Math 214A Notes

Ben West

Winter 2015

1 Chapter 1: Affine Varieties

1.1 Closed sets

Let k be an algebraically closed field. Let An = Ank = {a = (a1, . . . , an) : ai ∈ k} = kn be affine space. SoA0 is a point, and A1 = k. Let An = k[T1, . . . , Tn] be the k-algebra of polynomials. If f ∈ An, a ∈ An, thenf(a) ∈ k.

Suppose S ⊆ An be a subset. Then Z(S) = {a ∈ An : f(a) = 0,∀f ∈ S} ⊆ An. A subset Z ⊆ An iscalled closed if Z = Z(S) for some S ⊆ An.

Example 1.1. If n = 1, Z ⊂ A1 is closed iff Z is finite or Z = A1.

Some properties:

1. If S ⊂ S′ ⊂ An, then Z(S) ⊃ Z(S′).

2. S ⊂ An, and I an ideal in An generated by S. Then Z(S) = Z(I) = Z(√I), where

√I = {g ∈ An :

∃m, gm ∈ I}, the radical of I. If J =√I, then

√J = J , so J is a radical ideal in An.

3. Hilbert Basis Theorem: every ideal in An is finitely generated. Writing J = 〈f1, . . . , fk〉, so Z(S) =Z(J) = Z(f1, . . . , fk). So every closed set can be given by finitely many polynomials.

4. If Sα ⊂ An, then Z(∪Sα) =⋂Z(Sα), so the intersection of closed sets is closed.

5. Z(S · T ) = Z(S) ∪ Z(T ), so the union of finitely many closed sets is closed. Hence we get the Zariskitopology on An with closed sets Z(S). Going back to A1, the open sets are ∅, or the cofinite sets. Soevery two nonempty open sets intersect.

If X ⊂ An is a subset, then I(X) = {f ∈ An : f(x) = 0,∀x ∈ X}. We get a correspondence betweensubsets of An and subsets of An given by Z(−) and I(−). In fact, I(X) is a radical ideal in An.

Some properties of I:

1. If X ⊂ X ′, then I(X) ⊃ I(X ′).

2. IZ(S) ⊃ S, and ZI(X) ⊃ X.

3. IZI = I and ZIZ = Z.

Proof. Note IZI(X) = IZ(I(X)) ⊃ I(X). But ZI(X) ⊃ X, applying I, IZI(X) ⊂ I(X).

4. ZI(X) = X

Proof. We have ZI(X) ⊃ X, but ZI(X) is closed, so ZI(X) ⊃ X. But X = Z(S) for some S, andZI(X) ⊂ ZI(X) = ZIZ(S) = Z(S) = X. So ZI(X) = X iff X is closed.

5. If S ⊂ An, IZ(S) =√〈S〉.

1

Proof. It’s clear that IZ(S) ⊃ S is a radical ideal, so IZ(S) ⊃√〈S〉. Recall Hilbert Nullstellensatz: If

S ⊂ An, and f ∈ An such that f is zero on the set of all common zeroes of polynomials in S. So f ∈ IZ(S),so fm ∈ 〈S〉 for some m > 0, and this gives the other containment.

So we get a correspondence of the radical ideals in An and the closed sets in An, with inverse bijectionsgiven by the restriction of Z(−) and I(−). We have An ↔ ∅, and 0 ↔ An, and maximal ideals correspondto points. (Warning: We need k to be algebraically closed for this to work, as we use the Nullstellensatz.)

A subset X ⊂ An is quasi-affine if X = Z ∩ U where Z is closed, and U is open in An.

Example 1.2. In A1, X is quasi-affine iff X is closed or open.

Some properties of quasi-affine sets:

1. Closed and open subsets are quasi-affine.

2. The intersection of two quasi-affine sets is also quasi-affine. Every closed (open) subset of a quasi-affineset is quasi-affine.

3. A subset X is quasi-affine iff X is open in X.

Proof. If X = U ∩X, for U open in An, then X is quasi-affine. Conversely, if X is quasi-affine, thensince Z ⊃ X implies Z ⊃ X, we have X = Z ∩ U ⊃ X ∩ U ⊃ X, which implies X = X ∩ U , so X isopen in X.

1.2 Regular Functions

Let X ⊂ An be a closed subset. A function f : X → k is called regular if there exists F ∈ An such thatf(x) = F (x) for all x ∈ X. Denote by k[X], the k-algebra of all regular functions on X. We get a surjectivemap

An → k[X] : F 7→ F |Xwith kernel I(X), since this is precisely the functions which are 0 on X. So k[X] ' An/I(X) by the firstisomorphism theorem, and is a reduced k-algebra, i.e., has no nonzero nilpotent elements.

Let ti = Ti|X be the coordinate functions on X. So k[X] is generated by the ti, hence is finitely generatedby the coordinate functions.

Proposition 1.3. Every commutative finitely generated reduced k-algebra is isomorphic to k[X] for someclosed set X in An for some n.

Proof. Take A = k[a1, . . . , an]. Consider the surjective homomorphism An → A by Ti 7→ ai, with kernel J ,which is radical since the algebra is reduced, so A ' An/J . Let X = Z(J), so I(X) = IZ(J) =

√J = J . So

An/J = An/I(X) ' k[X].

Example 1.4. 1. k[A1] = An.

2. If X = {a} a point, and a ∈ An, so k[X] = k since all functions on a point are constant.

3. Let f ∈ An be irreducible polynomial, X = Z(f) ⊂ An a hypersurface. Then k[X] = An/√〈f〉 =

An/〈f〉, since 〈f〉 is prime, since f is irreducible, hence radical.

If X ⊂ An is closed, k[An] → k[X] is surjective. If S ⊂ k[X], define ZX(S) = {x ∈ X : f(x) = 0,∀f ∈S} ⊂ X. If T ⊂ k[An] and maps onto S ⊂ k[X], we have ZX(S) = ZAn(T ) ∩X, closed in X.

If Y ⊂ X, define IX(Y ) = {f ∈ k[X] : f(y) = 0,∀y ∈ Y }, a radical ideal in k[X]. This gives acorrespondence between closed sets in X and radical ideals in k[X] given by IX(−) and ZX(−). In fact,observe that closed sets in X are a subset of closed sets in An. We have a correspondence between closedsets in An and radical ideals in An given by I and Z, and we have an inclusion of radical ideals in k[X] intothe radical ideals in An by sending J 7→ ϕ−1(J), where ϕ : An is the surjection onto k[X]. It follows thatIX and ZX are inverse bijections. We have the correspondence between ∅ ↔ k[X], X ↔ 0, and {x} ↔ mmaximal ideals. Also, if S ⊂ k[X], then ZX(S) = ∅ ⇐⇒

√〈S〉 = k[X] ⇐⇒ 〈S〉 = k[X].

2

Suppose X ⊂ An is closed. For f ∈ k[X], define DX(f) = X \ ZX(f) = {x ∈ X : f(x) 6= 0}, called aprincipal open set in X. If f, g ∈ k[X], then DX(fg) = DX(f) ∩DX(g). If U ⊂ X is open, X \ U = ZX(S)is closed, then U = X \ ZX(S) =

⋃f∈S DX(f). S can be chosen to be finite, so principal open sets form a

basis for the Zariski topology on X.Recall that a quasi-affine set is of form Z ∩ U for Z closed and U open.

Example 1.5. Set X = A1 \ {0}, T ∈ A1 = k[A1], and t = T |X . Suppose t(x) 6= 0 for all x ∈ X, so1t : X → k should be regular.

If X ⊂ An is quasi-affine, f : X → k and x ∈ X, then f is regular at x if there exists a neighborhood U

of x and G,H ∈ An such that H(y) 6= 0 for all y ∈ U and f(y) = G(y)H(y) for all y. First, there exist G,H ∈ An

such that H(x) 6= 0, f = HG in a neighborhood of x. If X is closed, there exist g, h ∈ k[X], h(x) 6= 0, f = g

hin a neighborhood of x.

Proposition 1.6. Let X ⊂ An be closed, f : X → k. Then, f is regular iff f is regular at every point x ∈ X.

Proof. Suppose f is regular at every point. Then for all x ∈ X, there exists a neighborhood Ux 3 x suchthat f = gx

hxin Ux for some gx, hx ∈ k[X]. So gx = f · hx on Ux. Then x /∈ X \ Ux is closed, so there

exists `x ∈ k[X] such that `x(x) 6= 0, but `x|X\Ux ≡ 0. Also, gx`x = fhx`x holds in Ux and X \ Ux. Since

(hx`x)(x) 6= 0, we have f = gx`xhx`x

in Ux. Replace gx with gx`x and hx with hx`x. Then gx = fhx in k[X],and f = gx

hxin Ux.

Put S = {hx : x ∈ X}, hx(x) 6= 0 implies Z(S) = ∅, so that 〈S〉 = k[X]. So we can write

1 =∑x∈X

hxmx

for almost all mx ∈ k[X] equal to 0. Then

f =∑

fhx︸︷︷︸gx

·mx =∑

gxmx ∈ k[X]

so f is regular.

So if X ⊂ An is a quasi-affine subset, a function f : X → k is regular if f is regular at every x ∈ X. Thenk[X] is the k-algebra of all regular functions on X, and is reduced, but not finitely generated in general.

Example 1.7. Set X = A1 \{0}, and t coordinate function on X. So 1t ∈ k[X]. So k[t, t−1] ⊆ k[X], but not

generated by t. Also, consider the quasi-affine set X = Z(T1T2 − T3T4) \ Z(T2, T4) ⊂ A4. Let f = t1t4

= t3t2

,and for all x ∈ X, either t2(x) 6= 0 or t4(x) 6= 0. Then X = U2 ∪ U4, where U2 is the set t2 6= 0 and U4 isthe set t4 6= 0, so f is regular.

Properties of regular functions

1. Let Y be a quasi-affine subset of a quasi-affine set X, and f ∈ k[X]. Then f |Y ∈ k[Y ]. This followssince the restriction of polynomials to a neighborhood is still a polynomial.

2. Let X be a quasi-affine set, and f : X → k. Put X =⋃α Uα and open cover. Then f ∈ k[X] ⇐⇒

f |Uα ∈ k[Uα]. The direct statement follows from the above, and the converse follows from localstatement of regularity.

3. if f : X → k is regular at x ∈ X, then f is regular in a neighborhood of x.

4. Let f ∈ k[X] be such that f(x) 6= 0 for all x ∈ X. Then 1f is also regular.

Proof. Pick x ∈ X, in a neighborhood of x, f(y) = G(y)H(y) for polynomials G and H. But f(x) 6= 0 =⇒

G(x) 6= 0, so G is not zero in a nbhd of x. In a nhbd of x, f = GH , and G(y) 6= 0 for all y. So 1

f = HG

is regular at x.

3

5. First, we need a lemma.

Lemma 1.8. Let X =⋃α Uα be an open cover of a topological space X, and Y ⊂ X . Then Y is

closed in X iff Y ∩ Uα is closed in Uα for every α.

Proof. For the forward direction, easy. Conversely, we prove that X \ Y is open. Take x ∈ X \ Y ,so x ∈ Uα for some α. So there is a closed subset Zα ⊂ X such that Y ∩ Uα = Zα ∩ Uα. SetU := Uα ∩ (X \ Zα), which is open in X. Since x /∈ Y , x /∈ Zα. So x ∈ Uα ∩ (X \ Zα. We claimU ∩ Y = ∅. If y ∈ U ∩ Y , then y ∈ Uα ∩ Y = Uα ∩ Zα ⊂ Zα, but y /∈ Zα, a contradiction. So U is anbhd of x, and U ⊂ X \ Y .

Let f ∈ k[X]. Then the set ZX(f) = {x ∈ X : f(x) = 0} is closed in X.

Proof. Take x ∈ X, in a nbhd Ux of x, f = GxHx

, where Gx, Hx are polynomials. So

ZX(f) ∩ Ux = ZUx(f) = ZUx(Gx) = ZX(Gx) ∩ Ux

which is closed in Ux since ZX(Gx) is closed in Zariski topology since Gx is a polynomial. ThenX =

⋃x∈X Ux, so by lemma, ZX(f) is closed in X.

6. Let f ∈ k[X], then f : X → k = A1 is continuous.

Proof. Take Z ⊂ A1 closed, need to show f−1(Z) is closed in X. If Z = A1, then f−1(Z) = X is closedin X. If Z is finite, then

f−1(Z) =⋃z∈Z

f−1(z) =⋃z∈Z

ZX(f − z)

which is a finite union of closed sets (by property 5), and the finite union of closed sets is closed.

Let X be quasi-affine. We get maps between the closed subsets of X and radical ideals in k[X] given byI and Z. Let Z ⊂ X be closed, so Z = ZX(S). Then (ZI)(Z) = ZIZ(S) = Z(S) = Z. So ZI is the identity.So I is injective, and Z is surjective. So we have ”more” radical ideals in k[X] than closed subsets.

Example 1.9. Consider X = A2 \ {(0, 0)}. Then we will see k[X] = k[A2] = A2. If J1 = 〈1〉 is the unitideal in k[X], and J2 = 〈t1, t2〉. Then Z(J1) = ∅, but Z(J2) = ∅ since we’ve removed the only zero (0, 0), sothe map Z is not injective in general.

2 Regular Maps

Take X ⊂ An quasi-affine, and f : X → Am and write f(x) = (f1(x), f2(x), . . . , fm(x)), where fi : X → k.We say f is regular if fi ∈ k[X] for all i.

Example 2.1. 1. If f : X → k = A1, so f is regular as a function is equivalent to saying it’s regular asa map.

2. The inclusion X ↪→ An is regular, since it’s given by coordinate functions which are always regular.

Suppose X ⊂ An and Y ⊂ Am are quasi-affine sets. A map f : X → Y is regular if the composition

Xf−→ Y ↪→ Am is regular. So to give a regular map f : X → Y is to give f1, . . . , fm ∈ k[X] such that

(f1(x), . . . , fm(x)) ∈ Y for all x ∈ X.

Example 2.2. 1. Inclusions are regular.

2. f : X → Y is regular, and X ′ ⊂ X is a quasi-affine subset, then f |X′ : X ′ → Y is regular. Supposeim(f) ⊂ Y ′ is a quasi-affine subset of Y , then f ′ : X ′ → Y ′ is regular, because we compose withinclusion. So we can also shrink the target provided we know the target is quasi-affine.

4

3. Consider f : A2 → A2, and define f(x, y) = (xy, y), for x and y the coordinate functions. The imageis everything except the x-axis, but we keep the origin. The image is not a quasi-affine subset. CallIm(f) = X. But X = A2, and A2 \X = L \ {0}. If X is quasi-affine, X is open in A2, which impliesL \ {0} is closed in A2. But L \ {0} = L.

Proposition 2.3. Regular maps are continuous.

Proof. Consider f : X → Y ⊂ An regular. Take Z ⊂ Y closed. There exists S ⊂ Am such that Z = ZY (S).Then f−1(Z) =

⋂G∈S f

−1(ZY (G)) since ZY (S) =⋂G∈S ZY (G). It suffices to show f−1(ZY (G)) is closed

in X. But f−1(ZY (G)) = {x ∈ X : G(f1(x), . . . , fn(x)) = 0}. So G(f1, . . . , fn)(x) = 0, and G(f1, . . . , fn) isregular, so this set is closed as the zero set of a regular function.

Corollary 2.4. If f : X → Y is a regular map, Y ′ ⊂ Y is quasi-affine, then f−1(Y ′) is also a quasi-affinesubset of X.

Proof. Write Y ′ = Z ∩ U , for Z closed, U open in Y , so f−1(Y ′) = f−1(Z) ∩ f−1(U), which is quasi-affinein X as an intersection of closed and open sets, since f is continuous.

Take f : X → Y a map. If h : Y → k, take h ◦ f : X → k.

Proposition 2.5. Let X and Y be quasi-affine sets, f : X → Y a map. Then f is regular iff for everyh ∈ k[Y ], we have h ◦ f ∈ k[X]. So regular maps are those which take regular functions to regular functionsafter precomposing.

Proof. For the converse direction, let ti be a coordinate function on Y , and f = (f1, . . . , fm). Then ti◦f = fi,and ti is regular. By the condition, fi is regular, so f is regular.

For the forward direction, suppose h ∈ k[Y ]. Is h ◦ f regular? Take x ∈ X, and y = f(x) ∈ Y . In aneighborhood U of y, h = G

H for G,H polynomials, H 6= 0. On f−1(U), nbhd of x, then

h ◦ f =G(f1, . . . , fm)

H(f1, . . . , fm)= G(f1, . . . , fm) · 1

H(f1, . . . , fm)

with both maps on the far RHS are regular on f−1(U), so this is regular at x, hence h ◦ f is regular.

Proposition 2.6. If Xf−→ Y

g−→ Z are regular maps, then g ◦ f is also regular.

Proof. Take h ∈ k[Z]. Then h◦g ∈ k[Y ] since g is regular, and thus h◦g ◦f ∈ k[X] by the previous theorem.So g ◦ f is regular.

For f : X → Y regular, we have an induced map f∗ : k[Y ]→ k[X] by h 7→ h◦f , which is a homomorphism

of k-algebras. If Xf−→ Y

g−→ Z, then (g ◦ f)∗ = f∗ ◦ g∗.

3 Categories and Functors

A category A is a class of objects O(A), and for X,Y ∈ O(A), we have a set of morphisms MorA(X,Y ). Wehave an associative composition, and the identity morphism 1X ∈ MorA(X,X).

For example, QASets(k). Here objects are quasi-affine sets, and the morphisms are regular maps. An-other example is CAlg(k), where the objects are commutative k-algebras, and morphisms are k-algebrahomomorphisms. Also, CSet(k) is the category of closed sets, with morphisms the regular maps.

A category B is a full subcategory of A if O(B) ⊂ O(A), and MorB(X,Y ) = MorA(X,Y ) for X,Y ∈O(B). We denote this B ⊂ A. For example, CSet(k) ⊂ QASets(k).

If C is a category, the dual category Cop, where O(C) = O(Cop), but MorCop(X,Y ) = MorC(Y,X).Let C be a category, and f : X → Y . We say f is an isomorphism if there exists a morphism g : Y → X

such that g ◦ f = 1X and f ◦ g = 1Y . We write g = f−1. For example, a map f : X → Y of quasi-affine setsis an isomorphism in QASets(k) iff f is a bijection, and f−1 is regular.

A functor F between two categories A and B, F : A → B, takes an object X ∈ O(A) to F (X) ∈ O(B),and a morphism f : X → Y to F (f) : F (X)→ F (Y ) in B. Also, F (g◦f) = F (g)◦F (f), and F (1X) = 1F (X).

5

In particular, we have a functor QASets(f)∗ → CAlg(k) by the assignment X 7→ k[X], and f : X → Yto f∗ : k[Y ] → k[X]. Alternatively, we can say this is a contravariant functor if we wish to avoid the dualcategory.

If F : A→ B is a functor, and f : X → Y is an isomorphism in A, then F (f) is an isomorphism in B. Sofunctors map isomorphisms to isomorphism. Furthermore, F (f)−1 = F (f−1).

Example 3.1. 1. Let Y ⊂ A2 give by y2 = x3. Let X = A1. Take f : X → Y to be t 7→ (t2, t3), i.e.,x = t2, and y = t3). This is regular since the maps are given by polynomials. This is a bijection sincet = y

x . We have

f∗ : k[Y ] //

��

k[X]

��

k[x, y]/(y2 − x3) // k[t]

where the vertical maps are equality, and the bottom map is given by x 7→ t2 and y 7→ t3. But t /∈ Im(f∗)so f∗ is not an isomorphism and f is not isomorphism.

We have

f−1 =

{yx if x 6= 0,

0 if x = 0.

but is not regular.

2. Suppose char(k) > 0. Consider A1 → A1 : t 7→ tp, the Frobenius map. This is a bijection. Thenf∗ : k[t] → k[t] is given by t 7→ tp, but is not an isomorphism, since its image consists of polynomialsin tp. Note f−1(s) = s1/p, but is not regular since it’s not given by a polynomial.

Let F : A→ B be a functor, X,Y ∈ O(A). Then F yields a map of sets

ϕX,Y : MorA(X,Y )→ MorB(F (A), F (B)) : f 7→ F (f)

We say F is faithful if ϕX,Y is injective for all X,Y . We say F is full if ϕX,Y is surjective for all X,Y .

Example 3.2. Let A ⊂ B be a full subcategory. We have the embedding functor F : A ↪→ B by X 7→ X andf 7→ f . So F is fully faithful since these categories have the same morphisms.

Lemma 3.3. Let F : A→ B be a fully faithful functor, X,Y ∈ O(A). Then X ' Y iff F (X) ' F (Y ) in B.That is to say, fully faithful functors reflect isomorphism.

Remark 3.4. Consider the sets of isomorphism classes in A and B. If we have functor F : A→ B, we geta map between them. If F is fully faithful this map is injective.

Consider QASets(k)op → CAlg(k). Is it full or faithful? Take f = (f1, . . . , fm) : X → Y ⊂ Am to be aregular map of quasi-affine sets. We get a morphism f∗ : k[Y ] → k[X]. We have the coordinate functionsti ∈ k[Y ]. Then f∗(ti) = fi.

So consider f = (f1, . . . , fm) and g = (g1, . . . , gm). If f∗ = g∗, then fi = f∗(ti) = g∗(ti) = gi, so f = g.So the functor is always faithful. It is not full however. So the algebras in the algebraic category comingfrom the sets in the geometric category have special properties.

Proposition 3.5. Let X be a quasi-affine set and Y a closed set. Then the map

MorQASets(k)(X,Y )→ MorCAlg(k)(k[Y ], k[X])

is a bijection.

Proof. The map is given by f 7→ f∗. We have seen that it is always injective. Take α : k[Y ] → k[X] ak-algebra homomorphism. We need to construct a regular map f : X → Y such that α = f∗. We haveY ⊂ Am is closed. Let t1, . . . , tm be the coordinate functions on Y , so ti ∈ k[Y ]. Set fi = α(ti) ∈ k[X]. Putf = (f1, . . . , fm) : X → Am which is regular as the fi are regular. I claim Im(f) ⊂ Y .

6

Take F ∈ I(Y ) ⊂ Am. By definition, F |Y = 0. Then F (T1, . . . , Tm)|Y = 0. Since ti = Ti|Y , we haveF (t1, . . . , tm) = 0 in k[Y ]. So

0 = α(F (t1, . . . , tm)) = F (α(t1), . . . , α(tm)) = F (f1, . . . , fm)

so F (f1(x), . . . , fm(x)) = 0 for all F ∈ I(Y ). For any x ∈ X, (f1(x), . . . , fm(x)) ∈ ZI(Y ). But ZI(Y ) = Ybecause Y is closed. (This is why we must assume Y closed.) So Im(f) ⊂ Y .

So consider f : X → Y . Then X → Y ↪→ Am is regular, so f is regular. Now α(ti) = fi = f∗(ti), so αand f∗ agree on the generators ti of k[Y ]. Thus α = f∗.

If F : A→ B is a fully faithful functor, then F induces a injection on the isomorphism classes of objects.A functor F : A→ B is essentially surjective if every Y ∈ O(B) is isomorphic to F (X) for some X ∈ O(A).

A functor F : A → B is an equivalence if F is full, faithful, and essentially surjective. Such an F givesan isomorphism on classes of objects and on morphisms: MorA(X,Y ) ' MorB(F (X), F (Y )).

Remark 3.6. 1. Suppose A ⊂ B is a full subcategory such that every object of B is isomorphic to anobject of A. Then the inclusion functor is an equivalence.

In B, let Y ' Y ′ be isomorphic, and distinct. Set A = B \ {Y ′}. Then A and B are equivalent.

For example, let B be the category of finite dimensional vector spaces over a field k. Let A be thecategory O(A) = {kn : n ≥ 0}. So A ↪→ B is an equivalence. Moreover, let A′ = {n : n ≥ 0}, withMor(i, j) = Mi×j(k).

2. If H ⊂ G subgroup, and H ↪→ G an injection, then H is isomorphic to its image in G.

If A ⊂ B is full subcategory, then A ↪→ B is full and faithful. If F : A → B full and faithful, putO(B′) = {Y ∈ O(B) : Y ' F (X) for some X ∈ B}. Then A is equivalent to B′ ⊂ B.

Theorem 3.7. The functor CSets(k)op → CAlg(k) given by X 7→ k[X] and f 7→ f∗ is full and faithful. Thisfunctor gives and equivalence between CSets(k)op and the full subcategory of CAlg(k) consisting of all reducedand finitely generated algebras.

4 Products

We identify An × Am = An+m. If X = ZAn(S) is closed in An, and Y = ZAm(T ) is closed in Am, thenX × Y ⊂ An+m is equal to ZAn+m(S ∪ T ), so the product of closed sets is closed.

Suppose X ⊂ An and Y ⊂ Am are quasi-affine sets. Write X = X1 \ X2 for X1, X2 closed in An andY = Y1 \ Y2 for Y1, Y2 closed in Am. Then

X × Y = (X1 ×X2) \ (X1 × Y2 ∪X2 × Y1)

which are both closed, hence it is quasi-affine in An+m. Similarly, if X,Y open, then X × Y is open.Consider the maps p : X × Y → X and q : X × Y → Y which are regular as projections. They induce

p∗ : k[X] → k[X × Y ] and q∗ : k[Y ] → k[X × Y ]. By the universal property of the tensor product, we get amap

ϕ : k[X]⊗k k[Y ]→ k[X × Y ] : f ⊗ g 7→ ((x, y) 7→ f(x)g(y)).

Later, we will prove it is an isomorphism. For now,

Proposition 4.1. The map ϕ is injective.

Proof. Let {fi}i∈I be a basis for k[X]. Then every element in k[X]⊗k k[Y ] can be uniquely written in formh =

∑i∈I fi ⊗ gi for gi ∈ k[Y ] almost all zero. Then suppose

0 = ϕ(h)(x, y) =∑i∈I

fi(x)gi(y) ∀x, y

Fixing y, we get∑figi(y) = 0 in k[X], but the fi are linearly independent, so gi(y) = 0 for any y ∈ Y . So

all gi = 0, thus h = 0.

7

Corollary 4.2. The map ϕ is isomorphism if X and Y are closed.

Proof. This is because k[X × Y ] is generated by the coordinate functions, so ϕ is surjective.

Take X ⊂ An to be a quasi-affine set. Then X ×X ⊃ ∆X = {(x, x) : x ∈ X} = Z(ti − t′i). This is calledthe diagonal of X, and is closed in X ×X.

We have the diagonal map d : X → X ×X : x 7→ (x, x) which is regular and Im(d) = ∆X . Also, d givesan isomorphism X → X ×X because the inverse map is the projection, which is regular.

Remark 4.3. The Zariski topology on X × X is not the product topology. The diagonal ∆X is closed inX × X in the product topology iff X is Hausdorff, but the Zariski topology is not Hausdorff since any twononempty open sets intersect.

Consider f : X → Y and f ′ : X ′ → Y ′ to be regular maps of quasi-affine sets. Then f × f ′ : X ×X ′ →Y × Y ′ defined by (f × f ′)(x, x′) = (f(x), f ′(x′)). Write f = (f1, . . . , fm) and f ′ = (f ′1, . . . , f

′n). If p, q are

the projection maps, then f × f ′ = (f1 ◦ p, . . . , fm ◦ p, f ′1 ◦ q, . . . , f ′n ◦ q) which is also a regular map.Let f be a regular map. Take f ×1Y : X×Y → Y ×Y ⊃ ∆Y . Then (f ×1Y )−1(∆Y ) = {(x, y) ∈ X×Y :

f(x) = y}, which is closed in X × Y . This is the graph of f , denoted Γf . This is isomorphic to X under theprojection p : Γf → X and X → Γf : x 7→ (x, f(x)).

Proposition 4.4. (Universal Property of the Product) Let X,Y, Z be quasi-affine sets and f : Z → X andg : Z → Y be regular maps. Then there exists a unique regular map h : Z → X × Y such that p ◦ h = f andq ◦ h = g.

Proof. Assuming h exists, p(h(z)) = f(z) and q(h(z)) = g(z), so necessarily h(z) = (f(z), g(z)). This givesuniqueness.

On the other hand, if h is defined as such, Then

h : Zd−→ Z × Z f×g−−−→ X × Y

which is regular. Write h = (f, g).

Proposition 4.5. Let f, g : X → Y be two regular maps. If f and g agree on a dense subset of X, thenf = g.

Proof. Consider

h : Xd−→ X ×X f×g−−−→ Y × Y ⊃ ∆Y

then h−1(∆Y ) is closed in X. But this set contains a dense subset of X. So h−1(∆Y ) = X since it is aclosed set containing a dense set. So f = g.

5 Affine and Quasi-affine Varieties

Take X = Z(xy− 1) ⊂ A2 and Y = A1 \ {0} ⊂ A1 open. Then X is closed in A2, and Y is not closed in A1.But they are isomorphic under the maps X → Y : (x, y) 7→ x and Y → X : t 7→ (t, 1t ), which is regular since1t is regular since t 6= 0. So two sets can be isomorphic even though one is closed and the other is not.

Note A1 \ {0} = DA1(t) = {x ∈ A1 : t(x) 6= 0}.Take X a closed set, f ∈ k[X], and DX(f) = {x ∈ X : f(x) 6= 0}. Define Z ⊆ X×A1 by Z = Z(f · t−1),

where t is the coordinate function. I claim DX(f) ' Z. The maps are DX(f) → Z : x 7→ (x, 1f(x) ) and

Z → DX(f) : (x, t) 7→ x.Now k[DX(f)] ' k[Z] = k[X × A1]/

√〈ft− 1〉. But k[X × A1] ' k[X] ⊗ k[A1] = k[X] ⊗ k[t] = k[X][t].

Thus k[X × A1]/√〈ft− 1〉 = k[X][t]/

√〈ft− 1〉. Also

k[X][t]/(ft− 1) ' k[X](f)

is the localization of k[X] with respect to {fn}. Then k[X] is reduced so k[X](f) is reduced. So 〈ft− 1〉 =√〈ft− 1〉 since it is radical as the quotient is reduced, so finally k[DX(f)] ' k[X](f).

8

Example 5.1. 1. k[A1 \ {0}] = k[DA1(t)] = k[t](t) = k[t, t−1]

2. k[A2 \ {0}] = k[t1, t2]. If U ⊂ X is a dense subset, then k[X]→ k[U ], the restriction map, is injective.If f : U → X is the inclusion, the restriction map is f∗. It is injective because if they agree on U , theyagree on X since U is dense.

Let U1 = DA2(T1) ⊂ A2 and U2 = DA2(T2) ⊂ A2. Then U1 ∩ U2 ⊂ X = A2 \ {0} ⊂ A2. Thenk[A2] ↪→ k[X] ↪→ k[U1] ↪→ k[U1 ∩ U2] and likewise k[X] ↪→ k[U2] ↪→ k[U1 ∩ U2].

But k[U1] = k[T1, T2, T−11 ] and k[U2] = k[T1, T2, T

−12 ]. And k[U1∩U2] = k[DA2(T1T2)] = k[T1, T2, T

−11 , T−12 ]

so k[T1, T2] ↪→ k[X] ↪→ k[U1] ∩ k[U2] ⊂ k[U1 ∩ U2] = k[T1, T2, T−11 , T−12 ]. But k[T1, T2, T

−11 ] ∩

k[T1, T2, T−12 ] = k[T1, T2], so we get equality.

A quasi-affine variety is a quasi-affine set. An affine variety is a quasi-affine variety that is isomorphic toa closed set. Then QAVar(k) ⊃ AffVar(k) and we have a functor QAVar(k)op → CAlg(k) by X 7→ k[X] andf 7→ f∗.

If X is an affine variety, f ∈ k[X], then DX(f) is also affine.We have an embedding CSets(k) ↪→ AffVar(k) as a full subcategory, and this is an equivalence of categories.The functor AffVar(k)op → CAlg(k) is full and faithful. The image lives in the set of reduced, finitely

generated, commutative k-algebras.If X is an affine variety, and Z ⊂ X is closed, then Z is affine and k[X]→ k[Z] is surjective.Affine varieties X and Y are isomorphic iff k[X] ' k[Y ]. This follows from the equivalence of categories.

But X,Y live in both AffVar(k) ⊂ QAVar(k), but it doesn’t matter which category we view them in.

Example 5.2. X = A2 \ {0} is not affine.

1. The map Z from radical ideals in k[X] = k[t1, t2] to closed subsets in X is not injective, Z(k[t1, t2]) =Z(〈t1, t2〉) = ∅, but if the punctured plane were affine, this map would be an isomorphism.

2. The embedding f : X ↪→ A2 is not an isomorphism. Applying f∗ : k[A2]→ k[X] is an isomorphism asthey are both polynomomial rings in two variables. But the functor AffVar(k)op → CAlg(k) is full andfaithful.

3. View A1 \ {0} to be closed in X under the map g : t 7→ (t, 0). But g∗ : k[X] = k[T1, T2]→ k[A1 \ {0}] =k[t, t−1] is given by T1 7→ t and T2 7→ 0, so is not surjective.

Theorem 5.3. Let X be a quasi-affine variety, x ∈ X. Then x has an affine neighborhood.

Proof. We have X ⊆ An is a quasi-affine subset, so X = Z1 \ Z2, both closed sets in An. So x /∈ Z2, thusthere exists F ∈ An such that F |Z2

= 0, but F (x) 6= 0. Let f = F |Z1∈ k[Z1]. Consider the principal open

DZ1(f) = Z1 \ Z(f) ⊂ Z1 \ Z2 = X because Z2 ⊂ Z(F ). And x ∈ DZ1

(f) since f(x) 6= 0, so DZ1(f) is a

neighborhood of x in X, and DZ1(f) is affine.

Corollary 5.4. Every quasi-affine variety admits an affine open cover.

6 Irreducible Varieties

Lemma 6.1. Let X be a topological space. Then the following are equivalent:

1. Every sequence of closed sets Z1 ⊃ Z2 ⊃ Z3 ⊃ · · · is stable. I.e., there exists n such that Zn = Zm forall m ≥ n.

2. Every nonempty set A of closed subsets of X has a minimal element. That is, there exists Z ∈ A suchthat Z ′ 6⊂ Z for all Z ′ ∈ A distinct from Z.

We say X is Noetherian if X satisfies those conditions.

Example 6.2. An is Noetherian since there is a correspondence between closed subsets and radical ideals,which reverses the inclusion, and An is Noetherian as a ring. So every increasing sequence of ideals in Anstabilizes, so every decreasing sequence of closed sets in An stabilizes.

9

Lemma 6.3. Every subspace of a Noetherian space is Noetherian.

Corollary 6.4. Every subset of An is Noetherian.

Lemma 6.5. Every Noetherian space is compact.

Proof. We use the definition of compactness that if Zα ⊂ X is a family of closed subsets, and⋂α Zα = ∅,

then there exist α1, . . . , αn such that Zα1∩ · · · ∩ Zαn = ∅, i.e., the finite intersection property.

Let {Zα} be as above. Let A = {Zα1 ∩ · · · ∩Zαk} be the set of all finite intersections of members of thisfamily. So

⋂A ⊂

⋂α Zα = ∅. Also, A is a set of closed sets. Let Zα1 ∩ · · · ∩ Zαn be minimal in A. I claim

this intersection is empty. If not, there exists x ∈ Zαi for 1 ≤ i ≤ n. There exists αn+1 such that x /∈ Zαn+1

since⋂A = ∅. Then

x /∈n+1⋂i=1

Zαi (n⋂i=1

Zαi

a contradiction to minimality.

Proposition 6.6. Every quasi-affine variety admits a finite affine open cover.

Lemma 6.7. Let X be a topological space. TFAE:

1. If Z1 ( X and Z2 ( X are closed, then Z1 ∪ Z2 6= X.

2. If U1 ( X and U2 ( X are nonempty open subsets, then U1 ∩ U2 6= ∅.

3. Every nonempty open subset of X is dense in X.

Proof. (1) ⇐⇒ (2) just take complements. (2) =⇒ (3). Take U ⊂ X nonempty open, and U ∩(X \U) = ∅.By (2), X \ U = ∅, so X = U , so U is dense in X.

(3) =⇒ (2) Take U1, U2 ⊂ X to be nonempty, open. If U1∩U2 = ∅, then U1 ⊂ X\U2, so U1 ⊂ X\U2 6= X,so U1 is not dense in X.

A topological space X satisfying the above properties is called irreducible.We have the following Irreducibility Test: Suppose A ⊂ X. Then A is irreducible iff the following

condition holds: if A ⊂ Z1 ∪ Z2, for Z1, Z2 closed, then either A ⊂ Z1 or A ⊂ Z2.

Proposition 6.8. A quasi-affine variety X is irreducible iff k[X] is a domain.

Proof. Suppose X is irreducible. Take f, g ∈ k[X] such that fg = 0. But then X = Z(fg) = Z(f) ∪ Z(g).So X is a union of two closed sets, so X = Z(f) or X = Z(g), whence f = 0 or g = 0.

Conversely, suppose k[X] is a domain. Suppose X = Z1 ∪ Z2, both closed. Towards a contradiction,suppose they are both proper subsets. So there exists f1 ∈ k[X] such that f1 6= 0 but f |Z1 = 0, and f2 ∈ k[X]such that f2 6= 0 but f2|Z2

= 0. So f1f2|X = 0, hence k[X] has zero divisors, a contradiction.

Example 6.9. 1. An is irreducible since coordinate ring is a polynomial ring, hence a domain.

2. A1 \ {0} is irreducible, since k[A1 \ {0}] = k[t, t−1] which is a domain.

3. If f ∈ An is irreducible, then Z(f) ⊂ An is called a hypersurface, and is irreducible because k[Z(f)] =An/(f) which is a domain since (f) is prime.

4. Take X = Z(T1) ∪ Z(T2) ⊂ A2. This is Z(T1T2). Then k[X] = k[T1, T2]/(T1T2), which is not adomain since t1t2 ≡ 0 on the axes. Note Z(Ti) ' A1, hence both axes are irreducible, so X is a unionof irreducible parts. Actually, these components are unique.

5. Take X affine variety. We have a correspondence between the closed subsets of X and the radical idealsof k[X]. So if Y closed and I the corresponding ideal, k[Y ] ' k[X]/I under this correspondence. Thisimplies the irreducible closed subsets are in correspondence with the prime ideals in k[X].

Theorem 6.10. Let X be a Noetherian topological space. Then there are irreducible closed subsets Xi ⊂ X,i = 1, . . . , n such that X =

⋃iXi and Xi 6⊂ Xj for i 6= j. These subsets are unique.

10

Proof. Let Y be a minimal closed subset of X that has no decomposition Y =⋃Yi of closed subsets. Then

Y is not irreducible, therefore Y = Y ′ ∪ Y ′′ for Y ′, Y ′′ closed and strictly smaller than Y . But then Y ′ andY ′′ have decompositions, hence so does Y ′ ∪ Y ′′ = Y , a contradiction.

For uniqueness, suppose X1 ∪ X2 ∪ · · · ∪ Xn = X ′1 ∪ X ′2 ∪ · · · ∪ X ′m. Now Xi is contained in the RHS,and the irreducibility test says Xi ⊂ X ′j for some j. Then find k such that X ′j ⊂ Xk, so Xi ⊂ X ′j ⊂ Xk, soi = k. So Xi = X ′j .

The Xis are called the irreducible components of X.

Example 6.11. Take X = Z(y2 − xz, z2 − y3) ⊂ A3. Note

z3(z − x3) = z4 − x3z3 = z4 − y6 = y6 − y6 = 0

So X = ZX(z)∪ZX(z− x3). Are they irreducible? Well ZX(z) = ZA3(z, y2− xz, z2− y3) = ZA3(y, z) ' A1,hence irreducible.

On ZX(z − x3), note x4 = xz = y2, and z2 = x6, so

x4(y − x2) = x4y − x6 = y3 − x6 = y3 − z2 = 0.

So ZX(z − x3) = ZX(z − x3, x) ∪ ZX(z − x3, y − x2). But ZX(z − x3, x) = ZA3(x, y, z) = {0}, which is inZX(z) already.

Now k[ZX(z − x3, y − x2)] = k[x, y, z]/(z − x3, y − x2) ' k[x]. So ZX(z − x3, y − x2) ' A1, hence isirreducible. So X = ZA3(y, z)∪ZA3(z − x3, y − x2). The first is embedded as the x-axis, and the second is aline embedded under the parametrization x = t, y = t2, and z = t3. The first set contains (1, 0, 0), and thesecond contains (1, 1, 1).

Proposition 6.12. Let A be a subset of a topological space X. Then A is irreducible iff A is irreducible.

Proof. For forward direction, write A ⊂ Z1 ∪ Z2 as a union of closed sets. Then A ⊂ Z1 ∪ Z2, so byirreducibility test, A ⊂ Zi for some i, and thus A ⊂ Zi since A is closed.

Conversely, suppose A ⊂ Z1∪Z2. Then A ⊂ Z1∪Z2, so since A is irreducible, A ⊂ Zi for some i, whenceA ⊂ Zi for some i.

Corollary 6.13. Let X be an irreducible topological space, U ⊂ X a nonempty open subset. Then U is alsoirreducible.

Proof. Since U is dense in X, U = X is irreducible, so U is irreducible by the proposition.

Proposition 6.14. If X and Y are irreducible quasi-affine varieties, then so is X × Y .

Proof. If y ∈ Y , then X ' X × y ⊂ X × Y is a closed subset isomorphic to X. Suppose X × Y = Z1 ∪ Z2,for Zi closed. Then X × y ⊂ Z1 ∪ Z2, so by irreducibility test, X × y ⊂ Z1 or X × y ⊂ Z2.

Now set Yi = {y ∈ Y : X × y ⊂ Zi} , so Y = Y1 ∪ Y2. Claim that Yi is closed in Y . For any x ∈ X,consider (x× Y ) ∩ Zi = x× Yx where Yx ⊂ Y is closed as the intersection of two closed sets. So⋂

x∈XYx = {y ∈ Y : (x, y) ∈ Zi,∀x ∈ X} = Yi

so Yi is closed as intersection of closed sets. Since Y is irreducible, we must have Y = Yi. Thus X × Y =Zi.

Corollary 6.15. Let A and B be commutative k-algebras. If A and B are domains, then so is A ⊗k B.(Here k assumed to be algebraically closed.)

Proof. Assume A and B are finitely generated. Then A = k[X] and B = k[Y ] for X and Y irreducible affinevarieties. Then X × Y is irreducible, then k[X × Y ] = k[X]⊗k k[Y ] = A⊗k B is a domain.

Note this is not true if k is not algebraically closed, for instance, C⊗R C ' C× C.

11

7 Rational Functions and Maps

Let X be an irreducible quasi-affine variety. Consider the set of pairs (U, f) where U ⊂ X is a nonempty,open subset of X, and f is a regular function on U . We say (U1, f1) ∼ (U2, f2) if f1|U1∩U2 = f2|U1∩U2 . Notesince X is irreducible, U1 ∩ U2 6= ∅.

Suppose (U1, f1) ∼ (U2, f2) ∼ (U3, f3). We have f1|U1∩U2∩U3= f2|U1∩U2∩U3

. We need to show f1|U1∩U3=

f3|U1∩U3. But X ⊃ U1 ∩ U3 ⊃ U1 ∩ U2 ∩ U3 6= ∅. Where X and U1 ∩ U3 are irreducible, and U1 ∩ U2 ∩ U3 is

dense in U1 ∩ U3, so the functions agree on U1 ∩ U3, as they agree on a dense subset.Denote by k(X) the set of equivalence classes of pairs, denoted [U, f ]. If U ′ ⊂ U is open, then (U, f) ∼

(U ′, f ′) where f ′ = f |U . Thus [U, f ] = [U ′, f ′], so we can shrink open sets and replace functions by theirrestrictions.

These equivalence classes are called rational functions. We define operations

[U1, f1] + [U2, f2] = [U1 ∩ U2, f1|U1∩U2+ f2|U1∩U2

]

and[U1, f1] · [U2, f2] = [U1 ∩ U2, f1|U1∩U2

· f2|U1∩U2]

anda · [U, f ] = [U, af ]

This makes k(X) a commutative k-algebra. Also 0 = [U, 0] = [X, 0]. So if [U, f ] is nonzero in k(X), thenf 6= 0. Then note

[DX(f),1

f] · [U, f ] = [DX(f), 1]

which is the identity in k(X). So in fact k(X) is a field extension of k.We can embed the domain k[X] ↪→ k(X) : f 7→ [X, f ].We say X is irreducible iff k[X] is a domain. We say say X is irreducible if X 6= ∅ and X is not the union

of two proper closed sets. We know every X is a unique union of irreducible varieties.Let X be irreducible in An. We embed k[X] ⊂ k(X) as g 7→ [X, g].

Proposition 7.1. k(X) is a quotient field of k[X].

Proof. Let [U, f ] ∈ k(X), so U ⊂ X is open and nonempty, and f ∈ k[U ]. Pick x ∈ U . In a neighborhoodU ′ ⊂ U of x, we can write f = G

H , where G,H are polynomials, and H 6= 0 in U ′. Let g = G|U ′ and

h = H|U ′ . On U ′, we have f |U ′ = gh . So [U, f ] = [U ′, f |U ′ ] = [U ′,g]

[U ′,h] = [X,g][X,h] .

We write f for [X, f ] for f ∈ k[X] and [X, f ] ∈ k(X). So every f ∈ k(X) can be written as f = gh ∈ k[X],

h 6= 0.

Remark 7.2. If X ⊂ An, then k(X) is generated by the coordinate functions ti as a field. In particular,k(X)/k is finitely generated as a field.

Let X be irreducible. Take f ∈ k(X). So f = [U, g], where U ⊂ X is nonempty open, and g ∈ k[U ]. LetU =

⋃U , where the union is taken over all U ⊂ X open such that f = [U, g] for some g ∈ k[U ]. Then U is

open in X. If f = [U ′, g′], then g and g′ agree on U ∩U ′. Then there exists h ∈ k[U ∪U ′] such that h|U = gand h|U ′ = g′.

So there exists a unique g ∈ k[U ] such that f = [U , g]. We call U the domain of definition of f . We sayf is regular at every point x ∈ U (or defined at x ∈ U), and f(x) = g(x).

Note f is defined for all x ∈ X iff U = X iff f is regular.

Example 7.3. Take X = Z(x2 + y2 − 1) ⊂ A2. Then k[X] = k[x, y]/(x2 + y2 − 1). Let f = 1−yx ∈ k(X)

is defined at (x, y) with x 6= 0. But note that (1− y)(1 + y) = 1− y2 = x2, so 1−yx = x

1+y in k(X). This is

defined at (0, 1) if characteristic of k is not zero.But 1−y

x /∈ k[X] so the domain of definition of U = X \ {(0,−1)}.

12

Let X and Y be quasi-affine varieties, so X is irreducible. Consider pairs [U, f ] and U ⊂ X open, andf : U → Y be a regular map. We define (U, f) ∼ (U ′, f ′) if f and f ′ agree on U ∩U ′. A rational map X → Yis an equivalence class [U, f ] of a pair (U, f).

A rational function on X is a rational map X → A1. Suppose Y ⊂ Am, and [U, f ] : X → Y , thenf = (f1, . . . , fn) for fi ∈ k[U ], so we can view fi as rational functions on X. So to give a rational mapX → Y is the same as to give rational functions f1, . . . , fn ∈ k(X) such that (f1(x), f2(x), . . . , fn(x)) ∈ Yfor all x where all fi’s are defined.

Every rational map X → Y has the domain of definition U ⊂ X. Let f : X → Y defined on U ⊂ X, sof = [U , g] for g : U → Y regular. Define im(f) = im(g) ⊂ Y . We say f is dominant if im(f) is dense in Y .

Suppose

Xf// Y

g// Z

If im(f) ∩ Ug = ∅, we cannot define g ◦ f . But if f is dominant, im(f) is dense in Y , thus im(f) ∩ Ug 6= ∅and we can define g ◦ f as follows:

g ◦ f = [f−1(Ug) ∩ Uf , h]

such that h(x) = g(f(x)).

Take Xf// Y

g// A1 where g ∈ k(Y ) and X and Y both irreducible. Assume that f is dominant.

Therefore, g ◦ f ∈ k(X). We get a map

f∗ : k(Y )→ k(X) : g 7→ g ◦ f

which is a homomorphism of fields over k.

If Xf// Y

g// Z , then (g ◦ f)∗ = f∗ ◦ g∗. So we get a functor from the opposite category of

irreducible quasi-affine varieties and dominant rational maps to the category of field extensions of k withmorphisms homomorphisms of fields over k. It is given by X 7→ k(X) and f 7→ f∗.

Example 7.4. Let X be irreducible, and U ⊂ X a nonempty open set, so U is irreducible. We have theinclusion map U ↪→ X which is dominant because the image is U is every nonempty open is dense in anirreducible variety. I claim this is an isomorphism in the above category. We define an inverse map X → Uby [U, idU ]. So U ≈ X are “birationally isomorphic.” Therefore k(X) ' k(U).

We have a functor from the opposite category of irreducible varieties over k to the category of fieldextensions of k.

Proposition 7.5. This functor is full and faithful.

Proof. Take X and Y to be irreducible varieties. Let α : k(Y ) → k(X) be a homomorphism over k. Weneed to find f : X → Y such that α = f∗. Replace Y by a nonempty open affine subvariety. Such existsbecause Y is covered by open affine subvarieties, and Y is isomorphic to its open subvarieties since it isirreducible. So we may assume Y is affine. So Y ⊂ Am closed, with t1, . . . , tm coordinate functions on Y .Now ti ∈ k[Y ] ⊂ k(Y ).

Let gi = α(ti) ∈ k(X). All the gi are defined on some nonempty open set U ⊂ X. So the restrictions arein gi ∈ k[U ] ⊂ k(U) = k(X). We have

k(Y )α // k(X)

ti ∈ k[Y ]?�

OO

β// k[U ] 3 gi

?�

OO

Since Y is affine, there exists a regular map h : U → Y such that h∗ = β. Also h∗(ti) = gi. So h defines arational map f : X → Y given by [U, h]. So f∗ : k(Y )→ k(X) = k(U) and f∗(ti) = gi = α(ti). So α and f∗

agree on the ti, which generate F (Y ) (k(Y )?) so f∗ = α.We need to also show f is dominant. Now

Im(h) = h(Z(0)) = Z(f∗−1(0)) = ker f∗ = Z(kerβ) = Z(0) = Y.

13

since β is injective as a field homomorphism. It follows that h and f are dominant.Now we need to show that if f : X → Y and g : X → Y are such that f∗ = g∗, then f = g. Assume

Y ⊂ Am. Then f = (f1, . . . , fm) and g = (g1, . . . , gm), where fi, gi ∈ k(X), and fi = f∗(ti) = g∗(ti) = gi,and hence f = g.

Recall that the field k(X) is always finitely generated.

Corollary 7.6. The functor from the opposite category of irreducible quasi-affine varieties with dominantmorphisms over k to the category of finitely generated field extensions of k with homomorphism over k is anequivalence.

Proof. Suppose K/k is a finitely generated field extension. We need to find an irreducible quasi-affine varietyX such that k(X) ' K. Suppose K = k(f1, . . . , fn). Then let A = k[f1, . . . , fn] ⊂ K. Then A is a finitelygenerated domain as a subring of a field. Therefore, there exists an irreducible affine X such that k[X] ' Abecause of the equivalence between affine varieties and finitely generated reduced algebras. Then

K = Quot(A) = Quot(k[X]) = k(X).

We will write X ≈ Y if they are birationally isomorphic. Note that X ' Y implies X ≈ Y . However,if U ⊂ X is open, then U ≈ X, but it is not necessarily true that U ' X since A1 \ {0} 6≈ A1 since theircoordinate rings are k[t, t−1] and k[t], which are not isomorphic since any isomorphism must send constantsto constants.

Example 7.7. Consider X given by x3 = y2,

X ��

// A2

~~

A1

OO

where the maps are t 7→ (t2, t3) and (x, y) 7→ yx . Then X ≈ A1 but X 6' A1.

Proposition 7.8. Two irreducible varieties X and Y are birationally isomorphic if and only if there existnonempty open subsets U ⊂ X and V ⊂ Y such that U ' V .

Proof. For the backward direction, we have X ≈ U ' V ≈ Y . For the forward direction, suppose X ≈ Y bymaps f : U ′ → Y and g : V ′ → X for U ′ and V ′ open in X and Y , respectively. We get a map

f−1(V ′)f−→ V ′

g−→ X

and

g−1(U ′)g−→ U ′

f−→ Y

Both these maps are the identity, so these composites are inclusions, so g(f(x)) = x for all x ∈ f−1(V ′) andf(g(y)) = y for all y ∈ g−1(U ′).

But X ⊃ U := f−1(g−1(U ′)) ⊂ f−1(V ′) and Y ⊃ V := g−1(f−1(V ′)) ⊂ g−1(U ′). If x ∈ U ⊂ f−1(V ′),then g(f(x)) = x. Applying f yields f(g(f(x))) = f(x) ∈ V . But f(x) = f(g(f(x))) ∈ V ′ which implies theprevious sentence.

Similarly, if y ∈ V , then g(y) ∈ U , and f(g(y)) = y, so we get inverse isomorphisms f and g between Uand V .

Proposition 7.9. Every irreducible variety is birationally isomorphic to a hyperplane in some An.

14

Proof. Let X be an irreducible variety. Take K = k(X)/k which is a finitely generated field extension. Wecan find a sequence of extensions

K/k(t1, . . . , tm)/k

where K is finite and separable over k(t1, . . . , fm) (it is separable since k is algebraically closed, henceperfect), and the second purely transcendental. Denote E = k(t1, . . . , tm). So K = E(α) for some α ∈ K.Let f ∈ E[t] be the minimal polynomial of α. So we find g ∈ k[t1, . . . , tm, t] = R[t] where R = k[t1, . . . , tm]is a UFD where g has relatively prime coefficients in R. So f is primitive.

If g is irreducible in E[t], primitive in R[t] and R is a UFD, then this implies g is irreducible inF [t1, . . . , tm, t]. So ZAm+1(g) ⊂ Am+1 is irreducible. Also,

k(Y ) = Quot(k[Y ]) = QuotF [t1, . . . , tm, t]/(g) = QuotE[t]/(f) = QuotE(α) = E(α) = K = k(X)

and so X ≈ Y .

Example 7.10. Let X = Z(x2 + y2 − 1) ⊂ A2. Then X ≈ A1 via the maps t 7→(

2t1+t2 ,

t2−11+t2

)and

(x, y) 7→ 1+yx .

8 Local Ring of a Subvariety

Let X be a quasi-affine variety, Y ⊂ X a closed irreducible subvariety. A neighborhood of Y in X is an openset U ⊂ X such that U ∩ Y = ∅. If U1 and U2 are two neighborhoods of Y , then so if U1 ∩ U2, since Y isirreducible.

Let (U, f) denote U a neighborhood of Y , and f ∈ k[U ]. Define an equivalence relation such that(U1, f1) ∼ (U2, f2) if there exists a neighborhood W of Y in X such that W ⊆ U1 ∩ U2 and f1 and f2 agreeon W . The set of classes [U, f ] forms a k-algebra denoted OX,Y .

Example 8.1. If X is irreducible, then OX,X = {[U, f ]} = k(X).

Define a map OX,Yπ−→ k(Y ) : [U, f ] 7→ [U ∩ Y, f |U∩Y ]. Define kerπ = mX,Y , which is an ideal in OX,Y .

Proposition 8.2. 1. The map π is surjective.

2. The ring OX,Y is a local ring with maximal ideal mX,Y and residue field k(Y ).

Proof. Let g ∈ k(Y ), so g = [W,h] where W = U ∩ Y and U is open in X. Let y ∈ W . There exists anaffine neighborhood U ′ of y ∈ U ′ ⊂ U . So U ′ ∩ Y 6= ∅, and thus U ′ is a nieghborhood of Y in X. ThenW ′ = U ;∩Y is closed in U ′, so W ′ ⊂ U ′ is affine and closed. The map k[U ′]→ k[W ′] is surjective so thereexists f ∈ k[U ′] such that f |W ′ = h|W ′ . Then

π[U ′, f ] = [W ′, h|W ′ ] = g.

For the second claim, since k(Y ) is a field, mX,Y is a maximal ideal in OX,Y . Let [U, f ] ∈ OX,Y \mX,Y ,so f |U∩Y is not zero. Thus DU (f) ∩ Y 6= ∅, so DU (f) is a neighborhood of Y in X. So

[DU (f), f−1] · [DU (f), f ] = 1

but [U, f ] = [DU (f), f ], so [U, f ] is a unit. The claims follow.

Let g : Y ⊂ X → Y ′ ⊂ X ′ be a dominant regular map, with Y ⊂ X closed and irreducible. Consider themap g∗ : OX′,Y → OX,Y . Let [U ′, f ′], so f ′ ∈ k[U ′], and thus g ◦ f ′ ∈ k[g−1(U ′)]. Since g is dominant sothat g(Y ) is dense in Y ′, we have g(Y ) ∩ U ′ 6= ∅, so g−1(U ′) ∩ Y 6= ∅, so g−1(U ′) is a neighborhood of Y inX. Then

g∗[U ′, f ′] = [g−1(U ′), g ◦ f ′].

Also, g∗(mX′,Y ′ ⊂ mX,Y , g∗ is a local homomorphism of local rings.Let Y ⊂ X be a closed and irreducible subvariety. What is OX,Y . We have a map

α : k[X]→ OX,Y : f 7→ [X, f ].

15

We have mX,Y ⊂ OX,Y is a maximal ideal, and α−1(mX,Y ) = {f ∈ k[X] : f |Y = 0} = I(Y ) which is a primeideal in k[X].

If g ∈ k[X] \ I(Y ), then α(g) ∈ O×X,Y . By the universal property of localization, α extends to a uniquering homomorphism

α : k[X]I(Y ) → OX,Y :f

g7→ α(f)

α(g).

Proposition 8.3. α is an isomorphism.

Proof. Suppose α( fg ) = 0. Then 0 = [X, f ] ∈ OX,Y . So f |U = 0 for a neighborhood U of Y in X. We need

to show fg − 0 in k[X]I(Y ). Pick y ∈ Y ∩ U . Then X \ U is closed in X, and does not contain y. So there

exists h ∈ k[X] such that h|X\U = 0 but h(y) 6= 0. So h /∈ I(Y ). So fh = 0 in k[X], and thus fg = 0 in

k[X]I(Y ). This gives injectivity.Now take [U, f ] ∈ OX,Y , with U a neighborhood of Y in X, and f a regular function on U . So pick

y ∈ Y ∩ U . In a neighborhood W of y in U , f = gh where g, h ∈ k[X] h(y) 6= 0. But

[U, f ] = [W, f |W ] =[W, g|W ]

[W,h|W ]=

[X, g]

[X,h]=α(g)

α(h)= α

( gh

)∈ Im(α).

Example 8.4. 1. Let X = Z(xy) ⊂ A2, the union of the two axes. So X is reducible. Let Y = {0}. Butk[X] = k[x, y]/(xy) ⊃ (x, y)/(xy) = I(Y ). Then OX,Y = k[X]I(Y ) is not a domain since x · y = 0 butx 6= 0 and y 6= 0. This reflects the fact that there are two irreducible components passing through Y .

2. Let X = Z(y2−x3−x2) ⊂ A2. Put Y = {(0, 0)}. Then k[X] = k[x, y]/(y2−x3−x2), but y2−x3−x2is irreducible, so X is irreducible, and I(Y ) = (x, y)/(y2−x3−x2). So OX,Y = k[X]I(Y ) is a domain.This reflects the fact that their is one irreducible component passing through Y .

Let Y ⊂ X be closed, with both X and Y irreducible. Then we have

k[X] ⊂ OX,Y ⊂ k(X) = Quot(OX,Y )

so OX,Y = {f ∈ k(X) : f is regular in a neighborhood of Y in X}. This is the same as {f ∈ k(X) :f is regular at a point y ∈ Y }. If Y = {y}, then OX,Y = OX,y = {f ∈ k(X) : f is regular at y}.

Now suppose Y ⊂ Z ⊂ X with Y and Z both closed and irreducible. We have that

ker (OX,Y → OX,Z → k(Z))

is a prime ideal in OX,Y . The map is just given by [U, f ] 7→ [U, f ]. The ideal is {[U, f ] : f |Z = 0} = I(Z).Then the closed irreducible subsets Z in X containing Y is in correspondence with the prime ideals in

OX,Y given by Z 7→ I(Z).

Proposition 8.5. This map is a bijection.

Proof. Idea: Let U ⊂ X be a neighborhood of Y in X. Consider the collection of closed irreducible subsetsin U containing U ∩ Y . This is the same as the closed irreducible subsets Z in X containing Y . Thiscorresponds to prime ideals in OU,U∩Y , but OX,Y = OU,U∩Y .

Now choose U affine. We may assume that X is affine. Then the collection of closed irreducible Y ⊂Z ⊂ X are in correspondence with the prime ideals p ⊂ k[X] with p ⊂ I(Y ). By commutative algebra, theseprime ideals are in correspondence with the prime ideals in k[X]I(Y ) = OX,Y . The composition of these twomaps is checked to be the same as the map above.

This map is inclusion reversing, so in particular, Y corresponds to mX,Y . The irreducible components ofX containing Y correspond to minimal prime ideals in OX,Y . There is only one irreducible component of Xcontaining Y iff OX,Y is a domain, because in that case (0) is the only minimal prime.

Suppose X and Y are quasi-affine varieties. Suppose X,Y ⊂ An. Consider X ' X × {0} ⊂ An+1 andY ' Y × {1} ⊂ An+1. Then write X + Y , the disjoint union of X and Y , or sum of varieties, to be

X + Y = X × {0} ∪ Y × {1}.

Finally, k[X + Y ] = k[X]× k[Y ].

16

9 Quasi-projective varieties

9.1 Quasi-projective Sets

Recall that if we have a map from a compact space to a Hausdorff space, then its image is always closed.However, the projection X = Z(xy − 1) → A1 has image which is open but not closed. If X = X ∪ {∞},then we can map ∞ to the origin. These varieties are called projective. So projective varieties are “truecompact.”

We will construct a full and faithful functor QAVar→ QPVar.Consider An+1 with coordinates (t0, . . . , tn). A subset Y ⊂ An+1 is called a cone if the following holds:

If y ∈ Y and c ∈ k, then cy ∈ Y . So note ∅ and {0} are cones, and any nonempty cone must contain 0, so0 ∈ Y ⇐⇒ Y 6= ∅.

Definition 9.1. A polynomial F ∈ k[t0, . . . , tn] is called homogeneous of degree d if F (ct) = cdF (t) for everyc ∈ k. Equivalently, every monomial in F has degree d.

Definition 9.2. An ideal I ⊂ An+1 = k[t0, . . . , tn] is called homogeneous if F ∈ I implies Fi ∈ I for all i,where Fi is the degree i homogeneous component of F .

Proposition 9.3. 1. An ideal I ⊂ An+1 is homogeneous iff I is generated by homogeneous polynomials.

2. If I is homogeneous, then so is√I.

Lemma 9.4. A closed subset Y ⊂ An+1 is a cone iff I(Y ) ⊂ An+1 is homogeneous.

Proof. For the forward direction, suppose Y is a cone, and F ∈ I(Y ). Write F =∑Fi. We need to show

Fi ∈ I(Y ) for all i. We have F (cy) = 0 for all c ∈ k and all y ∈ Y , since cy ∈ Y . But

F (cy) =∑i

ciFi(y)

which implies Fi(y) = 0 for any i and any y ∈ Y , so Fi ∈ I(Y ). To see this, fix y ∈ Y . Then F (Xy) =∑i Fi(y)Xi ≡ 0 is the zero polynomial in the polynomial ring k[X], so necessarily Fi(y) = 0 for all i.Conversely, suppose y ∈ Y and c ∈ k. But I(Y ) is generated by a set of homogeneous polynomials. If F

is a homogeneous generator,F (cy) = cdF (y) = 0

which implies G(cy) = 0 for all G ∈ I(Y ) which implies cy ∈ ZI(Y ) = Y . So Y is a cone.

We get a correspondence between the collection of closed cones in An+1 and the category of radicalhomogeneous ideals in An+1 given by the maps I and Z. As usual, ∅ ↔ An+1 and {0} ↔ 〈t0, . . . , tn〉.

The unit group k× acts on An+1 \ {0} by c · (a0, . . . , an) = (ca0, . . . , can). Then

Pnk = Pn = (An+1 \ {0})/k×

is the projective space. The elements are classes [a0 : · · · : an] with not all ai are zero. So note [a0 : · · · :an] = [b0 : · · · : bn] iff there exists c ∈ k× such that bi = cai for all i. Equivalently, aibj = ajbi for any i andj.

We have a canonical map π : An+1 \ {0} → Pn defined by (a0, . . . , an) 7→ [a0 : · · · : an].If X ⊂ Pn, then π−1(X) is a cone in An+1.We say X ⊂ Pn is closed (resp. open) if π−1(X) ⊂ An+1 \ {0} is closed (resp. open). Let S ⊂ An+1 be

a set of homogeneous polynomials. Then Z(S) = {a ∈ Pn : F (a) = 0, for all F ∈ S}. All closed sets takethis form. If I ⊂ An+1 is a homogeneous ideal, then Z(I) = Z(S), where S is the set of all homogeneouspolynomials in I.

If F is a homogeneous polynomial, then D(F ) = DPn(F ) = {a ∈ Pn : F (a) 6= 0} are principal opens.We get a correspondence of closed sets in Pn and the radical homogeneous ideals in An+1 different from

the ideal corresponding {0}, that is, 〈t0, . . . , tn〉. Then ∅ ↔ An+1 and Pn ↔ 0.

Theorem 9.5. (Projective Nullstellensatz) Let I ⊂ An+1 be a homogeneous ideal. Then Z(I) = ∅ iff〈t0, . . . , tn〉m ⊂ I for some m > 0.

17

Proof. Note ZPn(I) = ∅ iff ZAn+1(I) ⊂ {0} = ZAn+1(〈t0, . . . , tn〉). This holds iff 〈t0, . . . , tn〉 ⊂√I.

We have Pn = π(An+1 \ {0}) is Noetherian and irreducible as the image of a Noetherian, irreduciblespace (the punctured space is an open subset of an irreducible space). So every subset of Pn is compact andadmits a unique decomposition into a union of irreducible closed subsets.

A quasi-projective subset in Pn has the form U ∩ Z for U open and Z closed. A subset X ⊂ Pn isquasi-projective iff X is open in X.

9.2 Regular functions

Note that if x ∈ Pn, then FG (x) ∈ k if F and G are homogeneous of the same degree.

So let X ⊂ Pn be a quasi-projective set, x ∈ X. A function f : X → k is called regular at x if there arehomogeneous polynomials F and G in An+1 of the same degree such that G(x) 6= 0 and f(y) = F

G (y) for ally in a neighborhood of x. (So G(y) 6= 0 for all y in this neighborhood.) We say f is regular if f is regularat every x ∈ X. We write k[X] for the k-algebra of all regular functions on X.

Example 9.6. We have k[Pn] = k. We need to show every regular function is constant. Suppose f ∈ k[Pn] isnot constant. So there exists x ∈ Pn such that f = F

G in a neighborhood U of x and F and G are homogeneousnonconstant polynomials of the same degree. Assume F and G are coprime. Then Z(G) 6⊆ Z(F ) becauseapplying I(−) implies

√〈G〉 ⊃

√〈F 〉, but F /∈

√〈G〉 since they are coprime. For if Fm ∈ 〈G〉, then G | Fm,

contradiction. So there exists x′ ∈ Pn such that G(x′) = 0 but F (x′) 6= 0. In a neighborhood U ′ of x′, write

f = F ′

G′ where F ′ and G′ are homogeneous (these are not derivatives!).

So in the dense set U ∩ U ′ (since Pn is irreducible), we have FG = F ′

G′ so FG′ = F ′G holds as equality ofpolynomials. So F (x′)G(x′) = F ′(x′)G(x′) = 0. But F (x′) 6= 0, and G′(x′) 6= 0, so we have a contradiction.

Some properties:

1. If f ∈ k[X] and Y ⊂ X is a quasi-projective set, then f |Y ∈ k[Y ].

2. if X =⋃Uα is an open cover, and f : X → k, then f ∈ k[X] ⇐⇒ f |Uα ∈ k[Uα] for all α.

3. if f : X → k is regular at x ∈ X, then f is regular in a neighborhood of x.

4. If f ∈ k[X], then {x ∈ X : f(x) = 0} is closed. So f is continuous as a map X → A1.

5. If f ∈ k[X] and f(x) 6= 0 for all x ∈ X, then 1/f ∈ k[X].

Suppose f : X → Y ⊂ Pm. Take x ∈ X. We say f is regular at x if there exists a neighborhood U ofx and functions f0, f1, . . . , fm ∈ k[U ] such that fi is nowhere 0 on U for some i, with the property thatf(x′)] = [f0(x′) : f1(x′) : · · · : fm(x′)] for all x′ ∈ U . We say f is regular if f is regular at every x ∈ X.

As before, f is regular if and only if Xf−→ Y ↪→ Pm is regular.

We can write fi = GiH at a neighborhood of x, where Gi and H are homogeneous polynomials of the same

degree d. We don’t subscript H since we can find a common denominator. Then

f =

[G0

H:G1

H: · · · : Gm

H

]= [G0 : · · · : G1 : · · · : Gm].

Example 9.7. 1. If f : X → Y is regular, X ′ ⊂ X and Y ′ ⊂ Y such that f(X ′) ⊂ Y ′, then f |X′ : X ′ →Y ′ is also regular.

2. If X ⊂ Y , then X ↪→ Y is regular.

Example 9.8. 1. Let L1, L2, . . . , Lk be linearly independent linear forms on Pn. Then Z := Z(L1, L2, . . . , Lk) ⊂Pn, define

Pn \ Z → Pk−1 : x 7→ [L1(x) : · · · : Lk(x)]

This is called the projection with center Z. When k = n, then Z is a point, so if we take Li = ti thecoordinate function, then Z = {[1 : 0 : · · · : 0]}. Thus we get a projection Pn \ {∗} → Pn−1.

18

2. Fix m,n. Consider the monomials tα00 · · · tαnn where

∑αi = m. There are

(n+mm

)such monomials.

Write α = (α0, α1, . . . , αn). Introduce new variables Sα. Define

vn,mPn → Pdn,m

where dn,m =(n+mn

)− 1, where vn,m([t0 : t1 : · · · : tn]) = [Sα], where Sα = tα = tα0

0 tα11 · · · tαnn , where

deg(α) =∑αi = m. Note if α = (0, . . . ,m, · · · 0) then Sα = tmi . If ti 6= 0, then Sα 6= 0. This is the

Veronese map.

Put Z = Z(SαSβ−SγSδ : α+β = γ+δ) ⊂ Pdn,m . So Im(vn,m) ⊂ Z. Claim: vn,m gives an isomorphismPn → Z. We will construct g : Z → Pn.

Take g([Sα]) = [S(β))+1,β1,...,bn) : S(β0,β1+1,...,βn) : · · · : S(β0,...,βn+1) where deg(β) = m− 1. We need to

show that is independent of the choice of β, and for every [Sα] ∈ Pdn,m , there exists β of degree m− 1such that one of the coordinates is not zero.

Consider the case where n = 1 and m = 2. The monomials are t20, t0t1, t21. We get a map v1,2P1 → P2

defined by [t0, t1] 7→ t20 : t0t1, t21] =: [s0 : s1 : s2]. So Z = Im(v1,2) = Z(s21 − s0s2) ⊂ P2. This is a conic

in P2 given by s21 = s0s2, so P1 ' Z. The map g : Z → P1 sends [s0 : s1 : s2] 7→ [s0 : s1] = [s1 : s2].

Proposition 9.9. A regular map f : X → Y ⊂ Pm is continuous. It suffices to show that f−1(Z(G))is closed in X where G is a homogeneous polynomial since any closed subset is an intersection of closedsets given by one polynomials, and preimages behave well with intersections. Take x ∈ f−1(Z(G)). In aneighborhood U of x, f = [f0 : f1 : · · · : fm], where fi ∈ k[U ], where one is nowhere zero on U . So

f−1(Z(G)) ∩ U = {y ∈ U : G(f0, . . . , fm)(y) = 0}

where G(f0, . . . , fm) is regular on U , hence this set is closed in U . So f−1(Z(G)) is locally, therefore locallyclosed.

Corollary 9.10. If f : X → Y is regular, Y ′ ⊂ Y is quasi-projective, then f−1(Y ′) is also quasi-projective.

Lemma 9.11. Let f : X → Y be a regular map, g ∈ k[Y ]. Then g ◦ f ∈ k[X].

Proof. Take x ∈ X, y = f(x). Since g is regular, g = FG in a neighborhood W of y in Y , where G is nowhere

zero on W . In a neighborhood U of x in X, f = [f0 : · · · : fn], with fi ∈ k[U ], one is nowhere zero on U . InU ∩ f−1(W ) is a neighborhood of X, and

g(f(y)) =F (f0, . . . , fn)(y)

G(f0, . . . , fn)(y)= F (f0, . . . , fn)(y) · 1

G(f0, . . . , fn)(y)

which is regular on U ∩ f−1(W ). Hence this is regular in a neighborhood of x, hence is regular.

So f : X → Y regular gives a map f∗ : k[Y ]→ k[X] by f∗(g) = g◦f , which is a k-algebra homomorphism.

Proposition 9.12. Let f : X → Y ⊂ Pm be a map of quasi-projective sets. Then f is regular if and only iffor every open V ⊂ Y and every g ∈ k[V ], the composition g ◦ f |f−1(V ) ∈ k[f−1(V )].

Proof. For the forward direction, f |f−1(V ) : f−1(V )→ V is regular, hence g ◦f |f−1(V ) is regular. Conversely,take x ∈ X and y = f(x) = [y0 : · · · : ym] with yk 6= 0. We may assume that yk = 1. Let s0, . . . , sm bethe coordinates of Pm. Then V = DY (sk) is a neighborhood of y in Y . Write f = [f0 : f1 : · · · : fm] in aneighborhood U of x in X. Dividing all by fk, we may assume that fk ≡ 1 in that neighborhood U . We needto show that all fi are regular at x. Take g = Si

Sk∈ k[V ]. Therefore, g ◦ f |f−1(V ) is regular by assumption.

In f−1(V ) ∩ U , g ◦ f = fi so is regular at x.

Corollary 9.13. The composition of two regular maps is regular.

We will construct a fully faithful functor

θ : QAffVar(k)→ QProjSets(k).

19

We have An with coordinates T1, . . . , Tn and Pn with coordinates S0, . . . , Sn. We define U = DPn(S0) ⊂Pn open.

We construct a map An g−→ U by (T1, . . . , Tn) 7→ [1 : T1 : · · · : Tn]. We get an inverse map by [S0 : S1 :

· · · : Sn] 7→(S1

S0, . . . , SnS0

). I claim that g is a homeomorphism.

Take F ∈ k[T1, . . . , Tn] and G = F(s1s0, . . . , sns0

)· sk0 for sufficiently large k so that G is a polynomial.

Then g(ZAn(F )) = ZU (G).Conversely, given G homogeneous in k[S0, S1, . . . , Sn], define F = G(1, T1, . . . , Tn) ∈ k[T1, . . . , Tn]. Then

g−1(ZU (G)) = ZAn(F ). If V ⊂ An is an open set and h : g(V )→ k a function, then h is regular if and onlyif h ◦ (g|V ) : V → k is regular. Locally, we may write h = K

L on g(V ), where K and L are homogeneous ofthe same degree. Then

h ◦ g =K(1, T1, . . . , Tn)

L(1, T1, . . . , Tn)

is regular. If h ◦ g = FH locally, where F,H ∈ k[T1, . . . , Tn], then

h =F ( s1s0 , . . . ,

sns0

) · sk0H( s1s0 , . . . ,

sns0

) · sk0

is regular.Then θ is the assignment given by X ⊂ An 7→ g(X) ⊂ U ⊂ Pn. If f : X → Y , with X ⊂ An and Y ⊂ Am,

we have gn(X) ⊂ Un ⊂ Pn, and gm(Y ) ⊂ Um ⊂ Pm. We have a diagram

θ(f) : gn(X)� _

��

g−1n // X� _

��

f// Y � _

��

gm // gm(Y )� _

��

Un� _

��

An Am Um� _

��

An Am

We claim θ is full and faithful. We have

f : Xgn // gn(X)

θ(f)// gm(Y )

g−1m // Y

So QAffVar(k) is a full subcategory of QProjSets(k).

Definition 9.14. A quasi-projective variety is a quasi-projective set. A quasi-affine variety is a quasi-projective variety isomorphic to a quasi-affine set. A projective variety is isomorphic to a closed set inPn.

We have the inclusionsAffVar ⊂ QAffVar ⊂ QProjVar

Also, if ProjVar lies in QProjVar and ProjVar ∩ AffVar consists of finite sets.Some properties:

1. Every quasi-projective variety admits an open affine finite cover.

Proof. We have Pn =⋃ni=0DPn(Si). But DPn(Si) ' An, so is affine. In general, if X ⊂ Pn is

quasi-projective, then

X =

n⋃i=0

(DPn(Si) ∩X)

and each member of this intersection is a quasi-affine variety, and every quasi-affine variety admits anopen affine finite cover.

20

2. A map f = (f1, . . . , fn) : X → An is regular if and only if all fi are regular functions.

3. k[DPn(Si)] = k[ s0si , · · · .si−1

si, si+1

si, . . . , snsi ]. To generalize, let L be a nonzero linear form on si, so

L =∑aisi for ai ∈ k. Then

k[DPn(L)] = k[s0L,s1L, . . . ,

snL

]

with∑aiSiL = 1.

Let F ∈ An+1 be homogeneous, and nonconstant, so deg(F ) > 0. Consider

DPn(F ) = {x ∈ Pn : F (x) 6= 0} = Pn \ Z(F ).

If F is linear, then DPn(F ) ' An, which is affine.

Proposition 9.15. The variety DPn(F ) is always affine.

Proof. Consider the Veronese map vn,m : Pn → Pdn,m . Then Z = Im(vn,m) ⊂ Pdn,m is closed, and Pn ' Z.Recall if α = (α0, . . . , αn), then |α| =

∑αi, and Sα = Sα0

0 · · ·Sαnn . Let

F =∑|α|=m

aαSα

for aα ∈ k. Recall vn,m([S0 : · · · : Sn]) = ([Sα]). If Tα are the coordinates of Pdn,m , then this map isTα = Sα. Let L =

∑|α|=m aαTα, which is linear on Pdn,m . Then

vn,m(DPn(F )) = DZ(L)

and moreover,DPn(F ) ' DZ(L) = Z ∩DPdn,m (L).

Since Z is closed, this intersection is closed in DPdn,m (L) ' Adn,m . Hence it is affine.

Example 9.16. Take X = Z(s1s2 − s20) ⊂ P2. We get a map

X'−→ P1 : [s0 : s1 : s2] 7→ [s0 : s1] = [s2 : s0].

Consider DX(s0). Let T1 = s1s0

and T2 = s2s0

, so T1T2 = 1. Then DX(s0) = ZA2(T1T2 − 1). Also,

ZX(s0) = X \DX(s0) = {[0 : 1 : 0], [0 : 0 : 1]}.

Call these points x1 and x2.Let the coordinates of P1 be w0, w1. Then DP1(w0) = A1. The coordinate on the affine line is u = w1

w0.

We have the map[1 : T0 : T1] 7→ [1 : T0] ∼ T0 ∈ A1

so the map X → P1 restricts to the map DX(s0) → A1 = DP1(w0). This gives the projection map from thehyperbola to the affine line. But P1 \DP1(w0) = {[0 : 1]}, call this point ∞. So if f : X → P1 is the map, wehave f(x1) =∞ and f(x2) = 0 ∈ A1.

10 Rational Maps and Functions

Let X be an irreducible, quasi-projective variety. For a pair (U, f) with U ⊂ X nonempty and open, f ∈ k[U ],we say (U, f) ∼ (U ′, f ′) if f and f ′ agree on U ∩ U ′. Denote by F (X) the field of rational functions on X,which are equivalence classes of [U, f ].

As before, we get a correspondence between the opposite category of irreducible quasi-projective varietieswith morphisms dominant rational maps and the category of finitely generated extensions of k, by theassignment X 7→ F (X) and f 7→ f∗.

The category of irreducible quasi-affine varieties with dominant rational maps is an equivalence with boththe above categories, so the functor in the previous paragraph is actually an equivalence.

21

11 Local Ring of a Subvariety

Let X be a quasi-projective variety, and Y ⊂ X closed and irreducible subvariety. Consider (U, f) to be aneighborhood of Y (i.e., U ∩ Y 6= ∅), and f ∈ k[U ]. Denote by OX,Y the local ring of equivalence classes. IfU is a nieghborhood of Y in X, then OX,Y ' OU,Y ∩U , so we may reduce to the affine or quasi-affine casefor the study of local objects.

12 Product of Quasi-projective Varieties

The main point is that Pn × Pm 6' Pn+m, although An × An = An+m. Let Si 0 ≤ i ≤ n be the coordinatesfor Pn and Tj for 0 ≤ j ≤ m the coordinates for Pm. Consider also Pk where k = (n + 1)(m + 1) − 1 withcoordinates wij .

Define a mapϕ : Pn × Pm → Pk : ([Si], [Tj ]) 7→ [wij ]

where wij = SiTj . This is well-defined because if we scale all Si, we scale all wij , likewise if we scale the Tj .Write V = Z(wijwkl − wilwkj) ⊂ Pk, which is closed. These equations can be rewritten as SiTjSkTl −

SiTlSkTj = 0, so Im(ϕ) ⊂ V .

Lemma 12.1. The map ϕ : Pn × Pm → V is a bijection.

Proof. We construct ψ : V → Pn × Pm. Take [wij ] ∈ V . Send [wij ] 7→ ([w0l : w1l : · · · : wnl], [wk0 : wk1 : · · · :wkm]). This is independent of k and l. If wkl 6= 0 and wkl 6= 0 in both points above. Then ψ = ϕ−1.

Define the product An×Pm = V ⊂ P(n+1)(m+1)−1. For example, P1×P1 ⊂ P3. The map ϕ : Pn×Pm → Pkis called the Segre embedding. We have canonical projections to Pn and Pm, which are regular.

Let Fr(S, T ) be homogeneous in S and in T , of possibly different degrees. Then Z(Fr(S, T )) ⊂ Pn×Pm ⊂Pk is closed. Since wij = SiTj , if Fr is homogeneous in S and T of the same degree, then Fr(S, T ) = Gr(W )by replacing the SiTj with wij . Suppose that degS(Fr) = a and degT Fr = b with a < b. Replace Fr(S, T )with Sb−ai Fr(S, T ). These polynomials defined the same closed set since at least one of the Si is nonzero.

Consider Pn×Pm. Take X = Z(S) ⊂ Pn and Y = Z(T ) ⊂ Pm closed. Then X×Y = Z(S∪T ) ⊂ Pn×Pmis closed.

IF X ⊂ Pn and Y ⊂ Pm are quasi-projective, say X = Z1 \ Z ′1 and Y = Z2 \ Z ′2, then

X × Y = (Z1 × Z2) \ (Z1 × Z ′2 ∪ Z ′1 × Z2) ⊂ Pn × Pm

is quasi-projective. Similarly, if X ⊂ Pn and Y ⊂ Pm are open, then X × Y ⊂ Pn × Pm is open by takingZ1 = Pn and Z2 = Pm.

In fact, X × Y is the categorical product in QProjVar(k). If X,Y are quasi-affine, then they are quasi-projective so in either category, X × Y is well-defined.

Some properties

1. If X and Y are affine, then so is X × Y .

2. If X and Y are projective, then so is X × Y .

3. If X ′ ⊂ X and Y ′ ⊂ Y is closed (resp. open), then X ′ × Y ′ ⊂ X × Y is closed (resp. open).

4. If f : X → X ′ and g : Y → Y ′ are regular, then so is f × g : X × Y → X ′ × Y ′. The components of

f × g are X × Y p−→ Xf−→ X ′ and X × Y q−→ Y

Y ′−→. Now use the universal property.

5. If X and Y are irreducible, then so is X × Y . (Same proof.)

6. The natural map ϕX,Y : k[X]⊗k k[Y ]→ k[X × Y ] is an isomorphism.

22

Proof. We know that when X and Y are affine. The proof that ϕX,Y is injective is the same. Choosebasis {gi} a basis for k[Y ]. Every element in k[X] ⊗k k[Y ] can be uniquely written

∑i fi ⊗ gi, for

fi ∈ k[X]. Claim: Suppose X =⋃Uα is an open cover. If ϕUα,Y is an isomorphism for all α, so

is ϕX,Y . To see this, Take h ∈ k[X × Y ]. Now h|Uα×Y ∈ Im(ϕUα×Y implies there exists a uniquefα,i ∈ k[Uα] such that

h|Uα×Y (x, y) =∑i

fα,i(x)gi(y) = ϕUα,Y (∑

fα,i ⊗ gi)

for all α. But

h|(Uα∩Uβ)×Y (x, y) =∑i

fα,i|Uα∩Uβ (x)·gi(y) =∑i

fβ,i|Uα∩Uβ (x)·gi(y) =∑i

fα,i|Uα∩Uβ⊗gi =∑i

fβ,i|Uα∩Uβ⊗gi

By uniqueness, fα,i|Uα∩Uβ = fβ,i|Uα∩Uβ so there exists unique fi ∈ k[X] such that fi|Uα = fα,i. Then

ϕX,Y (∑i

fi ⊗ gi)|Uα×Y = ϕUα×Y (∑i

fi|Uα ⊗ gi) = ϕUα×Y (∑i

fα,i ⊗ gi) = hUα×Y

Thus ϕX,Y (∑i fi ⊗ gi) = h.

If X is anything, and Y is affine, then ϕX,Y is an isomorphism. Let X =⋃Uα be an open affine cover.

Then ϕUα,Y is an isomorphism. By the claim, ϕX,Y is an isomorphism.

If X and Y are arbitrary, take Y =⋃Wβ be an open affine cover. Then ϕX,Wβ

is an isomorphism. Bythe claim, ϕX,Y is an isomorphism.

7. Suppose X ⊂ Pn and Y ⊂ Pm. Then a function on X × Y is regular at (x, y) is equal to F (S,T )G(S,T ) is a

neighborhood U of (x, y), where F and G are homogeneous in S and T of the same degree, with G 6= 0everywhere in U .

8. The diagonal ∆X ⊂ X × X is closed. Note if X ⊂ Pn, then ∆X = ∆Pn ∩ (X × X). Now ∆Pn =Z(sis

′j − sjs′i) ⊂ Pn × Pn is closed. Here S and S′ are the coordinates on the factors of Pn in the

product.

9. Consider Pn and Am with coordinates si and tj . Every closed set in Pn×Am is given by the equationsF (s, t) = 0 where F is a polynomial, homogeneous in S. Embed Am ⊂ Pm. A closed set in Pn×Am =(Pn ×Am)∩Z where Z is closed in Pn × Pm which is given by G(S, T ) = 0, homogeneous in S and T .Do the substitution T0 = 1 and Ti = ti for i > 0.

10. If f : X → Y is regular Γ(f) = {(x, f(x)) ∈ X × Y } which is closed. The projection p : Γ(f) → X isan isomorphism.

Proposition 12.2. Let X be a quasi-projective variety, and Y,Z ⊂ X are subvarieties. Then if both Y andZ are affine, then so is Y ∩ Z.

Proof. NoteY ∩ Z ' (Y × Z) ∩∆X ⊂ X ×X

Observe (y, z) ∈ ∆X ⇐⇒ y = z ∈ Y ∩ Z. But (Y × Z) ∩ ∆X is closed in Y × Z, which is affine as theproduct of affine varieties, hence it is affine since any closed subset of an affine variety is affine.

13 Proper Maps

Lemma 13.1. The projection Pn × Am → Am takes closed sets to closed sets. (This fails if we replace theprojective space by affine space, think the projection of the hyperbola.)

23

Proof. Let S and t be the coordinates of Pn and Am, respectively. Let Z ⊂ Pn × Am so Z = Z(Gi(S, t))with Gi homogeneous in S. The image p(Z) ⊂ Am, we need to show it is closed.

Observe

p(Z) = {x ∈ Am : the system Gi(S, x) = 0 has a solution [S0 : S1 : · · ·Sn] ∈ Pn}.

Butp(Z) = {x ∈ Am : 〈Gi(S, x)〉 6⊃ 〈S0, . . . , Sn〉k for all k}

Let Ix = 〈Gi(S, x)〉 ⊂ k[S0, . . . , Sn]. For each k > 0, let

Yk = {x ∈ Am : Ix 6⊃ 〈S0, . . . , Sn〉k} ⊂ Am

Thus p(Z) =⋂k Yk. Is it closed? It suffices to show that every Yk is closed in Am.

Let Gi = (S, T ), and degS(Gi) = di, x ∈ Am, and Ix = 〈Gi(S, T )〉 ⊂ k[S0, . . . , Sn]. Is Yk = {x ∈ Am :Ix 6⊃ 〈S0, . . . , Sn〉} closed in Am? Note

Ix ⊃ 〈S0, . . . , Sn〉k ⇐⇒ Ix contains all monomials in S of degree k.

This holds iff every monomical of degree k is equal to∑iGi(S, x)︸ ︷︷ ︸

di

Fi(Z)︸ ︷︷ ︸k−di

.

So Fi(S) is a linear combination of monomials of degree k − di. The polynomials Gi(S, x), where N is amonomial of degree k − di, span the vector space Vk of all monomials of degree k.

Let Ax be the matrix of coefficients of all Gi(S, x)N , so rk(Ax) = dim(Vk). So x ∈ Yk iff rk(Ax) <dim(Vk) iff all dim(Vk) × dim(Vk) minors in Ax are zero. But the minors are polynomials in x, hence ink[t1, . . . , tm]. Therefore, Yk ⊂ Am is closed.

So the projection Pn × Am → Pn is a closed map.

Corollary 13.2. Let X be a projective variety and Y a quasi-projective variety. Then the projection p : X×Y → Y takes closed sets to closed sets.

Proof. Let X = Pn and Y be affine. So we may assume Y ⊂ Am is closed. So Z ⊂ Pn × Y ⊂ Pn × Am isclosed. By the lemma, p(Z) is closed in Am, but p(Z) ⊂ Y , so p(Z) is closed in Y .

Now let X = Pn and Y arbitrary. Let Y =⋃Uα, where Uα ⊂ Y are open and affine. Let Z ⊂ X × Y be

closed. So Z ∩ (X × Uα) is closed in X × Uα. By the first case, p(Z ∩ (X × Uα) is closed in Uα. This set isp(Z) ∩ Uα), so p(Z) is locally closed, hence closed.

Now suppose X is projective and Y is arbitrary. So suppose X ⊂ Pn closed. Then Z ⊂ X × Y ⊂ Pn× Yis closed. So p(Z) is closed in Y .

A regular map f : X → Y is called proper if for every quasi-projective variety T , the map f×IT : X×T →Y × T takes closed sets to closed sets.

Example 13.3. The map A1 → {p} is not proper. Consider A1 × A1 p2−→ {p} × A1 = A1. If X ⊂ A1 × A1

is the hyperbola, the image p(X) is A1 \ {0} is not closed.

Proposition 13.4. The map X → {p} is proper if and only if X is projective.

Proof. For the reverse direction, we have X × T p2−→ {p} × T = T , and by the corrolary p2 takes closed setsto closed sets.

Forward direction, let X ⊂ Pn, and let f : X → Pn be an embedding. Then Γf ⊂ X × Pn is closed. But

Γf = {(x, x) : x ∈ X}. Take the projection X × Pn π−→ Pn = {p} × Pn. Then π takes closed sets to closedsets. But π(X) = X is closed in Pn. Thus X is projective.

Some properties,

1. Closed embeddings are proper.

2. A composition of proper maps is proper.

24

3. If f : X → Y is proper, then so is f × 1Z : X × Z → Y × Z.

Proposition 13.5. Let f : X → Y be a regular map. If X is projective, then f is proper.

Proof. We can factor the map asX // X × Y // Y

where the first map is the closed embedding x 7→ (x, f(x)), and the second is the projection. The first isproper. The second map is proper by and Property 3. So by Property 2, the map is proper as the compositionof proper maps.

Corollary 13.6. 1. The image of a regular map X → Y with X projective is closed in Y .

2. A subvariety X ⊂ Pn is projective iff X is closed in Pn. We could define projective varieties as closedsubsets of Pn.

Proof. For the first statement, if X is projective, then f : X → Y is proper, and since X ⊂ X is closed, f(X)is closed in Y .

For the second, if X ⊂ Pn is closed, X is projective. If X is projective, consider the embedding f : X →Pn. By 1), f(X) = X is closed in Pn.

Recall k[Pn] = k. This generalizes to

Proposition 13.7. If X is irreducible and projective, then k[X] = k.

Proof. If f ∈ k[X], then f : X → A1 ↪→ P1. Then Im(f) is closed in P1, and Im(f) 63 ∞. But closed setsin P1 are P1 and finite sets of points. So Im(f) is finite, and irreducible, hence is a one point set. So f isconstant.

Proposition 13.8. If X is projective and quasi-affine, then X is finite.

Proof. Since X is a finite union of irreducible components, we may assume X is irreducible. Let fi : X ⊂Am → A1 be the ith coordinate function. Since X is projective and irreducible, so fi is constant. Thus Xis a point.

14 Dimension

Let X be an irreducible quasi-projective variety. Define dim(X) = trdeg(k(X)/k). If X is an arbitraryquasi-projective variety, hen X =

⋃Xi, for Xi irreducible. Then dim(X) = max dim(Xi). If X ≈ Y , then

k(X)/X ' k(Y ), so dim(X) = dim(Y ). So if U ⊂ X is open in X irreducible, then dim(U) = dim(X).

Example 14.1. 1. k(An) = k(t1, . . . , tn), so dim(An) = n = dim(Pn and An ⊂ Pn is open.

2. We have dim(X × Y ) = dim(X) + dim(Y ).

Proof. We may assume X and Y are irreducible and affine. We may find an intermediate fieldk(X)/E/k such that E/k is purely transcendental and k(X)/E is finite. Likewise we may findk(Y )/L/k such that L/k is purely transcendental, and k(Y )/L is finite. Then k[X×Y ] = k[X]⊗k k[Y ].So

k(X × Y ) = Quot k[X × Y ] = Quot(k[X]⊗k k[Y ]) = Quot(k(X)⊗k k(Y ))

We can write E = Quot(k[x1, . . . , xn]) and L = Quot(k[y1, . . . , ym]) so dim(X) = n and dim(Y ) = m.Then QuotE ⊗k L = Quot(k[x1, . . . , xn, y1, . . . , ym]) = k(x1, . . . , xn, y1, . . . , ym). So we have theextension k(X × Y ) = Quot(k(X)⊗k k(X))/Quot(E ⊗ L), which is finite.

Sodim(X × Y ) = trdeg(Quot(E ⊗ L) = n+m = dim(X) + dim(Y ).

25

3. We have dim(X) = 0 iff X is finite.

Proof. SupposeX is irreducible and affine. So ifX is finite, X is a point, and k(X) = k, so trdeg(k/k) =0. If dim(X) = 0, then k(X)/k is finite, but since k is algebraically closed, k(X) = k. Then k ⊂ k[X] ⊂k(X), so k[X] = k = k[point]. So X is a point by the correspondence between affine varieties andcoordinate rings.

4. Let F ∈ k[t1, . . . , tn] be irreducible. Set X = Z(T ) ⊂ An. Then dim(X) = n − 1. Since F is notconstant, assume t1 is a variable in F . Claim that t2, . . . , tn ∈ k(X) = Quot(k[t1, . . . , tn]/(F )) arealgebraically independent over k.

Suppose G(t2, t3, . . . , tn) = 0 on X. If G ∈ I(X), then Gm is divisible by F by Nullstellensatz, but Fdepends on t1, but G does not, a contradiction.

Then t1 is algebraic over k(t2, . . . , tn), so dim(X) = trdeg(k(X)/k) = n− 1.

If R is a commutative ring, then the Krull dimension dim(R) is the largest n such that there exists achain of prime ideals P0 ( P1 ( · · · ( Pn. It is a fact from commutative algebra that if R is a finitelygenerated domain over k, then dim(R) = trdeg(Quot(R)/k). So if X is affine, then dim(X) = dim(k[X]).

We say P0 ( P1 ( · · · ( Pn is maximal if we cannot properly insert other prime ideals in it.It is a fact that if R is a localization of a finitely generated k-algebra, then all maximal chains of prime

ideals in R have the same length. (= dim(R).)If X is affine, prime ideals in k[X] are in bijection with irreducible closed subvarieties of X. If P is a

prime ideal of k[X],Spec(k[X]/(P )) = {p′ ∈ Spec(k[X]) : p′ ⊃ p}

which can be interpreted as the irreducible closed subsets of Y .Also,

Spec(k[X]P ) = {P ′′ ∈ Spec(k[X]) : P ′′ ⊂ P} ↔ { irreducible closed ⊃ Y }

So let P0 ( P1 ( · · · ( Pm = P ( Pm+1 ( · · · ( Pm+k be maximal, so m + k = dim(k[X]) = dim(X).This gives a corresponding chain

Y0 ⊃ Y1 ⊃ · · · ⊃ Ym = Y ⊃ Ym+1 ⊃ · · · ⊃ Ym+k

So k = dim(k[X]/(P )) and m = dim(k[X]P ), hence

dim(k[X]) = dim(k[X]/P ) + dim(k[X]p)

anddim(Y ) = dim(k[X]/P ) = k.

Lastly, dim(Yi) = dim(X)− i, so we get subvarieties of every intermediate dimension.If Y ⊂ X is closed, then dim(Y ) ≤ dim(X). If Y ( X is closed, and X irreducible, then dim(Y ) <

dim(X). To see this, we have dim(X) = dim(Y ) + dim(k[X]P ) where P is the prime corresponding to X, soP 6= 0 since X 6= Y . So k[X]P at least contains 0 ( P · k[X], so dim(k[X]p) ≥ 1, and the claim follows.

Suppose Y ⊂ X is closed, and irreducible. Then dim(X) = dim(Y ) + dim(OX,Y ).

15 Smooth Varieties

Take F ∈ k[T1, T2, . . . , Tn] nonzero. Set X = Z(F ) ⊂ An. Take x ∈ X.Define the tangent space of X at x to be the space cut out by the linear equation

n∑i=1

∂F

∂Ti(x)(Ti − xi) = 0

We view this as an (affine) subspace of An.

26

Example 15.1. Let F = T 31 − T 2

2 . Then if x = 0,

∂F

∂T1(x)T1 +

∂F

∂T2(x)T2 = 0

but both partial derivatives are 0, so the tangent space at x is A2.

Consider X ⊂ An to be closed. For all F ∈ I(X) ⊂ An and x ∈ X, then the equations

n∑i=1

∂F

∂Ti(x) · (Ti − xi) = 0

define the tangent space TX,x ⊂ An. It suffices to consider only F s from a generating set of I(X).Take G ∈ An. Define the differential of G at x,

DxG =

n∑i=1

∂G

∂Ti(Ti − xi)

to be a linear form on TX,x. (If G ∈ I(X), then DxG = 0).Consider the map

k[T1, . . . , Tn]→ (TX,x)∗ : G 7→ DxG

where the latter space is the dual space. This is a k-linear map. This map vanishes on I(X). We haveanother map

dx : k[X]→ T ∗X,x : f 7→ dxf = DxF |TX,xfor F ∈ An with F |X = f . This map is called the differential at x.

We have dx(f + g) = dxf + dxg and dx(fg) = f(x)dxg + g(x)dx(f). Let

Ix = {f ∈ k[X] : f(x) = 0} ⊂ k[X]

be the maximal ideal. Then dx|Ix : Ix → T ∗X,x. Claim this map is still surjective. Every linear form on TX,xextends to a linear form

∑ni=1Ai(Ti − xi) = f ∈ k[X]. But f(x) = 0, so f ∈ Ix and dxf = f ∈ T ∗X,x.

Furthermore, ker(dx|Ix) = I2x. If f, g ∈ Ix, then dx(fg) = f(x)dx(g) + g(x)dx(f) = 0 + 0 = 0. SoI2x ⊂ ker(dx|Ix). For simplicity, assume x = (0, 0, . . . , 0). If F ∈ k[T1, . . . , Tn], then DxF |TX,x = 0 (f =F |X ∈ ker(dx|Ix). Then

DxF =

m∑i=1

aiDxFi, Fi ∈ I(X).

Set F ′ = F −∑aiFi, then DxF

′ = 0 on An. So F ′ has no constant and linear terms. Now I2x = 〈TiTj |x〉since Ix = 〈Ti|x〉. Then f = F |X = F ′|X ∈ I2x, so the other inclusion holds.

So Ix/I2x ' T ∗X,x. Note k[X]/Ix = k, and

OX,x = k[X]Ix = {fg

: g(x) 6= 0}.

But mx = { fg : f(x) = 0, g(x) 6= 0} is contained in OX,x. We get a map i : k]X] → k[X]Ix = OX,x taking

Ikx → mkx and Ix/I2x → mx/m

2x. Claim that this is an isomorphism.

To see surjectivity, fix fg ∈ mx, so f(x) = 0 but g(x) 6= 0. Let a = g(x) 6= 0, and consider f

a ∈ Ix. Then

i(f/a) = f/a, andf

a− f

g=f(g − a)

ag∈ m2

x

since ag(x) 6= 0, and f ∈ Ix, and g − a ∈ Ix. So i(f/a) ≡ f/g (mod m2x).

For injectivity, take f ∈ Ix, and suppose i(f) = f/1 = 0 in mx/m2x. Hence there exists g ∈ k[X] such

that g(x) 6= 0 and fg ∈ I2x. We need f ∈ I2x. Let a = g(x) 6= 0. Then

af = gf︸︷︷︸∈I2x

+ (a− g)f︸ ︷︷ ︸∈I2x

∈ I2x

27

so a 6= 0 implies f ∈ I2x. So T ∗X,x ' mx/m2x. So TX,x ' (mx/m

2x)∗. Note that mx/m

2x is a module over

OX,x/mx = k.Now let X be a quasi-projective variety. Pick x ∈ X. Then OX,x ⊃ mx and OX,x/mx = k. Then mx/m

2x

is a finite dimensional vector space over k. So define the tangent now as TX,x = (mx/m2x)∗. If U ⊂ X is

open containing x, then TU,x = TX,x, so the definition is local.If f : X → Y is a regular map, x ∈ X and y = f(x) ∈ Y , then f∗ : OY,y → OX,x taking mky → mkx,

so my/m2y → mx/m

2x. We get a linear map df : TX,x → TY,y called the differential of f . If f is a closed

embedding, then f∗ : OY,y → OX,x is surjective, so df is injective. Also, d(gf) = dg ◦ df .

Example 15.2. Let X = Z(F ) ⊂ An, for F irreducible, so dim(X) = n−1. By Nullstellensatz, if ∂F∂Ti|X = 0,

then F |(∂F∂Ti

)kimplies F | ∂F∂Ti , so ∂F

∂Ti= 0 since it’s of lesser degree. Claim that ∂F

∂Ti|X 6= 0 for some i.

Suppose otherwise, that ∂F∂Ti

= 0 for all i. If char(k) = p > 0, then∂Tpi∂Ti

= pT p−1i = 0.

If g = aTα11 Tα2

2 · · ·Tαnn , then ∂g∂Ti

= 0 for all i iff all αi are divisible by p. So F = Gp, a contradictionsince F is irreducible.

Then TX,x is given by∑ni=1

∂F∂Ti

(x)(Ti−xi) = 0. Then ∂F∂Ti6= 0 for some i. So this equation is not trivial

for all

{x :∂F

∂Fi(x) 6= 0} = DX(

∂F

∂Ti) 6= ∅.

So dim(TX,x) = n − 1 = dim(X) for all x in a nonempty open subset of X. For other x, then TX,x = An,so dim(TX,x) = n > dim(X).

We have some facts from commutative algebra: Let O be a Noetherian local ring, m a maximal ideal,and k = O/m. So m is a finitely generated ideal, so m/m2 is a k-vector space of finite dimension. ThendimKrull(O) ≤ dim(m/m2). We say O is regular if dim(O) = dim(m/m2).

Elements x1, . . . , xd ∈ m for d = dimO, then x1, . . . , xn ∈ m/m2 is a k-basis iff x1, . . . , xd generated m,and (x1, . . . , xn) are called local parameters.

Let x ∈ X, for X quasi-projective. Then dim(OX,x) = max dim(Xi) = dimxX, where we take themaximum over all Xi irreducible components containing x. If X is irreducible, then dimx(X) = dim(X).We have OX,x is a Noetherian local ring, and

dimxX = dim(OX,x) ≤ dim(mx/m2x) = dim(TX,x)

We say is x is a regular, aka nonsingular, point if dimxX) = dim(TX,x) iff OX,x is regular. If x ∈ U ⊂ X isopen, then UU,x = Ox,x, and TU,x = TX,x so x is nonsingular on X iff x is nonsingular on U . We say X issmooth if all points are nonsingular. Note if U ⊂ X is open and X is smooth, then U is smooth.

Proposition 15.3. Let X be an irreducible and quasi-projective variety. Then there is a nonempty opensubset U ⊂ X such that every point of U is nonsingular (on U and X).

Proof. There exists an irreducible F ∈ k[T1, . . . , Tn] such that X ≈ Z(F ) ⊂ An. We have U ⊂ X andW ⊂ Z(F ) nonempty opens which are isomorphic. Take i such that ∂F

∂Ti6= 0, so from a previous example,

for all x ∈ D(∂F∂Ti

), then dimTZ(F ),x = n− 1 = dimZ(F ) = dimx(Z(F )). So x is nonsingular on Z(F ).

All points in W ∩D(∂F∂Ti

)are nonsingular, so there is U ′ ⊂ U is isomorphic to W ∩D

(∂F∂Ti

).

Suppose x ∈ X ⊂ An is closed, and X = Z(S) for S = {F1, . . . , Fm}. Then TX,x is given by

n∑i=1

∂Fj∂Ti

(x) · (Ti − xi) = 0, 1 ≤ j ≤ m

The matrix(∂Fj∂Ti

(x))

is an n×m matrix called the Jacobi matrix. Suppose it has rank r, so

dim(TX,x) = n− r ≥ dimxX

and x is nonsingular iff n− r = dimxX.

28

Example 15.4. 1. Both An and Pn are smooth.

2. Consider the curve y2 = x2 + x3. Then 0 is singular since all partial derivatives vanish there. TheJacobi matrix is (

−2x+ 3x2 2y)

which has rank 1 for all other points, so all other points are nonsingular.

We have the following fact: Every regular local ring is a domain. The minimal prime ideals of OX,x arein correspondence with the irreducible components containing x.

There is exactly one irreducible component of X containing a nonsingular point x. Equivalently, if Xi

and Xj are distinct irreducible components of X, then every point of Xi ∩Xj is singular on X.Suppose X is irreducible, and x1, x2 ∈ X are nonsingular. In general, OX,x1

6' OX,x2. Recall k(X) =

Quot(OX,xi). If there exists an isomorphism OX,x1→ OX,x2

, then we get an automorphism of k(X)/k,which gives a birational isomorphism f : X → Y defined at x2 and f(x2) = x1.

But there are many varieties Autbirat(X) is finite.

Suppose O is a local ring with maximal ideal m. The completion is O = limnO/mn.

Example 15.5. 1. Let x ∈ X be a nonsingular point, so OX,x is a regular, with f1, f2, . . . , fm ∈ mx asystem of local parameters. We get a map

k[[x1, . . . , xn]]→ O : xi 7→ fi

If x1, x2 ∈ X are nonsingular, then it may be that OX,x1 6' OX,x2 but OX,x1 ' OX,x2 as they are bothisomorphic to the ring of power series.

2. Let X = Z(y3−x2−x3) and Y = Z(y2−x2), with char(k) = 0. Then OX,0 6' OY,0, but the completionsare since y2 = x2 + x3 = x2(1 + x) = f(x)2 for some f since (1 + x)1/2 =

∑∞i=0 aix

i. So we have anisomorphism of completions with x 7→ f(x) and y 7→ y.

Also from commutative algebra, every regular ring is a UFD.

Theorem 15.6. Let X be a irreducible quasi-projective variety, and Y ⊂ X a closed irreducible subvarietyof codimension 1. Let x ∈ Y be a nonsingular point. Then there is an affine neighborhood U of x such thatIU (Y ∩ U) is principal in k[U ].

Proof. Let I ( OX,x be the ideal of Y , prime since Y is irreducible. It is nonzero, so pick a ∈ I a nonzerounit. Since OX,x is a UFD, so there exists f ∈ OX,x such that f | a. Then Z(f) is irreducible in aneighborhood of x. So Y ∩U ⊂ ZU (f) ⊂ U . Since Z(f) ( X, then dimX − 1 = dimY ≤ dimZ(f) < dimXso dimY = dimZ(f), so Y = Z(f).

Then IX(Y ) = IX(Z(f)) = (f) in a neighborhood of X, so IX(Y ) is principal.

Example 15.7. Let X = Z(y2 − x3). So OX,0 ⊃ mx = 〈x〉 is not generated by one element. But ZX(y) =

{0}, but (y) ( 〈x, y〉 =√

(y).

Theorem 15.8. Let f : XtoY be a rational map of quasi-projective varieties. Let U ⊂ X be the domain of definition of f . Note that(X \U) ( X is closed, so codimX(X \U) ≥ 1. If X is smooth and Y is projective, then codimX(X \U) ≥ 2.

Proof. Write X \U =⋃Yi for Yi irreducible. We need to show that codimX(Yi) ≥ 2 for all i. We claim that

if Z ⊂ X is closed and irreducible of codimension 1, then Z ∩ U 6= ∅, i.e., f is defined at at least one pointin Z. Pick x ∈ Z ⊂ X to be a nonsingular point. In a neighborhood of x, the ideal IX(Z) = (g) for someg ∈ OX,x is prime.

Assume Y = Pn. Then f = [f0 : · · · : fn] where fi ∈ k(X) = Quot(OX,x). So we may assume fi ∈ OX,xand are relatively prime by clearing denominators and dividing by the greatest common divisor. Thereexists i such that g does not divide fi in OX,x. This implies Z(g) 6⊂ Z(fi). For if Z = Z(g) ⊂ Z(fi) in aneighborhood of x, then fi must be divisible by a power of g, hence by g. So there is a point z ∈ Z suchthat z /∈ Z(fi), so fi(z) 6= 0. Thus f is defined a t z.

Now suppose Y ⊂ Pn is closed. Let f ′ : Xf−→ Y ↪→ Pn. Claim that the domain of definition of f and f ′

are the same. Write U ′ for the domain of definition of f ′. So U ⊂ U ′ ⊂ X. Now U ⊂ U ′ is dense, so f(U) isdense in f ′(U ′). So f(U) = f ′(U ′). But f(U) ⊂ Y since Y is closed in Pn, so f ′(U ′) ⊂ Y . Thus U ′ = U .

29

Corollary 15.9. Every rational map from an irreducible smooth curve to a projective variety is regulareverywhere.

Corollary 15.10. Every birationally isomorphic irreducible smooth projective curves are isomorphic.

Some problems so far:

1. Previously, we had to assume k is algebraically closed. We would like to replace k with any commutativering.

2. We had to assume varieties X ⊂ Pn, but we would like to define them more abstractly as existingindependently of some projective space.

3. Consider the projection f : X → A1 : (x, y) 7→ y where X = Z(y − x2) is the parabola. Then f−1(a)is a pair of points. The map of coordinate rings i k[y] 7→ k[x, y]/(y − x2), so

k[f−1(a)] = k[x, y]/(y − x2, y − a) ' k[x]/(x2 − a) ' k × k

if a 6= 0. If a = 0, then k[x]/(x2). Thus x is nilpotent, hence k[x]/(x2) is not an algebra of regularfunctions. So we should allow nilpotence now.

16 Sheaves

Let C denote any one of the categories Rings, Ab or Sets. Let A be another category. A presheaf on A withvalues in C is a functor F : A op → C . So A→ A′ is sent to F(A′)→ F(A).

Example 16.1. Let X be a topological space, AX be the category with objects open sets of X and morphisms

Mor(U,U ′) =

{{∗} if U ⊂ U ′,∅ ifU 6⊂ U ′.

A presheaf on X is a presheaf on AX . For F(U) ∈ O(C ), and U ⊂ U ′,

F(U ′)→ F(U) : f 7→ f |U

is the restriction.

Example 16.2. 1. Let C ∈ C and A ∈ A . Let F(A) = C for all A and F(α) = 1C . This is the constantpresheaf.

2. Let X be quasi-projective over k, U ⊂ X open is set to k[U ]. And U ′ ⊃ U is sent to k[U ′] is thepresheaf of regular functions on X.

Let X be a topological space, and F a presheaf on AX (on X) with values in C . Suppose

1. F(∅) = ∗ a final object in C .

2. If U ⊂ X is open, U =⋃Uα is an open cover, such that a, b ∈ F(U) : a|Uα = b|Uα for all α implies

a = b.

3. If U ⊂ X is open, U =⋃Uα is an open cover, such that aα ∈ F(Uα) : aα|Uα∩Uβ = aβ |Uα∩Uβ for all

α, β thne there exists a unique a ∈ F(U) : a|Uα = aα for all α.

Then F is a sheaf.Recall a presheaf is a functor F : Oop → C . It assigns U ⊂ X open to an object F (U) and restriction

maps rUV from F (U)→ F (V ).

30

Definition 16.3. (Alternative) We have F a presheaf is a sheaf if for all U ⊂ X open and for all opencovers U =

⋃Uα, the sequence

F (U)→∏α

F (Uα)→∏α,β

F (Uα ∩ Uβ)

is an equalizer, i.e., F (U) = {(sα)α ∈∏

F (Uα) : r1(sα)α = r2(sα)α}, r1, r2 being the maps in the secondmapping.

Some remarks, if C = Ab, then a sheaf is equivalent ot r = ker(r1 − r2).Also F (∅) = {∗} because ∅ admits the empty cover, and then this follows since the empty product is the

final object.Let F be a presheaf on X.

Definition 16.4. The stalk of F at X is FX = lim−→x∈U F (U) for U open neighborhoods of x. Explicitly,

consider the set of pairs (U, a) where U 3 x is an open neighborhood, and a ∈ F (U). Define an equivalencerelation on such pairs by (U, a) ∼ (U ′, a′) iff there exists a neighborhood W 3 x such that a|W = a′|W . Then

Fx = {(U, a)}/ ∼

Elements of Fx are called germs of F at x.

Note if F is a sheaf of rings or abelian groups, then Fx is a ring or abelian group. IF U 3 x we have anatural morphism F (U)→ Fx by a 7→ [(U, a)] = ax.

Example 16.5. 1. Let X be a quasi-projective variety. Then the stalk of OX at x ∈ X is the local ringOX,x.

2. If F is a constant presheaf with value A ∈ C , then for all x ∈ X, Fx = A.

3. Let X be a domain in Rn, and F be the sheaf of real analytic functions on X, then a germ of F atx ∈ X is determined by a sequence of real numbers.

Proposition 16.6. Let F be a sheaf on X and U ⊂ X open, and a, b ∈ F (U). Then a = b iff for all x ∈ U ,ax = bx.

Proof. Suppose that for all x ∈ U , ax = bx. Choose for each x ∈ U an open neighborhood U ⊂Wx 3 x suchthat a|Wx

= b|Wx. This is possible by explicit description of the stalk. Then U =

⋃x∈U Wx. Then we have the

condition that a|Wx = b|Wx for all x, so by the first part of the sheaf condition, (that F (U)→∏x∈U F (Wx)

is injective, so a = b.

Definition 16.7. Let X be a topological space, and F and G (pre-)sheaves on X. A morphism F → G isa natural transformation of functors, i.e., for all U ⊂ X open, a morphism ϕU : F (U) → G (U) such thatrUV ◦ ϕU = ϕV ◦ rUV .

So clearly (pre-)sheaves on X with values in C form a category. An isomorphism of (pre-)sheaves is onesuch that for all U ⊂ X is open, ϕU is an isomorphism in C .

Let f : F → G be a morphism of presheaves x ∈ X. Then we get an induced morphism in C , fx : Fx →Gx : [(U, a)] 7→ [(U, fU (a))].

The assignment F → Fx for some fixed x is a functor from presheaves to C .

Proposition 16.8. Let F and G be sheaves on X, and f : F → G a morphism. Then f is an isomorphismiff for all x ∈ X, fx : Fx → Gx is an isomorphism.

Proof. Suppose each fx is an isomorphism. For injectivity, let a, b ∈ F (U) for U ⊂ X open. SupposefU (a) = fU (b). Then for all x ∈ U , fU (a)x = fU (b)x. Thus fx(ax) = fx(bx), so by hypothesis ax = bx for allx ∈ U , so a = b.

For surjectivity, let b ∈ G (U). We need to produce b ∈ F (U) such that fU (b) = b. Let x ∈ U , then thereis ax ∈ Fx such that fx(ax) = bx. Now each ax is represented by a pair (Ux, a

x) such that b|Ux = fUx(ax),and Ux 3 x, ax ∈ F (Ux). Note that U =

⋃x∈U Ux. Take x, y ∈ U .

31

We have fUx∩Uy (ax|Ux∩Uy ) = b|Ux∩Uy = fUx∩Uy (ay|Ux∩Uy ) So since fUx∩Uy is injective, so ax|Ux∩Uy =ay|Ux∩Uy . Since F is a sheaf, there exists a unique a ∈ F (U) such that for all x ∈ U , a|Ux = ax. Thus,fU (a)|Ux = fUx(ax) = b|Ux for all x ∈ U . By sheaf condition for G , fU (a) = b.

Example 16.9. Let X be a space admitting a disconnected open subset. Let F 6= {∗} to be a set. Take Fconstant presheaf with value F and G be the constant sheaf with value F , so G (U) = Maps(U,F ) which isthe locally constant F -valued functions. (Here F has the discrete topology.) We have an obvious morphismF → G which is not an isomorphism, but is an isomorphism on all stalks, since X is locally connected.

Let F be a presheaf on X. Take U ⊂ X open. Define another presheaf F |U on U by F |U (W ) = F (W )for W ⊂ U ⊂ X all open. This is called the restriction of F . If F is a sheaf, then F |U is also a sheaf.

Suppose f : X → Y is continuous. Define a presheaf f∗F on Y defined by f∗F (U) = F (f−1(U)) forU ⊂ Y open. This is the pushforward of F . If F is a sheaf, so if f∗F .

Consider (X,F ) for X a topological space, and F is a presheaf on X. Let these be the objects of acategory. A morphism (X,F )→ (Y,G ) is a pair (f, g) where f : X → Y is continuous, and g : G → f∗F isa morphism of presheaves on Y .

Note for U ⊂ Y open, we have maps G (U)→ f∗F (U) = F (f−1(U)). If x ∈ X, we have a map of stalks

Gf(x) → (f∗F )f(x) → Fx.

We get a category of ringed spaces.

Example 16.10. Let X be a quasi-projective variety over k. We have the structure ringed space (X,OX)where OX(U) = k[U ]. If f : X → Y is regular, we get a map (X,OX) → (Y,OY ) given by f and k[u] =OY (U)→ OX(f−1(U)) = k[f−1(U)] defined by h 7→ h ◦ (f |f−1(U).

If x ∈ X, then (OX)x = OX,x.

17 Spectrum of a Ring

Let X be an affine variety, and k[X] its coordinate ring. For x ∈ X, we have Ix = {f ∈ k[X] : f(x) = 0}.So Ix = ker(k[X]→ k : f 7→ f(x)), so k[X]Ix ' k, so Ix is maximal.

There is a correspondence between maximal ideals in k[X] and (the points of) X. So we recover X as acollection of maximal ideals.

Let R be a commutative ring. Note Spec(R) = ∅ iff R− 0. Let I ⊂ R be an ideal, and

V (I) = {P ∈ Spec(R) : I ⊂ P} ⊂ Spec(R).

Then V (0) = Spec(R), and V (I) = ∅ iff I = R. We have

1. If I ⊂ J , then V (J) ⊂ V (I).

2. V (∑Iα) =

⋂α V (Iα)

3. V (I ∩ J) = V (I) ∪ V (J) = V (IJ). Now IJ ⊂ I ∩ J so V (I) ∪ V (J) ⊂ V (I ∩ J) ⊂ V (IJ). WhenP ∈ V (IJ), so IJ ⊂ P , so I ⊂ P or J ⊂ P since P is prime, so either P ∈ V (I) or V (J). SoV (IJ) ⊂ V (I) ∪ V (J), and we conclude.

We have the Zariski topology on Spec(R), where the V (I) form the closed sets. Take T ⊂ Spec(R) asubset. If T = V (I), then T ⊂ V (I), so for all P ∈ T , P ∈ V (I), so I ⊂ P , and thus our desired I isI =

⋂P∈T P . So if T = {P}, then {P} = V (P ) = {P ′ ∈ Spec(R) : P ⊂ P ′}. Also, P is closed iff P is

maximal. So not all points are closed.For a ∈ R, D(a) = Spec(R) \ V (a). These are the principal open sets. So

D(a) = {P ∈ Spec(R) : a /∈ P}.

Some properties:

32

1. We have V (I) = V (∑a∈I Ra) =

⋂a∈I V (a), implying Spec(R) \ V (I) =

⋃a∈I D(a). So every open is

a union of principal open sets.

2. So D(0) = ∅, and D(1) = Spec(R).

3. D(a) ∩D(b) = D(ab), so the D(a) form a basis of the Zariski topology.

4. D(an) = D(a).

5.⋃D(aα) = Spec(R) iff 〈aα〉 = R. For this is equivalent to saying

⋂V (aα) = V (

∑aαR) = ∅, which

holds iff∑aαR = R.

6. Spec(R) is compact. To see this, suppose Spec(R) =⋃αD(aα) iff 〈aα〉 = R, so 1 =

∑ni=1 aαibi for

some bi ∈ R. Thus 〈aαi〉 = R iff Spec(R) =⋃nn=1D(aαi).

Example 17.1. 1. If R is a field, Spec(R) = {0}.

2. If R is a discrete valuation ring, then Spec(R) = {0,m}. The open sets are ∅, {0,m}, and {0}. Andthe closed sets are ∅, {0,m} and {m}. This is because D(a) is ∅ if a ∈ R×, is {0,m} if a = 0, and {0}otherwise. Also {0} = {0,m}.

3. If R = Z, then Spec(Z) = {0, pZ}, and again {0} = Spec(Z).

Suppose α : R→ S is a ring homomorphism, with α−1(P ) = Q for P a prime. Then we have an injectionR/Q ↪→ S/P , so R/Q is a domain, so Q is prime.

We hae the pullback α∗ : Spec(S)→ Spec(R) : P 7→ α−1(P ). We have

1. I ⊂ R an ideal implies (α∗)−1(V (I)) = V (α(I)S). For

P ∈ (α∗)−1(V (I)) ⇐⇒ α∗(P ) ∈ V (I) ⇐⇒ α−1(P ) ∈ V (I)

⇐⇒ I ⊂ α−1(P ) ⇐⇒ α(I) ⊂ P ⇐⇒ α(I)S ⊂ P ⇐⇒ P ∈ V (α(I)S).

2. (α∗)−1(D(a)) = D(α(a)).

So we get a correspondence from the opposite category of commutative rings and topological spaces givenby R 7→ Spec(R), and α 7→ α∗.

Let I ⊂ R be an ideal. Define

V (I) = {p ∈ Spec(R) : I ⊂ P} ⊂ Spec(R)

to be closed. We haveSpec(R/I) = {Q/I : Q ∈ Spec(R), Q ⊃ I} ' V (I)

are homeomorphic. Take S ⊂ R to be a multiplicative subset. We have the canonical map i : R → S−1R.We have

Spec(S−1R) ' {p ∈ Spec(R) : P ∩ S = ∅}

under the maps Q 7→ i−1(Q) and P 7→ S−1P = S−1R · i(P ).

Example 17.2. 1. Take S = {an}, S−1R = Ra. Then

{p ∈ Spec(R) : P ∩ {an} = ∅} = D(a)

so Spec(Ra) ' D(a).

2. If p ∈ Spec(R), S = R \ P , then S−1R = RP , and

Spec(RP ) ' {P ′ ∈ Spec(R) : P ′ ⊂ P}.

Also,Spec(R \ P ) ' {P ′ ∈ Spec(R) : P ′ ⊃ P}.

33

Recall ⋂p∈Spec(R)

p = N(R) = {a ∈ R : an = 0 for some n > 0}.

Note D(a) = ∅ iff a ∈ N(R) and V (I) = Spec(R) iff I ⊂ N(R). Take I ⊂ R, so N(R/I) =√I/I. So⋂

p⊃I p = N(I). Also, P ⊃ I iff P ⊃√I, so V (I) = V (

√I).

Proposition 17.3. We have V (I) ⊂ V (J) iff J ⊂√I.

Proof. For the forward direction, every p ∈ Spec(R) and p ⊃ I implies P ⊃ J . So J ⊂⋂I⊂p p =

√I.

Conversely, V (I) = V (√I) ⊃ V (J).

Corollary 17.4. Note V (I) = V (J) iff√I =√J .

Proposition 17.5. Let α : R→ S be a ring homomorphism, and J ⊂ B an ideal, and I = α−1(J). Then

α∗(V (J)) = V (I)

If X is a topological space, a point x ∈ X is called generic if {x} = X. Note this implies X is necessarilyirreducible, since {x} is irreducible, and hence its closure is as well.

Proposition 17.6. Let R be a commutative ring. The following are equivalent

1. Spec(R) is irreducible.

2. N(R) is a prime.

3. N(R) is prime and N(R) is a generic point of Spec(R).

4. There exists a unique generic point in Spec(R).

Proof. Suppose (1). Suppose ab ∈ N(R). Then ∅ = D(ab) = D(a) ∩D(b). So either D(a) = ∅ or D(b) = ∅,so either a ∈ N(R) or b ∈ N(R). Hence (2).

Now suppose (2). So N(R) ⊂ P for all P ∈ Spec(R). So N(R) = V (N(R)) = Spec(R). Hence (3).Now suppose (3). If P1 and P2 are generic points, then P1 ∈ {P1} = V (P2), so P2 ⊂ P1. By symmetry,

P1 = P2.Now suppose (4). We have Spec(R) = {∗} is irreducible, where ∗ is the generic point.

We have maps from the category of irreducible closed subsets of Spec(R) and Spec(R) sending Z =Spec(R/I) to a generic point of Z and P maps to {P} = V (P ).

Recall an ideal of R× S has form I × J for ideals I and J , and

(R× S)/(I × J) ' (R/I)× (S/J)

so I × J is prime iff I is prime and J = S or I = R and J is prime. So

Spec(R× S) = Spec(R) t Spec(S).

We construct a sheaf of rings on Spec(R). Take U ⊂ Spec(R) is open. Recall if p ∈ Spec(R), thenRP = { af : f /∈ P} and a

f makes sense in Rp for all P ∈ D(f). Denote OR to be the structure sheaf of rings

on Spec(R). Define

OR(U) = {ϕ : U →⊔P∈U

RP : ϕ(P ) ∈ RP and in a nbhd W of every P ∈ U, ϕ(P ) =a

f, f /∈ Q, ∀Q ∈W}.

The restriction maps are given by restriction of functions. So OR is a sheaf of rings. Take a ∈ R. ThenSpec(Ra) = D(a) ⊂ Spec(R). Then OR|D(a) = ORa .

34

Proposition 17.7. The map

R→ OR(Spec(R) : a 7→ (Pvarphi−−−−→ a

1∈ Rp)

is a ring isomorphism.

Proof. For injectivity, suppose a ∈ R such that a1 = 0 in RP for all P . So there exists bp ∈ R \ P such that

abp = 0. Then Ann(A) 6⊂ P for all P , so ann(A) = R, so a = 0.For surjectivity, let ϕ : Spec(R)→

⊔Rp be as above. There is an open cover Spec(R) =

⋃Ui such that

ϕ(P ) = aifi

for all P ∈ Ui, fi /∈ P ′ for all P ′ ∈ Ui. We may assume Ui = D(gi). Then D(fi) ⊃ D(gi). Then

V (fiR) ⊂ V (giR) implies gi =√fiR. Then gni = fibi. So

ϕ(P ) =aifi

=aibifibi

=aibigni

but Spec(R) =⋃D(gi) =

⋃D(gni ). Replacing gi by gni and ai by aibi, so we may assume in Ui = D(gi) and

ϕ(P ) = aigi

. We may assume the cover is finite by compactness.For i.j we have

D(gi) ∩D(gj) = D(gigj) = Spec(Rgigj )

We getRgigj ↪→ ORgigj (Spec(Rgigj ) = OR(D(gigj))

We have the functions α : D(gigj)→⊔Rp for α(P ) = ai

fiand β : D(fifj)→

⊔RP with β(P ) =

ajgj

.

Now ai/gi = aj/gj in Rgigj and (aigj − ajgi)(gigj)n = 0 for aigni g

n+1j = ajg

n+1i gj . So

aigi

=aig

ni

gn+1I

andajgj

=ajg

nj

gn+1j

so by replacing ai by aigni and aj by ajg

nj with gi by gni and gj by gnj , we may assume that aigj = ajgi. So

〈gj〉 = R so there exists hi ∈ R such that∑gihi = 1. Set a =

∑k akhk ∈ R, so a

1 = aigi

= ϕ(P ) in Rgi .Then

a · gi =∑k

akgihk =∑k

aigkhk = ai · 1 = ai

and thus a1 = ai

gi.

Corollary 17.8. We have OR(D(a)) = Ra.

Proof. NoteOR(D(a)) = ORa(D(a)) = Ra

and apply the statement to Ra.

Proposition 17.9. For every P ∈ Spec(R), (OR)P ' RP .

Proof. We have a map (OR)P → RP by (U,ϕ) 7→ ϕP ) ∈ RP .For injectivity, in a neighborhood D(f) of P in U , ϕ = a

f , with af = 0 in Rp, thus as = 0 for s ∈ R \ P

and af = 0 in Rsf implies ϕ|D(sf) =). Then

(U,ϕ) ∼ (D(sf), ϕ|D(sf)) = 0.

For surjectivity, take af ∈ RP . Take f ∈ R \ P . Then (D(f), ϕ) with ϕ = a

f maps to af ∈ RP .

Take U ⊂ Spec(R) to be open. Let ϕ ∈ OR(U). If P ∈ U , we have ϕP ∈ RP → RP /PRP = Quot(R/P ),so ϕ(P ) is in a field.

If U = Spec(R) and ϕ ∈ OR(U) = R, then ϕ(P ) = 0 for all P ∈ Spec(R) iff ϕ ∈ P for all P , iff ϕ ∈ N(R).So ϕ ∈ N(P ) iff ϕ(P ) = 0 for every P , but ϕP = 0 in RP for all P iff ϕ = 0.

35

Let α : R→ S be a ring homomorphism. It induces a continuous mpa α∗ : Spec(S)→ Spec(R). We wantto make a map from OR(U) to OS((α∗)−1(U)) by the diagram

(α∗)−1(U)

α

��

//⊔Q∈(α∗)−1(U)SQ

Uϕ

//⊔P∈U RP

OO

The morphisms are then (Spec(S),OS) → (Spec(R),OR). So we get a functor of ringed spaces from theopposite category of commutative rings to the category of ringed spaces sending R to (Spec(R),OR).

This functor is faithful. The morphism R → S goes to (Spec(S),OS) → (Spec(R),OR). Note R =OR(Spec(R)) 7→ OS(Spec(S)) = S, so different morphisms give rise to different morphisms.

Observe (Spec(R),OR) is a local ringed space since all stalks are local rings.A ringed space (X,F) is called lcoal if Fx is a local ring for all x ∈ X. A morphism f : (X,F)→ (Y,G) of

local ringed spaces is local if for all x ∈ X, Gf(x) → Fx is a local ring homomorphism. So (Spec(R),OR) is alocal ringed space and if α : R→ S is a local ring homomorphism, then α∗ : (Spec(S),OS)→ (Spec(R),OR)is local.

Example 17.10. Let R be a DVR, so Spec(R) = {0,m}, so S = Quot(R) and Spec(S) = {0}. We make amap f : Spec(S) → Spec(R) by sending 0 to m. The nonempty open sets in Spec(R) are {0} and Spec(R).We also need to make a map OR(U)→ OS(f−1(U)). But R ↪→ S is not a local homomorphism, so the mapof ringed spaces is not local, so f 6= α∗ for any α : R→ S.

So we adjust the later category to just be local ringed spaces. This functor is full.

Theorem 17.11. This functor is full and faithful.

Proof. Let f : (Spec(S),OS) → (Spec(R),OR) is local. Take Q ∈ Spec(S) and P = f(Q) ∈ Spec(R). Wehave

R = OR(Spec(R))

��

α // OS(Spec(S)) = S

��

RP = (OR)PfP // (OS)Q = SQ

So f−1P (QSQ) = PRP . Thenα∗(Q) = α−1(Q) = P = f(Q)

so α∗ = f as maps of topological spaces.Now take a ∈ R and b = α(a) ∈ S. then f−1(D(a)) = (α∗)−1(D(a)) = D(b). Then

R = OR(Spec(R)

��

α // OS(Spec(S)) = S

��

OR(D(a))fD(a)

// OS(D(b))

We need f(D(a))( xan ) = α(x)

bn . But fD(a)(a) = α(a) = b ∈ Sb and fD(a)(x) = α(x) so

f(D(a))(x

an) =

f(D(a))(x)

f(D(a))(an)=α(x)

bn

so f = α∗ on the structure sheaves.

36

18 Definition of a Scheme

We have the opposite category of commutative rings is a full subcategory of local ringed spaces. If (X,F) isa local ringed space, and U ⊂ X is open, then (U,F|U ) is a local ringed space. A local ringed space (X,F) isa scheme if there is an open cover X =

⋃Uα such that (Uα,F|Uα) ' (SpecRα,ORα) for some commutative

ring Rα for every α. So the category of schemes sits between the opposite category of commutative ringsand local ringed spaces.

A scheme X,OX) is affine if OX = (X,F) ' (SpecR,OR) for some R. In short, we write X ' SpecRand OX ' OR.

Suppose X is a scheme, and U ⊂ X is open, then U is a scheme. That is (U,OX |U ) is a scheme. Needto check it’s locally affine. Pick x ∈ X. There exists an open neighborhood W of x such that W ' SpecR.Then U ∩W ⊂ W is open and corresponds to an open subset of SpecR. There is some D(a) ⊂ SpecRcorresponding to U ∩W .

Then OX |D(a) ' OR|D(a) = ORa . Then U |D(a) = X|D(a) ' Spec(Ra) is the desired affine neighborhoodof x.

Suppose U ⊂ X is open, and f : U ↪→ X. A map (U,OU )→ (X,OX) is given by the inclusion f : U → Xand OX → f∗(OX |U ) by

OX(W )→ f∗(OX |U )(W ) = OX |U (f−1(W )) = OX |U (U ∩W ) = OX(U ∩W ).

If x ∈ X for X a scheme, then (OX)x = OX,x is the local ring at x. If f : X → Y is a morphism ofschemes, and for x ∈ X, y = f(x), we get a local homomorphisms OY,y → OX,x. This descends to a map ofthe residue fields. If x ∈ X, and x ∈ U ⊂ X is open, and g ∈ OX(U) is a “function” on U . Then gx ∈ OX,xand then gx ∈ OX,x 7→ g(x) ∈ k(x) is the value of g at x.

Let X and Y be schemes. Define a presheaf FY of sets on X: If U ⊂ X is open, then FY (U) =Mor(U, Y ) =: Y (U), called the U -values of Y or set of points of Y with values in U . If U =

⋃Uα is an open

cover, then

Mor(U, Y ) ↪→∏α

Mor(Uα, Y )→∏α,β

Mor(Uα ∩ Uβ , Y )

exact. Then FY is a sheaf of sets. So every scheme gives us a sheaf of sets.Let X be a scheme, and R a commutative ring. If f : X → Spec(R), we get R = OR(Spec(R))→ OX(X).

We get a mapMor(X,Spec(R))→ HomRings(R,OX(X)).

Theorem 18.1. This map is a bijection.

Proof. If X is affine, X = Spec(S), then Mor(SpecS, SpecR) → HomRings(R,S) But recall that oppositecategory of commutative rings is a full subcategory of ringed spaces, so this is a bijection.

In general, take X =⋃Uα to be an affine open cover. We get an exact sequence of sets

Mor(X,SpecR) ��

//� _

��

∏α Mor(Uα,SpecR) //

'��

∏α,β Mor(Uα ∩ Uβ ,SpecR)

� _

��

Hom(R,OX(X))� � //

∏Hom(R,OX(Uα)) //

∏α,β Hom(R,OX(Uα ∩ Uβ)

A standard diagram chase shows bijectivity of the leftmost vertical map.

Then we have a correspondence between opposite category of commutative rings and the category ofschemes given by R 7→ SpecR and X 7→ OX(X). So R 7→ Spec(R) is right adjoint, and the other functor isthe left adjoint.

The homomorphism R → OX(X) makes OX(X) an R-algebra. This induces an R-algebra structure onevery OX(U) is an R-algebra. So to give a morphism X → SpecR is equivalent to giving R-algebra structureon all OX(U).

When R = Z, there exists a unique morphism X → SpecZ, so SpecZ is the final object in the categoryof schemes. We say X is a scheme of a commutative ring R if a morphism X → SpecR is given. Everyscheme is a scheme over Z.

37

Define the category of schemes over R to have objects X → SpecR, and the morphisms to be a mapX → Y making the following commute

X //

��

SpecR

Y

;;

So the category of schemes is the same of schemes over Z.Note for SpecS to be a scheme over R, we need S to be an R-algebra.Let k be an algebraically closed field, and X a quasi-projective variety over k. Let X = {Z ⊂ X :

Z closed and irreducible }. So we may view S ⊂ X be the identification x 7→ {x}. We want Z ⊂ X closedto imply Z ⊂ X be closed for the topology on X. We have X ∩ Z = Z and the Zariski topology is inducedon X so that X ↪→ X is continuous.

For U ⊂ X open, if Z = X \ U , so that Z ⊂ X, we define

U := X \ Z = {W ⊂ U : W is irreducible , W ∩ U 6= ∅}.

These are the open sets in X.If Z ∈ X, then {Z} = Z, so {Z} is closed in X iff Z = {z} is a singleton. So X ⊂ X is a set of closed

points.Take U ⊂ X for some U ⊂ X open. Note we can recover U by U = U ∩X. We define

OX(U) = k[U ].

So (X,OX) is a ringed space. Suppose X is affine, and R = k[X]. We have Spec(R) ' X are homeo-morphic. We want to construct a map OX → OR by

k[U ] = OX(U)→ OR(U) = {ϕ : U →∏OX,Z}

Given g ∈ k[U ], take P ∈ OR(U) corresponding to z which maps to image of g under k[U ]→ OX,Z = OU,Z∩U .

This gives map (SpecR,OR) → (X,OX). It’s sufficient to check k[X]g = OX(D(g)) ' OR(D(g)) = Rgwhere g ∈ R = k[X].

Let X be arbitrary quasi-projective, and X =⋃Xα for Uα open affines. Then X =

⋃Uα, and OX |Uα =

OUα . So (Uα,OX |Uα) = (Uα,OUα) is an affine scheme and (X,OX) is a scheme.

Suppose f : X → Y is a regular map, and Z ⊂ X is a closed irreducible set. Then f(Z) is also closedand irreducible. We get a map

f : X → Y : Z 7→ f(Z)

So if W ⊂ Y is open and W ⊂ Y is open. We want to construct a map f∗ from OY (W ) = k[W ] to

OX(f−1(W )) = k[f−1(W )] since f−1(W ) = ˜f−1(W ).

So f : (X,OX)→ (Y ,OY ) and we say f : X → Y is a morphism of schemes.

We’ve constructed a functor θ : QProjVar(k)→ Schemes by the assignment X 7→ X and f 7→ f .

Theorem 18.2. The functor θ is fully faithful.

Lemma 18.3. If Z ∈ X, then (OX)Z ' OX,Z , and Quot((OX)Z) ' k(Z).

Proof. (Of Theorem) if f : X → Y and f : X → Y , then f the “restriction” of f implies θ is faithful.Take g : X → Y be a morphism of schemes over k. We need to find a regular map f : X → Y such that

g = f . We need g(X) ⊂ Y .Take x ∈ X, so {x} ∈ X. Then k(x) = k since x corresponds to a maximal ideal. Take Z ∈ Y such that

g({x}) = Z. This induces a map k(Z)/k → k(x) = k over k. This implies dim(Z) = 0, so Z is a point. Sog(X) ⊂ Y , so set f = g|X : X → Y .

We need to check that f is regular, and g = f on top space and on structure sheaves. The proof isomitted.

38

QPorjVar(k) ��

// Schemes/k

AffVar(k)?�

OO

� � // AffSchemes/k?�

OO

and the bottom inclusion is that of reduced finitely generated k-algebras into commutative k-algebras.Let k[An] = k[T1, T2, . . . , Tn] and Ank = Spec(k[T1, . . . , Tn]). So A1

k = Spec(k[T ]) is in bijection with theset of monic irreducible polynomials and 0, which are the closed points and the generic point. So if P = (h)and OX,P = k[T ]P and k(P ) = k[T ]/(h)/k is of degree deg(h).

Let V be an n-dimensional k-vector space. Then V ' kn and view V as Ank . It turns out V =Spec(Sym(V ∗)). For suppose x1, x2, . . . , xn are the coordinates, so xi ∈ V ∗. Then k[x1, . . . , xn] is thesymmetric algebra on x1, . . . , xn.

39