kernel and call it F , and get a short exact sequence.Γ(X,Or

X)→ Γ(X,OX)→ H1(X,F ) = 0 exact so (f1, . . . , fr) = (1).

Cech CohomologyLetX be a topological space andX = ∪i∈IUi. U = (Ui)i∈I , for i0, . . . , ip ∈ I,

set Ui0,...,ip = Ui0 ∩ . . . ∩ Uip . Let F be an abelian sheaf on X.

Definition 8.17 (Alternating Function). A function α : Ip+1 → sections ofF is called alternating if α(i0, . . . , ip) ∈ F (Ui0,...,ip) and if α(i0, . . . , it, it+1, . . . , ip) =−α(i0, . . . , it+1, it, . . . , tp) and if α(i1, i1, i2, . . . , ip) = 0.

Definition 8.18 (Cech Complex). Cp(U ,F ) = alternating α : Ip+1 → sec-tions of F and maps d : Cp → Cp+1 by dα(i0, . . . , ip) =

∑p+1k=0(−1)kα(i0, . . . , ik, . . . , ip+1)|Ui0,...,ip+1

.

Check: d2α = 0.Then we have Hp(U ,F ) = Hp(C∗(U ,F )).

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Lemma 8.33. H0(U ,F ) = Γ(X,F ).

Proof. C0(U ,F ) =∏i∈I F (Ui). If α ∈ C0(U ,F ), then dα(i, j) = α(j)|Ui,j −

α(i)|Ui,j , so dα = 0 iff the sections are all compatible which is true iff ∃σ ∈ F (X)such that σ|Ui = α(i).

Therefore, H0(U ,F ) = ker(C0 → C1) = F (X).

Example: X = P1 = Proj k[x, y]. F = ΩP1 = Ω. U = U, V with U =D+(x) = Spec k[t], t = y/x and V = D+(y) = k[t−1]. U ∩ V = Spec k[t, t−1].Then we denote Cp = Cp(U ,Ω).

C0 = Γ(U,Ω) ⊕ Γ(V,Ω) = k[t]dt ⊕ k[t−1]d(t−1), C1 = Γ(U ∩ V,Ω) =k[t, t−1]dt. C2 = 0.

d : C0 → C1 by dt 7→ −dt and d(t−1) 7→ −t−2dt. f(t)dt ⊕ g(t−1)d(t−1) ∈ker d iff −f(t)− t−2g(t−1) = 0 iff f = g = 0, so H0(U ,Ω) = Ω(X) = 0.

Im(d) = (−f(t)− t−2g(t−1))dt ⊆ k[t, t−1]dt. This is ∑ait

i|a−1 = 0, soH1(U ,Ω) ' k generated by t−1dt.

Let X = ∪i∈IUi, and U = (Ui)i∈I . Then we have defined Cp(U ,F ).

Definition 8.19. Define an abelian sheaf C p(U ,F ) = C p given by Γ(V,C p(U ,F )) =Cp(U , j∗(F |V )) where j is the inclusion V → X.

0 → Fε→ C 0 d→ C 1 → . . .. σ ∈ F (V ), then εσ ∈ C0(U , j∗F |V ) and

εσ(i) = σ|Ui∩V .

Lemma 8.34. This is an exact sequence of sheaves.

Proof. Exact at p = 0. 0 → Γ(X, j∗(F |V )) → C0 → C1 → . . .. Let p ≥ 1.x ∈ X, choose k ∈ I with x ∈ Uk. Let αx ∈ C p

x . Then α ∈ C p(V ) for x ∈ V .WLOG, V ⊆ Uk.

Define hα ∈ Cp−1 by hα(i0, . . . , ip−1) = α(k, i0, . . . , ip−1) ∈ F (Uk,i0,...,ip−1∩V ) = F (Ui0,...,ip ∩ V ).

Now define h : C px → C p−1

x by αx 7→ (hα)x. Check that (dh+ hd)α = α, sodh+ hd = id− 0 so id ∼ 0 and so Hp(C ∗x ) = 0, and so C ∗ is exact.

Proposition 8.35. If F is flasque, then Hp(U ,F ) = 0 for p > 0.

Proof. Let C p = C p(U ,F ). This is flasque for all p. So 0 → F → C 0 →C 1 → . . . is an exact sequence of flasque sheaves. Taking the global sectionfuncto we get Cp, but it is still exact because it is a flasque sequence. SoHp(U ,F ) = Hp(C∗) = 0.

Note: F is an abelian sheaf, 0 → F → I 0 → . . . an inejctive resolution,then we get a map Hp(U ,F )→ Hp(X,F ).

Theorem 8.36. If X is Notherian separated scheme, and U = (Ui)i∈I is anopen affine covering, and F is quasi-coherent OX-module, then Hp(U ,F ) =Hp(X,F ).

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Proof. 0→ F → G → R → 0 with G flasque and quasicoherent, then R is qua-sicoherent. If X is seprated, then Ui0,...,ip is affine, so we get 0→ F (Ui0,...,ip)→G (Ui0,...,ip)→ R(Ui0,...,ip)→ 0.

So we get the long exact sequence 0 → H0(F ) → H0(G ) → H0(R) →H1(F )→ 0 and an exact sequence H0(F )→ H0(G )→ H0(R)→ H1(F )→ 0,and so taking the vertical maps, for p = 1, this proves it.

We can proceed by induction obtaining 0 → Hp(R) → Hp+1(F ) → 0 and0 → Hp(R) → Hp(F ) → 0 and looking at the vertical maps, whcih must beisomorphisms.

Let A be a Notherian ring, S = A[x0, . . . , xr], and X = ProjS = PrA.Then let F be any OX module. Recall that Γ∗(F ) = ⊕n∈ZΓ(X,F (n)).Note: f ∈ Sm = Γ(X,O(m)) gives f : F (n)→ F (n+m) by σ 7→ σ ⊗ f .Thus, f : Hp(X,F (n)) → Hp(X,F (n + m)), so Γ∗(F ) is a graded S-

module.

Theorem 8.37. 1. S → Γ∗(OX) is an isomorphism of graded S-modules.

2. Hp(X,OX(n)) = 0 for 0 < p < r any n.

3. Hr(X,OX(−r − 1)) ' A

4. Perfect Pairing: H0(X,OX(n))×Hr(X,OX(−n−r−1))→ Hr(X,OX(−r−1)) = A.

Proof. F = ⊕n∈ZOX(n) is quasicoherent. ThenHp(X,F ) = ⊕n∈ZHp(X,OX(n)).

Let U = U0, . . . , Ur with Ui = D+(xi) ⊆ X. Then Ui0,...,ip = D+(xi0 . . . xip),so we look at Γ(Ui0,...,ip ,OX(n)) = S(n)(xi0 ...xip ), and so Γ(Ui0,...,ip ,F ) =Sxi0 ...xip .

C∗(U ,F ): ⊕i0Sxi0 → ⊕i0<i1Sxi0xi1 → . . .→ ⊕rk=0Sx0...xk...xr → Sx0...xr

1. ker(first map) = ∩Sxi = S.

2. Cp(U ,F )xr =∏i0<...<ip

Sxi0 ...xipxir =∏i0<...<ip

⊕n∈ZΓ(Ui0,...,ip∩Ur,OX(n)).

This equals∏

Γ(Ui0,...,ip ∩ Ur,F ) = Cp(U ′,F |Ur ) where U ′ = Ui ∩ Uris an affine open covering of Ur. Thus, Hp(U ,F )xr = Hp(U ′,F |Ur ) =Hp(Ur,F |Ur ) = 0. So every element of Hp(X,F ) is killed by xNr . It isenough to show that Hp(X,F ) xr→ Hp(X,F ) is an isomorphism. WLOG,r ≥ 2.

H = V (xr) ⊆ X, H ' Pr−1A . We have 0 → OX(−1) → OX → OH → 0.

As F is locally free, tensoring with it is flat, so we get 0→ F (−1)→ F →FH → 0, and FH = ⊕n∈ZOH(n), with the first map being multiplicationby xr. By induction on r, we get that Hp(X,FH) = Hp(H,FH) = 0 for0 < p < r − 1.

So we get a long exact sequenceH0(X,F )→ H0(X,FH)→ H1(X,F (−1))→H1(X,F ) → H1(X,FH), so we get that H1(X,F ) ' H1(X,F (−1)).

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For 1 < p < r − 1, we have Hp−1(X,FH) = 0 → Hp(X,F (−1)) 'Hp(X,F )→ Hp(X,FH) = 0.

So we now have Hr−2(X,FH) 0→ Hr−1(X,F (−1)) xr→ Hr−1(X,F ) →Hr−1(X,FH) δ→ Hr(X,F (−1)) xr→ Hr(X,F ) → 0. Note that if r = 2,we don’t get Hr−2(X,FH) = 0, but the first map is zero.

Ann(xr) ⊆ Hr(X,F (−1)) has basis x`00 . . . x`r−1r−1 x

−1r |`i < 0 andHr−1(X,FH)

has basis x`00 . . . x`r−1r−1 |`i < 0. T his tells us that δ is injective, and so

the map before it is zero and we have an isomorphism Hr−1(X,F (−1)) =Hr−1(X,F ).

3. Image(last map)= spanAx`00 . . . x`rr |`i ∈ Z with some `i ≥ 0, so we get

Hr(X,F ) = spanAx`00 . . . x`rr |`i < 0∀i. And so Hr(X,OX(−r − 1)) =

A(x0x1 . . . xr)−1 ' A.

4. Hr(X,OX(−n−r−1)) = spanAx`00 . . . x`rr |

∑`i = −n−r−1, `i < 0. So

look at H0(X,OX(n))×Hr(X,OX(−n− r − 1))→ Hr(X,OX(−r − 1)).map (xm0

0 . . . xmrr )× (x`00 . . . x`rr ) 7→ x`0+m00 . . . x`r+mr

r .

The right hand side 6= 0 ∈ Hr(X,OX(−r − 1)) iff mi + `i = −1 for all i,so `i = −mi − 1 for all i.

Theorem 8.38. X a projective scheme over SpecA, A Notherian and OX(1) isa very ample invertible sheaf relative to Spec(A). If F is a coherent OX-module,then

1. Hp(X,F ) is a finitely generated A-module for all p ≥ 0.

2. Hp(X,F (n)) = 0 for all p > 0 for n >> 0.

Proof. Let f : X → Pr, OX(1) = f∗OPr (1). f∗F is coherent on Pr. f∗(F (n)) =f∗(f∗OPr (n)⊗F ) = OPr (n)⊗ f∗F = (f∗F )(n).

Hp(X,F (n)) = Hp(Pr, f∗F (n)) = Hp(Pr, (f∗F )(n)).WLOG, X = Pr. Note: Theorem is true for F = OPr (n).The theorem is true for p > r, as all the cohomology vanishes. We proceed

by decreasing induction on p. There exists E = ⊕OPr (qi) with finitely manyterms with 0 → R → E → F → 0 exact, so 0 → R(n) → E (n) → F (n) → 0exact.

Thus, Hp(Pr,E (n)) → Hp(Pr,F (n)) → Hp+1(Pr,R(n)). The outer mod-ules are finitely generated by the previous theorem and inductions. As A isNotherian, the middle module is finitely generated.

If the outer modules are zero, then Hp(Pr,F (n)) = 0 for all n >> 0,p > 0.

Application: ifX is a nonsingular curve, defined genus to be g = dimk Γ(X,ΩX) <∞. With Serre duality (to be shown), we see that g = dimkH

1(X,OX)

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Euler CharacteristicLet X be a projective scheme over k. Let F be a coherent OX -module and

dim(X) = r.

Definition 8.20 (Euler Characteristic). The Euler characteristic of F is χ(F ) =∑p≥0(−1)p dimkH

p(X,F ).

Then 0→ F ′ → F → F ′′ → 0 gives a long exact sequence 0→ H0(X,F ′)→H0(X,F )→ . . .→ Hr(X,F )→ Hr(X,F ′′)→ 0 an exact sequence of k-vectorspaces.

So χ(F ) = χ(F ′)+χ(F ′′). In general, if we have 0→ F 0 → . . .→ F ` = 0,then

∑(−1)iχ(F i) = 0.

Proposition 8.39. X a proper scheme over Notherian Spec(A). L an in-vertible OX-module. Then L is ample iff ∀ coherent OX-modules F , we haveHp(X,F ⊗L n) = 0 for all p > 0, n >> 0.

Proof. ⇒: L ample, then L ⊗m is very ample relative to SpecA for somem > 0. Thus, there exists an immersion X ⊆ X ′ ⊆ PrA with L ⊗m = OX(1).X is proper, so X = X ′ is closed in PrA. Choose n0 > 0 such that Hp(X,F ⊗L i ⊗ OX(n)) = 0 for all p > 0 and for 0 ≤ i < m and n ≥ n0. Thus,Hp(X,F ⊗L n) = 0 for n ≥ m+mn0.⇐: F a coherent OX -module. Show: F ⊗L n generated by global sections

for all n >> 0. Let p ∈ X be a closed point. Then we get 0 → Ip → OX →k(P ) → 0 ses, so we have Ip ⊗F → F → F ⊗ k(p) → 0, and to get a shortexact sequence, we take 0→ IpF → F → F ⊗ k(P )→ 0.

So then we have 0→ IpF ⊗L n → F ⊗L n → F ⊗L n ⊗ k(p)→ 0. Forn ≥ n0 = n0(p), we get Γ(X,F⊗L n)→ Γ(X,F⊗L n⊗k(P ))→ H1(X,IpF⊗L n) = 0, so the second term is (F ⊗L n)p ⊗ k(p). By Nakayama, (F ⊗L n)pis generated by global sections from Γ(X,F ⊗L n). p ∈ U = Spec(B) ⊆ X, BNotherian.

So M = Γ(U,F ⊗ L n) is a finitely generated B-module. M ′ ⊆ M is thesubmodule generated by the imeage of Γ(X,F ⊗L n). Mp = Fp = M ′p. Thenthere exists f ∈ B \ p suc that M ′f = Mf .

Set Up,n = D(f) ⊆ U , then p ∈ Up,n and (F ⊗ L n)|Up,n is generated byglobal sections.

Special case: may choose n1 and open Vp 3 p such that L n1 |Vp is generatedby global sections. Set Up = Vp ∩ Up,n0 ∩ . . . ∩ Up,n0+n1−1. Then if n ≥ n0(p),we have F ⊗L n = (F ⊗L n′)⊗ (L n1)m where n0 ≤ n′ < n0 +n1 with m ≥ 0.Then (F ⊗L n)|Up is generated by global sections from Γ(X,F ⊗L n).

So X = Up1 ∪ Up2 ∪ . . . ∪ Up` with n ≥ max(n0(p1), . . . , n0(p`)), and we getF ⊗L n generated by global sections.

Applications: A Notherian, X = PrA = Proj(S) for S = A[x0, . . . , xr]. LetM be a finitely generated graded S-module. We have maps Mn → Γ(Pr, M(n)),a homomorphism of graded S-modules ϕ : M → Γ∗(M) = ⊕n∈ZΓ(M(n)).

Claim: Mn ' Γ(Pr, M(n)) for all n >> 0

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Injective: N = kerϕ ⊆ M . Show that Nn = 0 for n >> 0. As N is finitelygenerated, it is enough to show that ∀m ∈ N and 0 ≤ i ≤ r, there exists p > 0so that xpim = 0. This is true because m = 0 in Γ(D+(xi), M(n)) = M(n)(xi) ⊆Mxi .

Surjective: 0 → F ′ → F → M → 0, F = ⊕S(qi) finite sum. So then0 → F ′(n) → F (n) → M(n) → 0, and so for n >> 0, we get Γ(Pr, F (n)) →Γ(Pr, M(n))→ H1(Pr, F ′(n)) = 0, so we have a surjection, but this is really anonto map Fn →Mn, and must also give a surejction Mn → Γ(Pr, M(n)).

X is projective over a field, k. X ⊆ Prk. Then there is OX(1) very ample.F is a coherent OX -module.

Definition 8.21 (Hilbert Polynomial). We define the hilbert polynomial to bePF (n) = χ(F (n)). In particular, PX(n) = POX (n).

Claim: PF (n) ∈ Q[n].

Proof. Notherian induction on Y = Supp F . Y = ∅ ⇒ F = 0⇒ PF (n) = 0.

If Y 6= ∅, take f ∈ Γ(X,OX(1)) such that Y ∩Xf 6= 0. 0→ R → F (−1)f ·→

F →M → 0 is an exact sequence of coherent OX -modules.Taking the Euler characteristics at n, we get PF (n)−PF (n−1) = PM (n)−

PR(n). The closures of Supp R and Supp M ⊆ Y ∩ V (f) ( Y . By Notherianinduction, the functions on the right hand side are polynomials in Q[n], andtherefore PF (n) ∈ Q[n].

Note Pr = ProjS, S = k[x0, . . . , xr], M is a finitely generated graded S-module. If n >> 0, then Mn = Γ(Pr, M(n)) = χ(M(n)).

Note: PF (n) depends on OX(1) on OX(1), but PF (0) = χ(F ). AssumedimX = r. Then the arithmetic genus if pa(X) = (−1)r(χ(OX) − 1) =(−1)r(PX(0) − 1), so if X is a connected curve, then pa(X) = 1 − χ(OX) =1 − dimkH

0(X,OX) + dimk(H1(X,OX)). So then, as X is projective, this isdimkH

1(X,OX).Ext Groups and Sheaves(X,OX) a ringed space, F ,G are OX -modules. Remember that hom(F ,−)

is a functor from Mod(X)→ Ab is left exact.

Definition 8.22 (Ext). Extp(F ,−) = Rp hom(F ,−) from Mod(X)→ Ab.E xtp(F ,−) = RpH om(F ,−).

Proposition 8.40. 1. E xt0(OX ,G ) = G , E xtp(OX ,G ) = 0 for p > 0.

2. Extp(OX ,G ) = Hp(X,G )

Lemma 8.41. I ∈ Mod(X) injective, U ⊆ X open, then I |U ∈ Mod(U)injective.

Proof. j : U → X inclusion.

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j∗(I |U )

j∗F

I

j!G.............................................................................................................................

........................................................................................ ............⊂

............................................................................................................ ............

.............................................................................................................................

Proposition 8.42. U ⊆ X open. Then E xtpX(F ,G )|U = E xtpU (F |U ,G |U ).

Note: 0→ G ′ → G → G ′′ → 0 gives a long exact sequence on Ext(F ,−)

Proposition 8.43. 0 → F ′ → F → F ′′ → 0 gives a long exact sequence0→ hom(F ′′,G )→ hom(F ,G )→ hom(F ′′,G )→ Ext1(F ′′,G )→ . . ..

Proof. 0 → G → I 0 → . . .. Then as hom(−,I P ) is exact and contravariant,we get 0→ hom(F ′′,I ∗)→ hom(F ,I ∗)→ hom(F ′,I ∗)→ 0, and so we geta long exact sequence in the first variable.

Definition 8.23 (Locally Free Resolution). A locally free resolution of F is aresolution . . . → E2 → E1 → E0 → F → 0 exact with Ep locally free of finiterank.

Proposition 8.44. If E∗ → F → 0 is locally free resolution, then E xtp(F ,G ) =Hp(H om(E∗,G )).

Proof. Both sides are δ-functors. RHS: 0 → G ′ → G → G ′′ → 0, Ep is locallyfree implies that H om(Ep,−) is exact, and so we get 0 → H om(E∗,G ′′) →H om(E∗,G )→H om(E∗,G ′)→ 0, so get long exact sequence.

Agree for p = 0: E1 → E0 → F → 0, so 0→H om(F ,G )→H om(E0,G )→H om(E1,G ) → . . . is exact, and so H0(H om(E∗,G )) = H om(F ,H ) =E xt0(F ,G ).

Universality: Both sides vanish when G is injective.

Note: E is locally free OX -module of finite rank, then E ∨ = H om(E ,OX)islocally free.

Exercise: H om(F ,G ) ⊗ E ' H om(F ,E ⊗ G ) ' H om(F ⊗ E ∨,G ) byϕ⊗ e 7→ [f 7→ e⊗ ϕ(f)] and ψ 7→ [f ⊗ e∨ 7→ (e∨ ⊗ 1)(ψ(f))]

Special Case: E = H om(O,E ) = H om(E ∨,OX) = E ∨∨.

Lemma 8.45. I is an injective OX-module, E is a locally free OX-module,then I ⊗ E is injective.

Proof. hom(−,I ⊗ E ) = hom(−⊗ E ∨,I ) is exact.

Proposition 8.46. E locally free of finite rank, then Extp(F ,E⊗G ) = Extp(F⊗E ∨,G ) and E xtp(F ,G )⊗ E = E xtp(F ,E ⊗ G ) = E xtp(F ⊗ E ∨,G ).

Proof. Everything is a δ-functor. They are the same for p = 0.They vanish for G injective.

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Note: If A is a ring, M an A-module, then ExtpA(M,−) = Rp homA(M,−).(pt, A) is a ringed space, and modules on it and A-mod are the same thing.If E is locally free of finite rank, then H om(E ,F )x = homOX,x(Ex,Fx).

Proposition 8.47. X a Notherian Scheme, F a coherent OX-module and Gany OX-module. Then E xtp(F ,G )x = ExtpOX,x(Fx,Gx) for all x ∈ X, p ≥ 0.

Proof. WLOG, X is affine. There exists a locally free resolution E∗ → F → 0,so there exists a free resolution of OX,x-modules E∗,x → Fx → 0.

E xtp(F ,G )x = Hp(H om(E∗,G ))x = Hp(H om(E∗,G )x) = Hp(hom(E∗,x,Gx)) =ExtpOX,x(Fx,Gx).

Proposition 8.48. X a projective scheme over Notherian Spec(A) with OX(1)is very ample relative to A. F ,G coherent OX-modules, p ≥ 0. Then Extp(F ,G (n)) =Γ(X,E xtp(F ,G (n))) for all n >> 0.

Proof. p = 0: hom(F ,G (n)) = Γ(X,H om(F ,G (n))) is true for all n.F is lcoally free of finite rank. Extp(F ,G (n)) = Extp(OX ,F∨ ⊗ G (n)) =

Hp(X,F∨ ⊗ G (n)) = 0. E xtp(F ,G (n)) = E xtp(OX ,F∨ ⊗ G (n)) = 0 for allp > 0 for all n, so it is ture for all n >> 0.

F coherent: 0 → R → E → F → 0 where E is locally free. 0 →H om(F ,G (n)) → H om(E ,G (n)) → H om(R,G (n)) → E xt1(F ,G (n)) →E xt1(E ,G (n)) = 0 and we can take the twisting outside.

So for n >> 0, we have 0→ hom(F ,G (n))→ hom(E ,G (n))→ hom(R,G (n))→Γ(X,E xt1(F ,G (n)))→ 0.

We also get 0 → hom(F ,G (n)) → hom(E ,G (n)) → hom(R,G (n)) →Ext1(F ,G (n)) → Ext1(E ,G (n)) = 0 for n >> 0. As the first three are thesame, the last one must be.

For p ≥ 2, we have long exact 0→ Extp−1(R,G (n))→ Extp(F ,G (n))→ 0for n >> 0. Also, for all n, 0→ E xtp−1(R,G (n))→ E xtp(F ,G (n))→ 0.

By induction, we get vertical maps, which prove the theorem.

Let A be a ring, M an A-module,∧p

M = M ⊗ . . .⊗M/ < . . . x⊗ x . . . >.0→ M → N → R→ 0 a ses of free modules of ranks m,n, r, then

∧nN '∧m

M ⊗A∧r

R is a natural isomorphism. This is important for sheafification.Let F be an OX -module,

∧p F = (U 7→∧p F (U))+.

Then if 0 → M → N → R → 0 a short exact sequence of locally freeOX -modules of ranks m,n, r, then

∧n N '∧m M ⊗

∧r R.Let X be a nonsingular variety over a field k. Then ΩX is locally free of

rank n.

Definition 8.24 (Canonical Sheaf of X). The canonical sheaf on X is ωX =∧n ΩX .

On Pnk , we have 0 → ΩPn → O(−1)⊕n+1 → OPn → 0. So ωPn =∧

ΩPn =∧n ΩPn ⊗∧1 OPn =

∧n+1(O(−1)⊕n+1) = O(−1)⊗n+1 = O(−n− 1).

Theorem 8.49 (Serre Duality, Version I). X = Pnk , k a field.

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1. Hn(X,ωX) ' k.

2. F coherent OX-module implies that hom(F , ωX)×Hn(X,F )→ Hn(X,ωX) =k is a perfect pairing.

3. Extp(F , ωX) ' Hn−p(X,F )∗, a natural isomorphism of functors in Ffrom Coherent Sheaves on X to k-modules.

(If V is a k-vector space, V ∗ = homk(V,Hn(X,ωX)))

Proof. 1. Hn(X,ωX) = Hn(X,O(−n− 1)) ' k.

2. Let F = O(q). Then hom(F , ωX)×Hn(X,F ) = Γ(X,O(−n− 1− q))×Hn(X,O(q)) → Hn(X,O(−n − 1)) = k is a perfect pairing. So nowconsider the case where F is coherent. Then ∃E1 → E0 → F → 0 exact,Ei = ⊕O(qij) a finite sum. Note that Hn(X,−) is right exact (end of thelong exact sequence). So we get Hn(X,E1)→ Hn(X,E0)→ Hn(X,F )→0, which gives us 0→ Hn(X,F )∗ → Hn(X,E0)∗ → Hn(X,E1)∗. We alsohave 0→ hom(F , ωX)→ hom(E0, ωX)→ hom(E1, ωX), with the last twoon each being equal. Thus, hom(F , ωX) = Hn(X,F )∗.

3. Both sides are contravariant δ-functors from Coh(X) → Mod(k), ie δ-functors Coh(X)op →Mod(k). They agree for p = 0. It remains to showthat they are universal: we will show that both sides are erasable. That is,for all coherent F , there exists an epi u : E → F such that both functorsvanish on u when p > 0.

So there is an epi E → F when E is a finite ⊕O(qi). WLOG, qi <0 as O(q − 1)⊕n+1 → O(q). So Extp(E , ωX) = ⊕Extp(O(qi),O(−n −1)) = Hp(X,OX(−n − 1 − qi)) = 0 for p > 0. Also, Hn−p(X,E ) =⊕Hn−p(X,O(qi)) = 0 for p > 0.

Definition 8.25 (Dualizing Sheaf). Let X be a proper scheme of dimension nover a field k. A dualizing sheaf for X is a coherent OX-module ωX togetherwith a k-linear trace map t : Hn(X,ωX) → k such that forall coherent F weget perfect pairings hom(F , ωX)×Hn(X,F )→ Hn(X,ωX) t→ k.

Proposition 8.50. Let (ω, t) and (ω′, t′) be dualizing sheaves for X. Then thereexists a unique isomorphism ϕ : ω → ω′ such that t = t′ ϕ∗ : Hn(X,ω) →Hn(X,ω′)→ k.

Proof. The perfect pairing requirement gives hom(ω, ω′)×Hn(X,ω)→ Hn(X,ω′) t′→k, and so t : Hn(X,ω) → k is given by some ϕ ∈ hom(ω, ω′). That is, thereexists a unique ϕ : ω → ω′ with t = t′ ϕ∗. By symmetry, there exists a uniqueψ : ω′ → ω such that t′ = t ψ∗. ψϕ : ω → ω satisfies t = t (ψϕ)∗, and theidentity does this, and so ψϕ = id and similarly for ϕψ.

Assume that X is projective over k, choose a closed embedding X ⊆ PNk .Set r = codim(X,PN ).

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Definition 8.26. ω∗X = E xtrp(OXωPN ).

Lemma 8.51. E xtiP (OX , ωP ) = 0 for i < r.

Proof. E xtiP (OX , ωP ) is coherent as an OX -module. Thus, E xtiP (OX , ωP )(q) isgenerated by global sections for all q >> 0.

Γ(P,E xtiP (OX(−q), ωP )) ' HN−i(P,OX(−q)) = ExtiP (OX(−q), ωP ). Thisis zero if N − i > dimX iff i < r.

Note: Assume 0 → M ′ → M → M ′′ → 0 ses. Then M ′ injective impliessplit exact, so hom(M,M ′)→ hom(M ′,M ′) surjective. If M ′,M injective, thenM ′′ injective. And 0 → M0 → M1 → . . . → Mr → 0 with M i injective fori < r, then Mr injective.

Lemma 8.52. There is a natural isomorphism homX(−, ωX) ' ExtrP (−, ωP )of functors from Mod(X)→Mod(k).

Proof. 0→ ωP → I ∗ an injective resolution inMod(P ). Set Jm = H omP (OX ,Im) ⊆Im the subsheaf of sections killed by the ideal sheaf of X.

Note: F an OX -module, ϕ : F → Im an OP -hom, then Im(F ) ⊆Jm.Thus, Jm is an injective OX -module. So homX(−,Jm) = homP (−,Im)

is exact.Hi(J ∗) = E xtiP (OX , ωP ) = 0 for i < r. So we have exact sequence 0 →

J 0 → . . . → J r−1 → J r, and we get J r−1 → J r1 → 0 with J r

1 =Im d(dr−1) ⊆ J r. The Note implies that J r

1 is injective, so we get 0 →J r

1 →J r →J r2 → 0 with J r = J r

1 ⊕J r2 , so all are injective.

So the exact complex can be called J ∗1 and J r

2 → J r+1 → . . . can becalled J ∗

2 .ωX = E xtrP (OX , ωP ) = Hr(J ∗) = ker(J r

2 → J r+1), so we get 0 →ωX →J r

2 →J r+1 which gives, for F an OX -module, 0→ homX(F , ωX)→homX(F ,J r

2 ) → homX(F ,J r+1), so ExtrP (F , ωP ) = Hr(homP (F ,I ∗)) =Hr(homX(F ,J ∗)) = Hr(homX(F ,J ∗

2 )) = homX(F , ωX).

Proposition 8.53. ωX is a dualizing sheaf for X.

Proof. F is any coherent OX -module. n = N − r = dim(X). homX(F , ωX)×Hn(X,F ) = ExtrP (F , ωP )×HN−r(P,F )→ HN (P, ωP ) ' k is a perfect pair-ing, by Serre Duality for PN .

Take F = ωX . Then homX(ωX , ωX)×HN (X,ωX)→ HN (P, ωP ) = k is per-

fect pairing. id ∈ homX(ωX , ωX) corresponds to t : Hn(X,ωX) → HN (P, ωP ),

the trace map. Because everything has been natural, we can use t to factor theperfect pairing.

Let X be a proper scheme over k of dimension n.

Definition 8.27 (Dualizing Sheaf). A dualizing sheaf for X is a coherent OX-module ωX with a k-linear trace map t : Hn(X,ωX) → k such that for allcoherent OX-mod F , we have a perfect pairing homX(F , ωX) × Hn(F ) →Hn(X,ωX) t→ k.

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Thus, it must be unique. ωPN =∧N ΩPN is the dualizing sheaf for PN . IF

X ⊆ PN is a closed subscheme of codimension r, then ωX = E xtrPN (OX , ωPN )dualizing for X.

Notice: A choice of trace map t gives a natural isomorphism θ : homX(−, ωX)→Hn(X,−)∗ of functors Coh(X)op →Mod(k).

Corollary 8.54. ∃ natural transformations θi : ExtiX(−, ωX)→ Hn−i(X,−)∗.

Proof. We must show that both sides are δ-functors that agree for i = 0 andthat ExtiX(−, ωX) is erasable in Coh(X)op for i > 0.

Choose OX(1) very ample. F coherent implies that there exists a surjectionO(−q)⊕N → F for q >> 0. ExtiX(O(−q)⊕N , ωX) = ExtiX(OX(−q), ωX)⊕N =Exti(OX , ωX(q))⊕N = Hi(X,ωX(q))⊕N = 0 for q >> 0, so done.

Definition 8.28 (Regular Sequence). Let A be a local ring, m ⊆ A maximalideal, and f1, . . . , fr ∈ m. Then f1, . . . , fr is a regular seqeunce if each fi is anonzero divisor on A/(f1, . . . , fi−1).

Definition 8.29 (Cohen-Macaulay Ring). A is Cohen-Macaulay (CM) if ∃ aregular sequence f1, . . . , fr ∈ m such that dim(A) = r.

Facts:

1. A regular local ring is CM

2. IfA is CM then f1, . . . , fr ∈ m is a regular sequence iff dimA/(f1, . . . , fr) =dimA− r.

3. A is CM and f1, . . . , fr ∈ m a regular sequence, then A/(f1, . . . , fr) is CM.

4. A CM, f1, . . . , fr ∈ m a regular sequence, then I = (f1, . . . , fr) ⊆ Aimplies that (A/I)[t1, . . . , tr]→ grI(A) = ⊕n≥0I

n/In+1 by ti 7→ fi ∈ I/I2

is an isomorphism, so I/I2 ' (A/I)⊕r.

Koszul ComplexA a ring, f1, . . . , fr ∈ A and A⊕r has basis e1, . . . , er. Define Kp =

∧p(A⊕r)for 0 ≤ p ≤ r. Basis ei1 ∧ . . . ∧ eip.

d1 : K1 → K0 by ei 7→ fi. dp : Kp → Kp−1 by dp(ei1 ∧ . . . ∧ eip) =∑pj=1(−1)j−1fijeij ∧ . . . ∧ eij ∧ . . . ∧ eip .So K∗ = K∗(f1, . . . , fr, A) is this complex.

Proposition 8.55. A a local ring, f1, . . . , fr a regular sequence for A. Then0→ K∗ → A/(f1, . . . , fr)→ 0 is exact.

Definition 8.30 (Cohen-Macaulay). A scheme X is Cohen-Macaulay if OX,Pis CM for all P ∈ X.

Definition 8.31 (Local Complete Intersection). If Y is a nonsingular varietyover a field k and X ⊆ Y is a closed subscheme of codim r, and IX ⊆ OYthe ideal sheaf of X. Then X is a local complete intersection in Y if IX,P isgenerated by r elements, ∀P ∈ X.

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Note: X ⊆ Y a local complete intersection implies thatX is Cohen-Macaulayand IX/I 2

X is a locally free OX -module of rank r, the ”Conormal Sheaf” of Xin Y .

Definition 8.32 (Normal Sheaf). NX/Y = (IX/I 2X)∨ is the Normal Sheaf of

X in Y .

Recall that X ⊆ Y nonsingular gives 0→ I /I 2 → ΩY ⊗OX → ΩX → 0 isexact, so dualizing we get 0→ TX → TY ⊗ OX → NX/Y → 0

Theorem 8.56. X ⊆ PNk local complete intersection of codim r, then ωX 'ωPN ⊗

∧r NX/Pn .

Proof. Let U = Spec(A) ⊆ PN open affine such that I(X ∩U) ⊆ A is generatedby f1, . . . , fr ∈ A. Then let P ∈ U ∩X. m = I(P ) ⊆ A is a maximal ideal.

ThenAm = OPN ,P is regular local, so CM. dimAm/(f1, . . . , fr) = dim OX,P =dimX = N − r. So f1, . . . , fr is a regular sequence.

So we look at 0 → K∗(f1, . . . , fr, Am) → Am/(f1, . . . , fr) → 0. Replace Uwith some Uh, so WLOG, 0→ K∗(f1, . . . , fr;A)→ A/(f1, . . . , fr)→ 0 is exact.

Thus, 0→ K∗ → OX∩U → 0 is locally free resolution on U .Therefore, ωX |U = E xtrPN (OX , ωPN ) = E xtrU (OX∩U , ωU ) = Hr(H omU (K∗, ωU )).H om(K∗, ωU ) gives the sequence . . . → ω⊕RU → ωU → 0 by (σ1, . . . , σr) 7→∑

(−1)j−1fjσj .Thus, ωX |U ' ωU/(f1, . . . , fr)ωU = ωU ⊗ OX .Note: f1, . . . , fr form a basis of IX/I 2

X , then f1 ∧ f2 ∧ . . .∧ fr is a basis of∧r(IX/I 2X). Thus (f1 ∧ . . . ∧ fr)−1 is a basis of

∧r(NX/Pn).We now define γ : ωX |U = ωU/(f1, . . . , fr)ωU → ωU ⊗ NX/PN by taking

σ 7→ σ ⊗ (f1 ∧ . . . ∧ fr)−1.Claim: γ is independent of choices, and so defines a global isomorphism

ωX → ωPn ⊗NX/PN .Let g1, . . . , gr ∈ I(X ∩ U) ⊆ A be another set of generators. Write gi =∑cijfj with cij ∈ A. Define a map ϕ : A⊕r → A⊕r by ϕ(ei) =

∑cijej . As

∧pis a functor, we get a map on exterior powers.

So we get 0 → K∗(g1, . . . , gr;A) → A/I → 0 and 0 → K∗(f1, . . . , fr;A) →A/I → 0, with vertical maps given by exterior powers of ϕ and the identityfrom A/I to itself, with everything commuting.

As∧r

ϕ = det(cij) : Kr(g)→ Kr(f) and so det(cij) : H omU (Kr(f), ωU )→H omU (Kr(g), ωU ) gives a map ωU → ωU .

Tus we get map ωU (f1, . . . , fr)ωUdet(cij)→ ωU/(g1, . . . , gr)ωU and maps from

each to ωU ⊗∧r NX/PN giving a commutative diagram.

Corollary 8.57. If X is a nonsingular projective variety then ωX = ωX =∧n ΩX for n = dimX.

Proof. X ⊆ PNk closed subvariety. So 0 → IX/I 2X → ΩPN ⊗ OX → ΩX → 0.

Thus, ωPN ⊗ OX =∧N (ΩPPN ⊗ OX) =

∧r IX/I 2X ⊗ ωX .

Thus, ωX = ωPn ⊗∧r NX/PN

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9 Curves

Let k = k be an algebraically closed field.

Definition 9.1 (Curve). A curve is a complete connected non-singular varietyof dimension 1 over k.

Fact: All curves are projective.Div(X) = D =

∑p∈X mpP and deg(D) =

∑mp.

L an invertible OX -module, s ∈ Γ(U,L ), U ⊆ X, then (s) =∑P∈X vP (s)P

gives Pic(X) ' C`(X) by L 7→ (s), s ∈ Γ(U,L ). The inverse map takes D toOX(D) by Γ(U,OX(D)) = f ∈ k(X)|vP (f) ≥ −mp where D =

∑mPP

Fact: s ∈ k(X)∗, then s ∈ Γ(U,OX), deg(s) = 0. So deg : C`(X) → Z iswell-defined.

ωX = ΩX is the canonical sheaf.Serre Duality: Hi(X,L −1 ⊗ ωX) = ExtiX(L , ωX) = H1−i(X,L )∗.

Definition 9.2 (Genus). The geometric genus is pg(X) = dimkH0(X,ωX).

The arithmetic genus is pa(X) = (−1)dimX(χ(OX)− 1).

When X is a curve, pa(X) = dimkH1(X,OX). By Serre Duality, we have:

Proposition 9.1. For a curve X, pa(X) = pg(X).

Proof. Set i = 0 then Serre Duality says that H0(X,L −1⊗ωX) = H1(X,L )∗.Then if L = OX , we get H0(X,ωX) = H1(X,OX)∗.

As ωX ∈ Pic(X), it corresponds to a divisor K = KX ∈ C`(X) the CanonicalDivisor, with ωX = OX(KX).

If D ∈ Div(X), we can define `(D) = dimkH0(X,OX(D)). So, for example,

`(KX) = g, the genus of X.

Lemma 9.2. 1. `(D) 6= 0⇒ deg(D) ≥ 0.

2. If `(D) 6= 0 and deg(D) = 0, then D ∼ 0 (ie, D = 0 in C`(X))

Proof. 1. ∃0 6= s ∈ Γ(X,OX(D)), and (s) =∑P∈X vP (s)P and so deg(D) =

deg(s) ≥ 0.

2. If deg(D) = 0, then (s) = 0, so then D ∼ 0.

Theorem 9.3 (Riemann-Roch). Let D ∈ Div(X). Then

`(D)− `(K −D) = deg(D) + 1− g

Proof. `(K − D) = dimkH0(X,OX(D)−1 ⊗ ωX) = dimkH

1(X,OX(D)). So`(D)− `(K −D) = dimkH

0(X,OX(D))− dimH1(X,OX(D)) = χ(OX(D)).If D = 0, then `(0)− `(K) = deg(0) + 1− g, that is, 1− g = 1− g.ETS that RR holds for D iff RR true for D + P (P a point).

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Let k(P ) = OX/IP and note that IP = OX(−P ). So we get 0 →OX(−P )→ OX → k(P )→ 0 ses.

Tensor with OX(D + P ) and get 0 → OX(D) → OX(D + P ) → k(P ) → 0ses. As Euler Characteristic is additive, χ(OX(D + P )) = χ(OX(D)) + 1.

Examples:

1. g − 1 = `(K)− `(K −K) = deg(K) + 1− g, so deg(K) = 2g − 2.

2. deg(D) > 2g − 2, then `(K −D) = 0, so `(D) = deg(D) + 1− g.

3. Assume g(X) = 0. Then deg(K) = −2. Let P 6= Q ∈ X, D = P − Q ∈Div(X). `(D) = `(D)−`(K−D) = 0+1−0 = 1. So then D ∼ 0, so P ∼ Q.Thus, X = P1 from last semester because f ∈ k(X) with (f) = P − Qgives a map f : X → P1, which turns out to be an isomorphism.

Fact: A morphism f : X → Y of curves is finite or constant.

Definition 9.3 (Degree of a Map). deg(f) = [k(X) : k(Y )].

If P ∈ X and Q = f(P ) ∈ Y with Y a nonsingular curve, then mQ = (t) ⊆OY,Q and f∗(t) ∈ OX,P .

Define eP = vP (f∗(t)).This number is called the ramification index of f at P . If eP = 1, then f is

unramified at P . If eP > 1, then f is ramified at P . In this case, Q is a branchpoint.

A ramification point is called tame if char(k) = 0 or if char(k) 6 |eP . If it isnot tame, then it is wild.

Recall: f∗ : Div(Y ) → Div(X) by f∗(Q) =∑P∈f−1(Q) ePP . This map in

fact gives a map f∗ : C`(Y ) → C`(X). This is becuase D ∈ Div(Y ), thenf∗(OY (D)) ' OX(f∗D)

Definition 9.4 (Separable Map). f : X → Y is separable if it is finite andf∗ : k(Y ) ⊆ k(X) is a separable extension of fields.

Proposition 9.4. f : X → Y is a separable map of curves, then 0→ f∗ΩY →ΩX → ΩX/Y → 0 is ses.

Proof. We already know everything except that the map f∗ΩY → ΩX is injec-tive. Note that these are both invertible OX -modules on an irreducible variety.

ETS that the map is nonzero.Let P0 ∈ X be the generic point. It is enough to show that (ΩX/Y )P0 = 0.

This stalk is equal to Ωk[X]/k[Y ] localized at 0. Taking modules of differen-tials commutes with localization, do we get Ωk(X)/k(Y ) = 0 as the extension isseparable.

Let f : X → Y be finite morphism of curves. P ∈ X has image Q = f(P ) ∈Y . mP = (u) ⊆ OX,P and similarly mQ = (t) ⊆ OY,Q. Then du ∈ ΩX,P is agenerator, and similarly, dt ∈ ΩY,Q is a generator. f∗(dt) ∈ (f∗ΩY )P ⊆ ΩX,P .So ∃h ∈ OX,P with f∗(dt) = hdu. We will set the notation dt/du = h. Thisjust means that f∗(dt) = dt/du · du ∈ Ω(X,P ).

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Proposition 9.5. Let f : X → Y be separable map of curves.

1. supp(ΩX/Y ) = P ∈ X|eP > 1, which is a finite set.

2. lengthOX,P (ΩX/Y,P ) = vP (dt/du).

3. If f is tamely ramified at P , then length(ΩX/Y,P ) = ep − 1. If f is wildlyramified, then length(ΩX/Y,P ) > ep − 1.

Proof. 1. ΩX/Y,P = 0 iff ΩX,P generated by f∗(dt) iff f∗t generates mP ⊆OX,P iff eP = 1.

2. (ΩX/Y )P ' Ω(X,P )/f∗ΩY,Q = ΩX,P /(f∗(dt)) ' OX,P /(dt/du).

3. Set e = eP and t = f∗(t). Then t = aue with a ∈ OX,P a unit. Thendt = daue + aeue−1du. If the ramification is tame, then e 6= 0 ∈ OX,P , so,as da = hdu for some h, vP (dt/du) = e− 1.

If wild, then e = 0 ∈ OX,P , so dt = ueda, so vP (dt/du) = vP (ueh) >≥e > e− 1.

Assume from now: f : X → Y separable finite morphism of curves.

Definition 9.5 (Ramification Divisor). R =∑P∈X length(ΩX/Y,P )P ∈ Div(X).

Proposition 9.6. KX ∼ f∗KY +R.

Proof. s ∈ Γ(V,ΩY ), then KY = (s) ∈ Div(Y ). f∗(s) ∈ Γ(f−1(V ), f∗ΩY ) ⊆Γ(f−1(V ),ΩX). KX = (f∗s) ∈ Div(X).

Let (KX ;P ) denote the coefficient of P inKX . Then (KX ;P ) = lengthOX,P (ΩX,P /(f∗s)) =length(f∗ΩY,P /(f∗s)) + length(ΩX/Y,P ) = (f∗KY ;P ) + (R;P ).

Theorem 9.7 (Hurwitz Theorem). 2g(X)− 2 = n(2g(Y )− 2) + deg(R) wheren is deg(f).

Example: f : X → Y any finite morphism of curves, then g(X) ≥ g(Y ).Why? Because g(X) = g(Y ) + (n− 1)(g(Y )− 1) + 1

2 deg(R).The ramification divisor has even degree.

Theorem 9.8 (Lyroth’s Theorem). If k ⊆ L ⊆ k(t) and L is a field, then k = Lor L = k(u) for some u ∈ L.

Proof. k 6= L, then k has transcendence degree 1, so let X = CL. Then L→ k(t)gives a map f : P1 → X a finite map of curves. As g(P1) ≥ g(X) and g(P1) = 0,g(X) = 0, so X ' P1. Thus, L ' k(u) for some u ∈ L.

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