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LECTURES ON ALGEBRAIC GEOMETRYMATH 202A

KIYOSHI IGUSABRANDEIS UNIVERSITY

Contents

Introduction 1Overfield 11. A�ne varieties 21.1. Lecture 1: Weak Nullstellensatz 21.2. Lecture 2: Noether’s normalization theorem 31.3. Nullstellensatz 41.4. Noetherian induction 81.5. Generic points 101.6. Morphisms 121.7. Algebraically closed case 131.8. Lecture 6: Review 161.9. First look at schemes 162. Projective varieties 202.1. Lecture 7. Basic properties of projective space 202.2. Segre embedding 242.3. Irreducible algebraic sets in P

n

24References 25

Introduction

These are lecture notes for Math 202a: Algebraic Geometry. The basic goal is to dovarieties over a nonalgebraically closed field and simultaneously lay the groundwork forschemes (reduced of finite type). The lectures will expand on the very terse notes for thecourse and give full details.

Overfield

We will assume that k is a field and ⌦ is an “overfield”. This is a field extension of kwhich has infinite transcendence degree and is algebraically closed.

⌦ = k(X1, X2, X3, · · · )

We also want to allow the number of variables to be more than countably infinite so thatk = Q,⌦ = C is an example.

Date: September 25, 2014.

1

2 KIYOSHI IGUSA BRANDEIS UNIVERSITY

I went over the basic properties of the overfield starting with clarifying the definition oftranscendence degree.

Definition 0.0.1. Suppose thatK is a field extension of k or any integral domain which con-tains k. We say that x1, · · · , xn 2 K, n � 1, are algebraically independent if they do not sat-isfy any polynomial equations over k. This means: given any polynomial f(X1, · · · , Xn

) 2k[X1, · · · , Xn

], if f(x1, · · · , xn) = 0 then f is zero as a polynomial. When n = 1 we saythat x1 is transcendental over k.

The transcendence degree of K over k is defined to be the largest number d so that Kcontains d algebraically independent elements x1, · · · , x

d

over k. (d can be infinite.)

One of the basic theorems is the following. The Noether Normalization Theorem is arefined version of this basic theorem.

Theorem 0.0.2. 1) If K is a field extension of k generated by x1, · · · , xn then a maximal

algebraically independent subset of this set of generators is a transcendence basis.

2) Any two transcendence bases for K over k have the same number of elements.

A basic property of the overfield ⌦ is the following.

Theorem 0.0.3. For any finitely generated field extension L = k(x1, · · · , xn) of k, there is

an embedding � : L ,! ⌦ over k. (over k means � is the identity on k)

Proof. Choose a transcendence basis x1, · · · , xk

for L over k. Let �(xi

) = Xi

for i =1, · · · , k. Since L is an algebraic extension of k(x1, · · · , x

k

) and ⌦ is algebraically closed,the embedding � : k(x1, · · · , x

k

)! ⌦ extends to an embedding � : L! ⌦. ⇤The class had no problem deriving the following consequence.

Corollary 0.0.4. Any finitely generated domain over k embeds in L. ⇤

1. Affine varieties

One of the basic objects of study will be subsets of Cn defined by polynomials equationswith coe�cients in Q. These are a�ne varieties over C but they also give a�ne schemesover Q. More generally we take k to be any field and ⌦ to be an “overfield”: an algebraicallyclosed field extension of k of infinite transcendence degree as discussed above.

The first theorem that we want to prove is the weak Nullstellensatz which identifies themaximal ideals in the ring k[X1, · · · , Xn

].

1.1. Lecture 1: Weak Nullstellensatz.

Lemma 1.1.1. Let a 2 ⌦n

. Then the set of all f(X) 2 k[X] so that f(a) = 0 is a prime

ideal which we denote ⇡(a) ✓ k[X]. (I.e., ⇡ is a mapping of sets ⇡ : ⌦n ! Spec(k[X]).)

For now, Spec(R) is the set of prime ideals in any commutative ring R. Students werewell aware of the fact that an ideal a in any ring R (always commutative) is prime i↵ R/ais a domain and a is a maximal ideal i↵ R/a is a field.

Proof. This is a prime ideal since it is the kernel of the k-algebra homomorphism

eva

: k[X]! ⌦

which sends f(X) to f(a). The image of this homomorphism is a subring of ⌦ and thus adomain. I also use the notation '

a

= eva

: “evaluation at a”. ⇤

LECTURES ON ALGEBRAIC GEOMETRY MATH 202A 3

In my lecture I emphasized the fact that f(x) as a function of x 2 ⌦n does not have easyto understand properties. It is a polynomial function. However, the function f 7! f(x) forany fixed x 2 ⌦n gives a homomorphism of k-algebras. (This means ring homomorphismwhich is the identity on k.) The lemmas being proved here use this nice interpretation off(x) as a function of f .

Lemma 1.1.2. ⇡ : ⌦n ! Spec(k[X]) is surjective.

We will give ⌦n a topology and Spec(k[X]) the quotient topology.

Proof. Take any prime p ⇢ k[X]. Then k[X]/p is a domain which contains k.Claim: There is an embedding ' : D = k[X]/p ,! ⌦.

Pf: D is finitely generated by the images xi

of the variables Xi

: D = k[x1, · · · , xn]. Byrenumbering the x

i

we may assume that x1, · · · , xk

are algebraically independent. EmbedD in its field of fractions Q(D) = {a

b

| a, b 2 D, b 6= 0}. Then Q(D) = k(x1, · · · , xn). Sincex1, · · · , x

k

are algebraically independent, we have an isomorphism:

L = k(x1, · · · , xk

) ⇠= k(X1, · · · , Xk

) ✓ ⌦

Since xk+1, · · · , xn are algebraic over L, the embedding L ,! ⌦ extends to an embedding

Q(D) ,! ⌦ and we get an embedding

' : D = k[X]/p ,! Q(D) ,! ⌦

This proves the claim.

Let ai

= '(Xi

) 2 ⌦. Let a = (a1, · · · , an) 2 ⌦n. Then f(a) = '(f) = 0 i↵ f 2 p. So,⇡(a) = p. So, ⇡ : ⌦n ! Spec(k[X]) is onto. ⇤Theorem 1.1.3. ⇡(a) is a maximal ideal if and only if all coordinates a

i

2 ⌦ of a are

algebraic over k.Proof. (We will go over this again after reviewing the Normalization Theorem.) Let '

a

:k[X]! ⌦ be given by evaluation at a so that ⇡(a) = ker'

a

. Then 'a

(k[X]) = k[a1, a2, · · · , an].Since this is a subring of ⌦, it is an integral domain finitely generated over k. So, the nor-malization theorem applies and this is a field i↵ all a

i

are algebraic over k. ⇤The usual weak Nullstellensatz follows.

Corollary 1.1.4. If k is algebraically closed then the only maximal ideal in k[X] are ⇡(a)where a 2 kn. (⇡(a) = (X1 � a1, · · · , Xn

� an

)).

1.2. Lecture 2: Noether’s normalization theorem. We need the following criticallyimportant theorem from commutative algebra.

Theorem 1.2.1 (Normalization Theorem). Let R be a finitely generated domain over a

field k with transcendence degree d. Then R contains transcendental elements x1, · · · , xd

over k so that R is an integral extension of the polynomial ring S = k[x1, · · · , xd

].

Recall the definitions: If S is a subring of R an element a 2 R is called integral over Sif there is a monic polynomial

f(X) = Xn + c1Xn�1 + c2X

n�2 + · · ·+ cn

2 S[X]

so that f(a) = 0. This is equivalent to the statement that S[a] is finitely generated as anS-module.

We say that R is an integral extension of S if S ✓ R and every element of R is integralover S. R is called integrally closed if the only integral extension of R is R itself.


Exercise 1.2.2. The proof in Mumford’s book gives an algorithm for finding x1, · · · , xd

.Use this algorithm to find S = k[x1] in the case when R = k[X2, X3] ✓ k[X].

We also need, for now, the following lemma used in the proof of the Going-Up theorem.

Lemma 1.2.3. Let R be an integral extension of S. If R is a field then S is a field.

Proof. Take any a 6= 0 2 S. Then R contains b = 1/a. Since R is integral over S, there isa monic polynomial f(X) 2 S[X] so that f(b) = 0:

bn + c1bn�1 + · · ·+ c

n

= 0, ci

2 S

Multiply by an�1 to get:

b+ c1 + c2a+ c3a2 + · · ·+ c

n

an�1 = 0

So, b = 1/a 2 S and S is a field. ⇤Corollary 1.2.4. Suppose R = k[a1, · · · , an] is a domain. Then R is a field if and only if

every ai

is algebraic over k.

Proof. If the ai

are algebraic over k then k[a1, · · · , an] = k(a1, · · · , an) is a field. Conversely,if some are transcendental, then, by the normalization theorem, R is an integral extensionof a polynomial ring S = k[X1, · · · , X

d

] with d > 0. But S is not a field. By the lemma, Ris not a field. ⇤

1.3. Nullstellensatz. Let’s review one construction from Commutative Algebra. Supposethat R is a ring and f 2 R is not nilpotent. I.e., fn 6= 0 for all n � 0. Then M ={1, f, f2, · · · } is a multiplicative set which means that M is closed under multiplication,contains 1 and does not contain 0. In general R

M

is defined to be the ring of all fractions

RM

= {ab| a 2 R, b 2M}

with the usual addition and multiplication rule for fractions and the identity a

b

= ac

bc

forall c 2 M . In particular, a

b

= 0 i↵ there is a c 2 M so that ac = 0 in R. WhenM = {1, f, f2, · · · } for f not nilpotent, the notation is: R

f

= RM

which is given byinverting just f :

Rf

=

⇢

a

fn

�

where a

f

n = 0 i↵ afm = 0 for some m. In particular 1f

6= 0. Another description of Rf

is:

Rf

= R[X]/(fX � 1).

We verified this in detail in class using the universal properties of Rf

and R[X].The universal property of R

f

is the following. Let ' : R! S be any ring homomorphismwhich sends f to a unit in S (unit means invertible element). Then there is a unique ringhomomorphism ' : R

f

! S making the following diagram commute:

R

' // S

Rf

'

>>

Let S = k[X]/(Xf � 1). Then X = 1f

in S, so the homomorphism R ,! R[X] ⇣ S factorsuniquely through R

f

.


The universal property of R[X] is the following. Given any ring homomorphism : R!S and any element x 2 S, there is a unique homomorphism : R[X]! S which is on Rand sends X to x. In this case we have a homomorphism : R[X]! R

f

which sends eacha 2 R to a

1 2 Rf

and sends X to 1f

2 Rf

. Then (Xf) = 1. So, Xf � 1 is in the kernel of . So, there is an induced homomorphism

R[X]/(Xf � 1)! Rf

So we have mapping in both directions Rf

$ R[X]/(Xf � 1) and the compositions are theidentity maps by universality.

1.3.1. Zariski topology. For any subset S of k[X], let V (S) be the set of all x 2 ⌦n so thatf(x) = 0 for all f 2 S. Two extreme cases are: V (;) = ⌦n and V (1) = ;.

Definition 1.3.1. A subset Z ✓ ⌦n is called an algebraic set defined over k if Z = V (S)for some S ✓ k[X].

Example 1.3.2. SL(2,⌦) is the algebraic subset of ⌦4 defined over any k by

SL(2,⌦) = {(a, b, c, d) 2 ⌦4 : ad� bc = 1}GL(2,⌦) is defined by the inequality ad�bc 6= 0. This does NOT define an algebraic subsetof ⌦4. However, GL(2,⌦) is “equivalent” to the algebraic subset of ⌦5 (defined over any k)given by

GL(2,⌦) ⇠= {(a, b, c, d, e) 2 ⌦5 : (ad� bc)e = 1}.The point is: GL(2,⌦) does not satisfy the definition of an algebraic subset of ⌦4. However,it is in bijection with a true algebraic subset of ⌦5. We will use this to define the structureof a variety on GL(2,⌦) to be the one given by this structure on this algebraic subset of ⌦5.One of the consequences of this concept is that the function 1

ad�bc

, which is not a polynomial

function on GL(2,⌦) ⇢ ⌦4, is a polynomial function on the corresponding subset of ⌦5. So,1

ad�bc

is a “regular function” on GL(2,⌦).

The basic property of algebraic sets is:

Theorem 1.3.3. The sets V (S) form the closed subsets of a topology on ⌦n

. (This is called

the Zariski topology or the k-topology.)

Proof. Recall that a collection of subsets C of a set X form the closed sets in a topology i↵they satisfy the following. The standard terminology is “A is closed” means A 2 C.

(1) ; is closed.(2) X is closed.(3) Any intersection of closed sets is closed: C

↵

closed )T

C↵

is closed.(4) The union of two closed sets is closed: A,B closed ) A [B is closed.

The definition of the set V (S) can be parsed as follows:

V (S) =\

f2SV (f).

So V (S) is an intersection of the hypersurfaces V (f) given by single equations f(a) = 0.So,

T

V (S↵

) = V ([S↵

). This proves (3): The collection of sets {V (S)} is closed underarbitrary intersections. If S, T are two subsets of k[X] and ST = {fg : f 2 S, g 2 T} then

V (ST ) =\

f2S,g2TV (fg) =

\

f2S,g2T(V (f) [ V (g)) =

\

f2SV (f) [

\

g2TV (g) = V (S) [ V (T ).


This proves (4): The collection of sets {V (S)} is closed under finite unions.Since V (;) = ⌦n and V (1) = ;, the collection of sets {V (S)} satisfies the axioms for the

closed sets in a topology on ⌦n as listed above. ⇤For any subset Y of ⌦n we take the induced topology. Thus a subset A ✓ Y is closed i↵

A = Y \ V (S) for some S.

Exercise 1.3.4. If Y = V (S) is a closed subset of ⌦n show that the (relatively) open setsYf

= {x 2 Y : f(x) 6= 0} form a basis for the Zariski topology on Y .

Proof. (Using things that come later.) Let U be an open subset of Y with complement Z.Since k[X] is Noetherian, the ideal I(Z) is finitely generated by, say, f1, · · · , fm. If y 2 Uthen f

i

(y) 6= 0 for some i. So y 2 Yfi ✓ U . So, the Y

f

are basic open sets in Y . ⇤1.3.2. Nullstellensatz. For any S ⇢ k[X], let (S) denote the ideal generated by S. (Weallow ideals to be the whole ring. An ideal which does not contain 1 is called a proper

ideal.) For any ideal a in k[X], the radical r(a) of a is defined to be the set of all f 2 k[X]so that some power of f lies in a. The following statement is clear:

Proposition 1.3.5. For any subset S ✓ k[X]:

V (S) = V ((S)) = V (r(S)).

Proof. What is clear is that

V (S) ◆ V ((S)) ◆ V (r(S))

since S ✓ (S) ✓ r(S). So, we need to show that V (S) ✓ V ((S)) ✓ V (r(S)).

(1) Proof that V (S) ✓ V ((S)):The ideal I = (S) generated by the set S in the ring k[X] is the set of all linear combi-

nations of elements of S:

(S) =n

X

f↵

g↵

| f↵

2 S, g↵

2 k[X]o

.

If a 2 V (S) then f↵

(a) = 0 for all f↵

2 S. But this impliesP

f↵

(a)g↵

(a) = 0. So,a 2 V ((S)) which shows that V (S) = V ((S)).

(2) V ((S)) ✓ V (r(S)): (Lecture 3 starts here.)I did a complicated proof of this in class but, after class, Mac showed me a quick proof:

f(a) = 0 , fk(a) = 0.

Why does this prove (2)? (After some discussion we decided this is “obvious”.) ⇤For any subset Z ✓ ⌦n let I(Z) be the set of all f(X) 2 k[X] so that f(z) = 0 for all

z 2 Z. Since f(z) = 0 i↵ f 2 ⇡(z), this is equivalent to:

I(Z) =\

z2Z⇡(z)

Therefore, I(Z) is an ideal in k[X]. Also, being an intersection of prime ideals, it is aradical ideal. (Recall that, if an ideal I is contained in a prime ideal p then r(I) ✓ p.Since each ⇡(z) = ker ev

z

: k[X] ! ⌦ is a prime ideal, r(I(Z)) ✓ ⇡(z) for each z. So,r(I(Z)) ✓

T

⇡(z) = I(Z). So, r(I(Z)) = I(Z). So, I(Z) is a radical ideal, i.e., an idealequal to its radical.)

Hilbert’s Nullstellensatz states:

Theorem 1.3.6. For any ideal a in k[X], I(V (a)) = r(a) .


Proof. (from the red book [4]) It is clear that a ✓ I(V (a)) and therefore r(a) ✓ r(I(V (a))) =I(V (a)). So, suppose g is not in r(a). Then we will show that g /2 I(V (a)) by finding apoint in x 2 V (a) so that g(x) 6= 0. (And x 2 V (a) means f(x) = 0 for all f 2 a.)

Step 1. Since g is not in r(a), g is not nilpotent in k[X]/a. So, we can invert g in k[X]/awithout making the ring trivial:

✓

k[X]

a

◆

g

6= 0.

The reason is that M = {1, g, g2, · · · } is a multiplicative set in k[X]/a which does notcontain 0. [This is also proved in the last paragraph of the proof of Theorem 1.3.6 in [4].]

Step 2a. Reformulate the equation:

0 6=✓

k[X]

a

◆

g

=k[X]

g

(a)=

R/(1�Xn+1g)

(a)=

R

(a, 1�Xn+1g)

where R = k[X][Xn+1] = k[X1, · · · , Xn+1] and (a) is the ideal in R generated by a ✓

k[X] = k[X1, · · · , Xn

]. The first equality follows from the universal properties: the ringhomomorphism k[X] ! k[X]

g

/(a) has a in its kernel. So, it induces a homomorphismk[X]/a! k[X]

g

/(a). This sends g to a unit. So, there is a unique induced homomorphism(k[X]/a)

g

! k[X]g

/(a). Similarly, we get a map the other way and, by the universalproperties defining these maps, both compositions are the identity.

The second equality comes from the equation k[X]g

⇠= k[X][Xn+1]/(Xn+1g�1) discussed

earlier. The third equality is clear.Step 2b. Let J = (a, 1 � X

n+1g). Since this R/J 6= 0, J is a proper ideal in R. So,J is contained in some maximal ideal m. Recall that ⇡ : ⌦n+1 ! Spec(R) is surjective.So, there is a point a 2 ⌦n+1 with ⇡(a) = ker ev

a

= m. By the weak Nullstellensatz, allcoordinates a

i

of a are algebraic over k.Step 3: Reinterpretation: Let x = (a1, · · · , an) 2 ⌦n. (So, a = (x, a

n+1).)a ⇢ J ⇢ m = ⇡(a) implies that f(a) = 0 for all f 2 a. But f(a) = f(x) since

f 2 a ✓ k[X] is a polynomial in X1, · · · , Xn

. So, the point x 2 ⌦n lies in V (a).But 1�X

n+1g is also in J ✓ m = ker eva

. So, 1 = an+1g(x) making g(x) = 1/a

n+1 6= 0as required to prove the theorem. ⇤

Since ai

are algebraic over k, ⇡(x) = ker'x

is a maximal ideal in k[X] which containsa but does not contain g. Since g was an arbitrary element of k[X] not contained in r(a),this proves the following.

Corollary 1.3.7. For any ideal a in k[X], r(a) is the intersection of all maximal ideals

containing a. Consequently, the Jacobson radical of any finitely generated algebra over any

field is equal to it nilpotent radical N = r(0).

Recall that the Jacobson radical of any ring is by definition the intersection of all maximalideals. There is a related theorem which says that the nilradical (r(0)) is the intersection ofall prime ideals. For finitely generated algebras over any field these are the same and this isthe key fact that allows us to study varieties by looking only at the k points in the variety.

Example 1.3.8. Let k = R and a = (X2 + Y 2). Then k = C is algebraically closed but ⌦is much bigger. The algebraic set V (a) is the set of all (x, y) 2 ⌦2 so that x2 + y2 = 0.


R2

V (a) \ R2

C2

V (a) \ C2

(1, i)

V (a) \ R2 is just one point. But V (a) \ C2 is two lines given by y = ix and y = �ix sinceX2 + y2 = (Y � iX)(Y + iX).

⇡(x) ✓ R[X,Y ] is a maximal ideal i↵ x 2 C2. For example,

⇡(1, i) = {f 2 R[X,Y ] : f(1, i) = 0} = (X � 1, Y � i) \ R[X,Y ] =⇤ (X � 1, Y 2 + 1)

where (⇤) needs proof. This is a maximal ideal since

R[X,Y ]

⇡(1, i)=

R[Y ]

(Y 2 + 1)⇠= C

is a field. Corollary 1.3.7 states that

J = I(V (a) \ C2) = r(a) =Nullst.

I(V (a)) ✓ R[X,Y ]

since V (a) \ C2 is the intersection of all maximal ideals containing a. Thus J/a is theJacobson radical of R[X,Y ]/a.

Question: Show geometrically that V (a) is irreducible in this case.

Corollary 1.3.9. There is an order reversing bijection between the set of radical ideals in

k[X] and the set of Zariski closed subsets of ⌦n

defined over k.

Proof. This follows from the formulas that we proved.

radical ideals ! closed subsets of ⌦n

start with a = r(a) �! V (a)

I(V (a)) =⇤ r(a) = a � V (a)

start with any closed set:

I(Z) = I(V (S)) =⇤ r(S) � Z = V (S) = V (r(S))

I(Z) �! V (I(Z)) = V (r(S))

where (⇤) is the Nullstellensatz. ⇤Exercise 1.3.10. Show that the closure (in the Q-topology) of the point (e, e2) 2 C2 is theparabola {(z, z2) | z 2 C}.

1.4. Noetherian induction. Recall that k[X] is Noetherian, i.e., the ideals in k[X] satisfythe ACC: any ascending chain of ideals is eventually constant: Given

I1 ✓ I2 ✓ I3 ✓ · · ·Then eventually I

n

= In+1 = · · · for some n. Therefore, the closed subsets of ⌦n satisfy

the DCC: any descending chain of closed subsets is eventually constant:

Z1 ◆ Z2 ◆ Z3 ◆ · · ·implies that, for some n, Z

n

= Zn+1 = · · · . This allows for the following argument called

Noetherian induction: For every statement about closed subsets of ⌦n which is not true,there is a minimal counterexample.


Proposition 1.4.1. Every closed subset of ⌦n

is a union of finitely many irreducible closed

subsets. (Irreducible means not the union of two proper closed subsets.)

Proof. Suppose not. Then, by Noetherian induction, there is a minimal counterexample,say Y . So, Y is not the union of finitely many irreducible closed subsets. In particular, Y isnot irreducible. So, Y is the union of two proper closed subsets Y = Y1[Y2. By minimalityof Y , each of Y1, Y2 is a union of finitely many irreducible closed subsets. So

Y = Y1 [ Y2 =[

Y j

1 [[

Y j

2

where each Y j

i

is irreducible. So, Y is a finite union of irreducible closed sets. ⇤

The proof of the next proposition uses the following argument. If X is irreducible andX ✓

S

Yi

then X ✓ Yi

for some i. The reason is:

X = X \[

Yi

=[

X \ Yi

Since X is irreducible, it must be equal to one of these: X = X \ Yi

which implies X ✓ Yi

.

Proposition 1.4.2. The decomposition of a closed set Y as a union of irreducible closed

subsets Y =S

Yj

is unique.

The unique pieces Yj

of a closed set Y are the components of Y .

Proof. If an irreducible X ✓S

Yi

then X is contained in one of the Yi

. So, ifS

Xi

=S

Yj

then X1 ✓ Yj

for some j. By the same argument, Yj

✓ Xk

for some k. But, X1 ✓ Xk

implies that X1 = Xk

and k = 1 (otherwise the decomposition X =S

Xi

has redundantsummands). So, X1 = Y

j

. Similarly, X2 = Yk

where j 6= k (since X1 6= X2). Do this for allthe X

i

to prove the theorem. (Any Yj

which are left over are redundant summands.) ⇤

Lecture 4 starts here:

Theorem 1.4.3. The prime ideal correspond to the irreducible subsets under the correspon-

dence of Corollary 1.3.9.

Proof. One direction is clear: Y not irreducible ) I(Y ) not prime:

Y = Y1 [ Y2, Yi

( Y ) I(Y ) = I(Y1) \ I(Y2), I(Y ) ( I(Yi

)

But a prime ideal cannot be written as an intersection of two larger ideals. [Pf: If p = I1\I2and p ( I

i

then there are a 2 I1\p, b 2 I2\p. But ab 2 I1I2 ✓ I1\I2 = p. Then either a or bis in p, a contradiction.] So, I(Y ) is not prime. Equivalently, I(Y ) prime) Y irreducible.

Conversely, suppose I(Y ) is not prime. Then, by the lemma below, I(Y ) is an intersectionof prime ideals:

I(Y ) = p1 \ p2 \ · · · \ pm

, I(Y ) ( pi

.

Then Y is a union of the corresponding closed sets (which are irreducible by the first partof the proof):

Y = V (p1) [ V (p2) [ · · · [ V (pm

), V (pi

) ( Y.

So, I(Y ) not prime ) Y not irreducible. So, Y irreducible , I(Y ) prime. ⇤

Lemma 1.4.4. In any Noetherian ring A, any radical ideal is the intersection of finitely

many prime ideals.


The proof uses the Primary Decomposition Theorem which says that, in any Noetherianring R, any ideal a is a finite intersection of primary ideals:

a = q1 \ q2 \ · · · \ qm

.

Recall that an ideal q in any ring R is primary if it satisfies the condition that a, b 2 q,a /2 q implies that bn 2 q for some n. In other words, b 2 r(q). The important property forus is: The radical of any primary ideal is prime. Proof: Suppose that a, b 2 R, ab 2 r(q)but a /2 r(q). Then (ab)n = anbn 2 q for some n. But a /2 r(q) means that an /2 q. Since qis primary, this implies (bn)m 2 q for some m. But then b 2 r(q). So, r(q) is a prime ideal.

Proof of Lemma 1.4.4. By the Primary Decomposition Theorem, any ideal in a Noetherianring is a finite intersection of primary ideals: a =

T

qi

. Since a is a radical ideal we have

a = r(a) = r⇣

\

qi

⌘

=⇤\

r(qi

).

Proof of the equality =⇤: The inclusion (✓) is clear sinceT

qi

✓ qi

implies r(T

qi

) ✓ r(qi

)for each i. So, suppose a 2

T

r(qi

). Then a 2 r(qi

) for each i. Then, for each i there is aninteger n

i

so that ani 2 qi

. Let n be the maximum of these ni

. Then an 2T

qi

= a. So,a 2 r(a) and a =

T

r(qi

). The lemma follows since each r(qi

) is prime. ⇤

This uses only the existence part of the Primary Decomposition Theorem which is easy:

Lemma 1.4.5. In any Noetherian ring A, every ideal is a finite intersection of primary

ideals.

Proof. Suppose not. Then there is a maximal counterexample, a. This ideal cannot bewritten as an intersection of two larger ideals. Otherwise, each of these is an intersection ofprimary ideals making a an intersection of these primary ideals. We claim that a is primary.

To show this, take two elements a, b 2 A so that ab 2 a but no power of a or b lies ina. For each n � 1 let I

n

= (a : an) be the set of all c 2 A so that anc 2 a. Then In

isan ascending chain of ideals so I

n

= In+1 for some n. Let J = (a, an), K = (a, b). Since

an, b /2 a these ideals properly contain a. We will show that J\K = a giving a contradiction(showing that a is primary).

Let c 2 J \ K. Since c 2 J , c = anx + y where y 2 a. Since aK ✓ a, ac 2 a. So,an+1x 2 a. So, x 2 I

n+1 = In

. So, anx 2 a. So, c 2 a. ⇤

1.5. Generic points.

Theorem 1.5.1. The closure of any point y 2 ⌦n

is an irreducible algebraic set defined

over k. Conversely, for any irreducible algebraic subset Y ✓ ⌦n

, there is an element y 2 Y(called a generic point of Y ) so that Y = y.

Proof. The first statement is easy: Suppose that y = Y1 [Y2. Then y must be in one of thetwo sets. Say, y 2 Y1. Since Y

i

are closed, we get y ✓ Y1. But Y1 cannot be bigger than y.So, Y1 = y. So, y is not the union of two smaller closed subsets.

The second statement follows from Lemma 1.1.2. Namely, Y , being irreducible corre-sponds to a prime ideal p = I(Y ). By Lemma 1.1.2, there is a point y 2 ⌦n so that ⇡(y) = p.Then y = V (p) = Y . We went through the proof of this in class:

For any ideal a we have an equivalence:

y 2 V (a) , (8f 2 a)f(y) = 0 , a ✓ p = ⇡(y) = ker evy


But the closure of y is the intersection of all closed sets containing y and all closed sets havethe form V (a) for some ideal a:

y =\

y2V (a)

V (a) =\

a✓p

V (a) = V (p) = Y.

⇤

Corollary 1.5.2. Two point x, y 2 ⌦n

map to the same prime ideal ⇡(x) = ⇡(y) if and

only if x = y.

Exercise 1.5.3. (1) Show that x ⇠ y i↵ x = y is an equivalence relation on ⌦n.(2) Show that x 2 y i↵ ⇡(x) ◆ ⇡(y).(3) Show that x is a finite set i↵ ⇡(x) is a maximal ideal.(4) When is x 2 ⌦n a closed point? (What if k is not perfect?)

I went through an example of generic points, but the example was a little too advanced.We will discuss this more later.

Definition 1.5.4. The Krull dimension d of any ring R is the maximum length of a chainof prime ideals.

p0 ( p1 ( · · · ( pd

For example, the dimension of Z is d = 1 since the longest tower is:

0 ( (p)

for any prime number p. Later we will discuss the theorem that the Krull dimension ofk[X1, · · · , Xn

] is n. For example, when n = 2, the longest chain of prime ideals looks likethis:

0 ( p ( m

where m is a maximal ideal. The prime ideal p in the middle is not zero and not maximal.It is called a height 1 prime. (0 has height 0, maximal ideals have height 2 in k[X1, X2].)

For such a chain of prime ideals we have the corresponding irreducible closed subsets:

V (0) ) V (p) ) V (m).

Each of these has a generic point. Conversely, for any point y 2 ⌦2, y is one of theseirreducible closed sets by Theorem 1.5.1. So, ⌦2 has three kinds of points:

(0) y is a finite set. Equivalently, ⇡(y) is a maximal ideal. (This is not obvious andrequires proof.)

(1) y = V (p) is a curve in ⌦2.(2) y = ⌦2. Equivalently ⇡(y) = 0.

The case y = V (p) occurs when the coordinates of y are algebraically dependent on eachother but one of them is not algebraic over k. For example, y = (T1, T

31 ). (Recall that ⌦

is the algebraic closure of k(T1, T2, · · · ), the field of rational functions on infinitely manyvariables.)

Exercise 1.5.5. Show that the closure of the point y = (T1, T31 ) is the set

y = V ((X2 �X31 )) = {(x, x3) |x 2 ⌦}.


1.6. Morphisms. Let Y ✓ ⌦n be an irreducible closed subset defined over k. Then aregular function f : Y ! ⌦ is any mapping which is the restriction to Y of a mapping⌦n ! ⌦ given by a polynomial in k[X1, · · · , Xn

]. The coordinate ring �(Y ) of Y is the ringof all regular functions Y ! ⌦. If two polynomials give the same regular function on Ythen their di↵erence is zero on Y and conversely. So,

�(Y ) ⇠= k[X]/I(Y ).

To clarify this: By definition, we have an epimorphism:

I(Y ) ,! k[X] ⇣ �(Y )

with kernel I(Y ) since every regular function is defined to be given by a polynomial.To set the notation, we denote polynomials by f(X1, · · · , Xn

) and the correspondingregular function on Y by f(x1, · · · , xn) using lower case letters to denote the functionsxi

: Y ,! ⌦n ! ⌦ which are projection to the ith coordinate. For example, the polynomialf = XY 3 � 5X gives the function f(x, y) = xy3 � 5x.

Since Y is irreducible, I(Y ) is a prime ideal and �(Y ) is an integral domain. If y 2 Y isa general point then f : Y ! ⌦ is uniquely determined by f(y). So, �(Y ) is isomorphic tothe image of y under k[X].

Given Y, Z irreducible closed subsets of ⌦n,⌦m resp, amorphism Y ! Z is a mapping f :Y ! Z given by polynomial equations. Thus, there exist m polynomials f

i

2 k[X1, · · · , Xn

]so that f(y) = (f

i

(y)) for all y 2 Y . There is an induced map:

f⇤ : �(Z)! �(Y )

given by composition of polynomial mappings:

f⇤(g : Z ! ⌦) = g � f : Y ! Z ! ⌦.

For any g 2 �(Z) ✓ k[T1, · · · , Tm

], f⇤(g) = g � f : Y ! ⌦ is given on each y 2 Y ✓ ⌦n by

f⇤(g)(y) = g(f(y)) = g(f1(y), · · · , fm(y)).

Theorem 1.6.1. Hom(Y, Z) ⇠= Homk�alg

(�(Z),�(Y )).

Proof. Easy part: Given f : Y ! Z, we get f⇤ : �(Z)! �(Y ) as described above.Hard part: Given any ' : �(Z) ! �(Y ) we need to construct a polynomial mapping

f'

: Y ! Z.The ring �(Z) is generated by the images of the generators T

i

of k[T1, · · · , Tm

] which arethe projection maps t

j

: Z ,! ⌦m ! ⌦ to the jth coordinate. (The coordinates of any pointz 2 Z are z

j

= tj

(z).) So, the homomorphism ' is determined by the images '(tj

) : Y ! ⌦of these projection maps. We define f

'

: Y ! Z to be the mapping given by

f'

(y) = ('(t1)(y), · · · ,'(tm)(y)) 2 ⌦m

for all y 2 Y . We need to verify that this is an element of Z. We use the fact that Z =V (I(Z)). In other words, a point x 2 ⌦m lies in Z i↵ g(x) = 0 for all g(T ) 2 I(Z) ✓ k[T ].

When we say that ' : �(Z)! �(Y ) is a homomorphism of k-algebras we mean:

' (linear combination of products of functions �j

: Z ! ⌦)

= the same linear combination of products of the functions '(�j

) : Y ! ⌦.

In other words, for any polynomial g(�j

) of (any) functions �j

: Z ! ⌦ we have:

'(g(�j

)) = g('(�j

)) : Y ! ⌦.


Now take �j

= tj

, the projection maps Z ,! ⌦m ! ⌦. Then:

'(g(t1, · · · , tm)) = g('(t1), · · · ,'(tm)) : Y ! ⌦.

For any g(T ) 2 I(Z) the function g(t⇤) = g(t1, · · · , tm) = 0 since this is the image of g(T )in �(Z). So, '(g(t⇤)) = g('(t⇤)) = 0 : Y ! ⌦. So, for any y 2 Y :

g('(t1)(y), · · · ,'(tm)(y)) = 0.

Since this holds for all g(T ) 2 I(Z) we get: ('(t1)(y), · · · ,'(tm)(y)) 2 Z as claimed.

Lecture 5 starts here.

We are left with two more easy steps: Verify that f⇤'

= ' and f'

= f if ' = 'f

.The map (f

'

)⇤ : �(Z)! �(Y ) is given on the generators tj

: Z ! ⌦ of �(Z) by

(f'

)⇤(tj

) = tj

� f'

= tj

(f'

) = tj

('(t1), · · · ,'(tm)) = '(tj

).

In other words, f⇤'

(tj

) = '(tj

). Since the tj

generate �(Z), f⇤'

= '.Conversely, if we start with f : Y ! Z and ' = f⇤ then

f'

= ('(t1), · · · ,'(tm)) = (f⇤(t1), · · · , f⇤(tm

)) = (t1 � f, · · · , tm � f).This is equal to f since it is the function whose jth coordinate is t

j

� f , the jth coordinateof f . So, f

'

= f . ⇤Proposition 1.6.2. The ideal I(f(Y ))/I(Z) is equal to ker f⇤ ✓ �(Z).

Proof. Since f⇤(g) = g � f , g 2 ker f⇤ i↵ g � f = 0 i↵ g(f(Y )) = 0 i↵ g 2 I(f(Y )).But the polynomial representing a function Z ! ⌦ lies in I(f(Y )) i↵ the function lies inI(f(Y ))/I(Z). ⇤Proposition 1.6.3. Any morphism f : Y ! Z is continuous in the Zariski topology.

Proof. First, suppose that Y = ⌦n, Z = ⌦m. Then we need to show that the inverse imageof any close set V (a) ✓ ⌦m is closed in ⌦n. But, for any y 2 ⌦n,

y 2 f�1(V (a)) , f(y) 2 V (a)

, g(f(y)) = 0 8g 2 a

, y 2 V ({g � f | g 2 a})So, f�1(V (a)) = V ({g � f | g 2 a}) is closed.

In the general case a morphism f : Y ! Z if the restriction of a morphism F : ⌦n ! ⌦m.Let C ✓ Z be closed. Then C is closed in ⌦m. So, F�1(C) is closed in ⌦n and f�1(C) =F�1(C) \ Y is closed in Y . ⇤

1.7. Algebraically closed case. Let’s compare the definitions and the last theorem tothe standard setup. Assume in this subsection that k = k is algebraically closed and letY0 = Y \ kn and Z0 = Z \ km. These are the sets of closed points in Y, Z.

Lemma 1.7.1. Suppose k = k and y 2 ⌦n

. Then y = {y} (y is a closed point) i↵ y 2 kn.

Proof. (() Suppose a 2 kn. Then

⇡(a) = ker⇣

k[X]eva��! k

⌘

= (X1 � a1, · · · , Xn

� an

) = ma

By the weak Nullstellensatz, these are all of the maximal ideals in k[X].


Recall from the proof of Theorem 1.5.1 that y = V (⇡(y)). So,

a = V (ma

) = {y 2 ⌦n | f(y) = 0 8f 2 ma

}Since X

i

� ai

2 ma

this implies (Xi

� ai

)(y) = yi

� ai

= 0. So, yi

= ai

for all i and y = a.So, a = {a}.

()) Conversely, suppose that y = {y}. If y /2 kn then ⇡(y) = p is not maximal (by weakNullstellensatz). So, there is a maximal ideal m

a

◆ p. Then {a} = V (ma

) ✓ V (p) = y.So, a 2 y. Since a 6= y, y has at least two element. This contradicts the assumptiony = {y}. ⇤Theorem 1.7.2. If k = k and Y = V (p) is an irreducible closed subset in ⌦n

corresponding

to the prime ideal p ✓ k[X1, · · · , Xn

] then

Y0 = Y \ kn = {a 2 kn | p ✓ ma

}Proof. For all a 2 kn we have: p ✓ m

a

, V (p)=Y

◆ V (ma

)={a} , a 2 Y . ⇤Lemma 1.7.3. If Y ✓ ⌦n

is an irreducible algebraic set then Y = Y0.

Proof. Let p = I(Y ). By the theorem above, Y0 = Y \ kn is the set of all points a 2 kn sothat p ✓ m

a

. Since the Jacobson radical of k[X]/p is equal to its nilradical, we have:

I(Y0) =\

a2Y0

ma

=\

ma◆p

ma

= r(p) = p

(The Jacobson radical of k[X]/p isT

ma

/p and the nilradical of k[X]/p is r(p)/p. We provedthat these are equal, so the numerators are equal.) So,

Y0 = V (I(Y0)) = V (p) = Y.

⇤When k = k, we define �(Y0) to be the ring of all functions f : Y0 ! k which are given

by polynomials f(X) 2 k[X]. Thus, by definition, we have an epimorphism k[X] ⇣ �(Y0).The kernel if I(Y0) = I(Y ). So,

�(Y0) =k[X]

I(Y0)=

k[X]

I(Y )= �(Y ).

So, any polynomial function f : Y0 ! k extends uniquely to a regular function f : Y ! ⌦.A morphism f : Y0 ! Z0 is defined to be any set mapping which is the restriction to Y0

of a polynomial function kn ! km given by m polynomials in n variables.

Theorem 1.7.4.

Hom(Y0, Z0) = Hom(Y, Z) ⇠= Homk�alg

(�(Z),�(Y )) = Homk�alg

(�(Z0),�(Y0)).

We already know that Hom(Y, Z) ⇠= Homk�alg

(�(Z),�(Y )) = Homk�alg

(�(Z0),�(Y0)).So, we just need to check that Hom(Y0, Z0) = Hom(Y, Z). This implies the standardbijection: Hom(Y0, Z0) ⇠= Homk�alg

(�(Z0),�(Y0)).

Proof. Any morphism f : Y ! Z sends Y0 into Z0 since evaluation of a polynomial withcoe�cients in k at a point with coordinates all in k will give elements of k. So f(Y0) ✓km \ Z = Z0. So, we get a restriction map Hom(Y, Z)! Hom(Y0, Z0).

Conversely, any morphism f : Y0 ! Z0 has coordinates fj

: Y0 ! k. These extenduniquely to regular functions f

j

: Y ! ⌦. Together these give a morphism

f : Y ! ⌦m.


We need to verify that f(Y ) ✓ Z. Equivalently, we need to show Y ✓ f�1

(Z). But f is

continuous and Z is closed. So, f�1

(Z) is a closed set and Y0 ✓ f�1(Z0) ✓ f�1

(Z). Taking

the closure of both sides we get Y = Y 0 ✓ f�1

(Z). So, f(Y ) ✓ Z.If we extend a function Y0 ! Z0 to Y then restrict again to Y0 we get back the same

function. Also, the extension f is unique. So, restricting and extending must give back thesame function Y ! Z. So, Hom(Y0, Z0) = Hom(Y, Z). ⇤

The proof of the following is easy and we skipped it. I emphasized instead the two keyproperties of open sets:

(1) If U is a nonempty open subset of an irreducible set Y then U = Y . The reason is:Y = U [ Y \U . Since Y is irreducible and Y 6= Y \U , we must have Y = U .

(2) If U, V are nonempty open subsets of Y then U \ V is nonempty. The proof is: IfU \ V = ; then Y = (Y \U) [ (Y \V ), a contradiction.

Proposition 1.7.5. For any nonempty open subset U of Y , Y is the closure of U \ Y0.

Proof. The lemma implies that every nonempty open subset of Y meets Y0. If the closureof U \ Y0 is not equal to Y then its complement V is a nonempty open set in Y . Since Yis irreducible U \ V is nonempty. So, U \ V \ Y0 is nonempty. Contradiction. ⇤

For the next corollary, I emphasized the statement.

Definition 1.7.6. If U is an open subset of Y , a regular function on U is a mappingf : U ! ⌦ which is given locally by rational functions. This means that, for all y 2 Uthere is an open neighborhood V of y in U and polynomial functions g, h : V ! ⌦ so thath(x) 6= 0 for all x 2 V and

f(x) =g(x)

h(x)

for all x 2 V . The polynomials g, h may be di↵erent in di↵erent neighborhoods V as longas the functions f = g/h agree on the intersections of these V ’s.

When k = k, a regular function f : U0 = U \ kn ! k is defined to be a function whichis given locally by rational functions f = g/h. The corollary below says that the ring ofregular function f : U0 ! k is equal to the ring of regular functions U ! ⌦. In other words,f extends uniquely to f : U ! ⌦.

We will come back to the discussion of regular functions on open sets after we look atprojective varieties.

Corollary 1.7.7. For any open subset U of Y , any mapping U \ Y0 ! k given locally as

a rational function with coe�cients in k extends uniquely to a regular function U ! ⌦ and

all regular functions U ! ⌦ take U \ Y0 into k.

Proof. Let f : U \ Y0 ! k and, for each x 2 U \ Y0, suppose that there is a basic openneighborhood V = Y

g

⇢ U of x and h 2 �Y so that f(y) = h(y)/g(y)m for all y 2 V \ Y0.Then h/gm is one extension of f to V . Since V \Y0 is dense in V , this is the only extension.

The second statement is clear. ⇤These theorems show that we do not lose any information by restricting functions to the

k points Y0 when k is algebraically closed. However, we are throwing away important anduseful parts of the structure of the a�ne variety and of morphisms of a�ne varieties byrestricting to just the closed points.


1.8. Lecture 6: Review. Before discussing projective space, let us briefly review theconcepts.

(1) k is any field and ⌦ = k(T1, T2, · · · ) is a very large field containing k.(2) There is an order reversing 1-1 correspondence between radical ideals (ideals a =

r(a)) in k[X] and closed subsets Z of ⌦n given by

V (a) := {y 2 ⌦n : f(y) = 0 8f 2 a}a = I(Z) := {f 2 k[X] : f(Z) = 0}

(3) Under this bijection, the prime ideals p ( k[X] correspond to the irreducible subsetsY ✓ ⌦n (those not equal to a union of two proper closed subsets).

(4) Every irreducible set Y contains an element y called a generic point so that y = Y .

I(Y ) = p = I(y) = ⇡(y) = ker(evy

: k[X]! ⌦)

(If y 2 Y is not generic then y = Z ( Y is an irreducible proper subset.)(5) If Y ✓ ⌦n, Z ✓ ⌦m are irreducible, a morphism f : Y ! Z is a function given by

m polynomial equations in n variables.(6) Hom(Y, Z) ⇠= Homk�alg

(�(Z),�(Y )) where

�(Y ) =k[X]

I(Y )

is the ring of all polynomial functions f : Y ! ⌦.(7) For any open subset U ✓ Y , (Y irreducible), a regular function on U is defined to

be a mapping f : U ! ⌦ so that f is given locally by rational functions f(x) = g(x)h(x) .

These functions form a ring �(U).

1.9. First look at schemes. The idea is that all generic points of an irreducible set Yare the same. Each of them represents the “general point” of Y . We do this all the timewhen teaching calculus: Take the curve Y ⇢ R2 given by the equation y = x2. The “generalpoint” of this curve is the point (x, x2) where “x” is a “variable”. But x is a “dummyvariable” which means that (z, z2) also represents a general point on the parabola Y . Inalgebraic geometry, we replace the expression “Let x be a variable” with the more preciseexpression: “Let T1 be a transcendental element over R.” Then y = (T1, T

21 ) is a general

point in Y . Recall that ⌦ is the algebraic closure of the field k(T1, T2, · · · ). These extratranscendental elements are useful to write down other general elements, such as:

T1 T2

T3 T4

�

This is the “general 2⇥ 2 matrix” whose entries are “general elements” of k.The point is that y = (T2, T

22 ) is also a general point in Y . So, we should think of all

these points as being the same point expressed in terms of di↵erent “dummy variables”.Using the concept that dummy variables are equivalent we will make the transition fromvarieties to schemes.

Definition 1.9.1. For any (commutative) ring R let Spec(R) denote the topological spacewhose elements are the prime ideals p ( R. We use the notation [p] for the point in thespace Spec(R) corresponding to p. The basic open subsets of X = Spec(R) are the sets

Xf

= {[p] : f /2 p}for any f 2 R. Recall that a collection of subsets of X form a basis for a topology on X ifthe intersection of any two basic open sets is a union of basic open sets. In this definition


we have Xf

\ Xg

= Xfg

since f, g /2 p , fg /2 p. So the sets Xf

form a basis for sometopology on Spec(R). Two special cases: X0 = ;, X1 = X.

If we think of p as representing the ring homomorphism 'p : R ⇣ R/p then

f /2 p , 'p(f) 6= 0.

When R = k[X1, · · · , Xn

] and p = ⇡(y) = ker evy

then R/p ⇢ ⌦ and

(1.1) [p] 2 Xf

, 'p(f) = f(y) 6= 0 , y /2 V (f).

In the example that we have been discussing so far, R is a finitely generated domain overthe field k. This has the form

R =k[X]

I(Y )where Y is irreducible. (Finitely generated over k means there is an epimorphism

k[X1, · · · , Xn

] ⇣ R.

R being a domain means the kernel is a prime ideal p. We know that such ideals are givenuniquely by p = I(Y ) where Y ✓ ⌦n is irreducible.)

Lemma 1.9.2. Let R be a finitely generated domain over any field k. Let f 2 R be

represented by a polynomial f(X). Let X = Spec(R). Then, for any y 2 Y ,

⇡(y) = p 2 Xf

, y 2 Yf

.

(Recall that Yf

= {y 2 Y : f(y) 6= 0}.) In other words, ⇡�1(Xf

) = Yf

where

⇡ : Y ! Spec(R).

Proof. This follows from (1.1):

y 2 Yf

, y /2 V (f) , [p] 2 Xf

.

⇤Theorem 1.9.3. There is a continuous epimorphism

⇡ : Y ⇣ X = Spec(R) = Spec(k[X]/I(Y ))

sending each point y 2 Y to the prime ideal ⇡(y) = ker evy

: k[X] ! ⌦. Furthermore, ⇡ is

open (the image of any open set in Y is open in Spec(R)). So, Spec(R) has the quotient

topology with respect to ⇡ (V ✓ Spec(R) is open i↵ ⇡�1(V ) is open in Y ).

Proof. Prime ideals in R = k[X]I(Y ) are p

I(Y ) where p is a prime ideal in k[X] containing I(Y ).

But p = ⇡(x) for some x 2 ⌦n and I(Y ) ✓ p i↵ x 2 y = Y . Therefore, ⇡ : Y ⇣ X = Spec(R)is surjective.

The lemma shows that ⇡ is continuous.The lemma also shows that ⇡ is open since ⇡ sends every basic open set Y

f

in Y to anopen set X

f

in X = Spec(R). But any open surjection is a quotient map. So, Spec(R) hasthe quotient topology. ⇤

What is the di↵erence between di↵erent generic points of the same Y ? (These are allelements of ⇡�1(⇡(y)).)

Theorem 1.9.4. For any y 2 ⌦n

there is a 1-1 correspondence between ⇡�1(⇡(y)) (the set

of generic points of y = Y ) and the set of all k-algebra monomorphisms

' : R = k[X]/⇡(y) ,! ⌦.


Proof. Let p = ⇡(y). Then x 2 ⇡�1(⇡(y)) i↵ ⇡(x) = p. For each such x we have

' = evx

: k[X]/p ,! ⌦

Conversely, given any ' : k[X]/p ,! ⌦, we get back the point x since its coordinates aregiven by x

i

= '(Xi

). Since evx

(Xi

) = xi

, this gives a bijection x$ '. ⇤Example 1.9.5. Suppose that k = R. Then any R = R[X]/p has two kinds of maximalideals:

(1) m so that R/m = R. Since there is only one R-algebra homomorphism R ,! ⌦,⇡�1(m) = {x} is a closed point. These are the elements of Y \ Rn. (Y = V (p))

(2) m so that R/m = C. There are exactly two R-algebra monomorphisms C ,! ⌦ (oneis the inclusion map C ,! ⌦ and the other is the composition C ��! C ,! ⌦ wherethe first map � is complex conjugation). So, ⇡�1(m) has two points which lie inY \ Cn and must be complex conjugates of each other.

Everyone knows that the parabola Y = V (X2 �X21 ) is missing its point at infinity and

this is one of the reasons for introducing projective space. But most people don’t realizethat the circle is also missing a point at infinity:

Example 1.9.6. What is Spec(R) where

R =R[X1, X2]

(X21 +X2

2 � 1)?

Here k = R and Y = V (X21 +X2

2 � 1) ⇢ ⌦2. The real points of Y form a circle of radius 1.The complex points of Y form a hyperbola since the solution set of

X21 +X2

2 = (X2 + iX1)(X2 � iX1) = 1

in C2 is given by B = 1A

where

(A,B) = (X2 + iX1, X2 � iX1).

X1

X2 R2

Y \ R2

C2

X1

X2 AB

Y \ C2

Note that, in Y \ C2, A is an arbitrary nonzero complex number, B = 1A

and X1, X2 aregiven by X2 =

12(A+B) and X1 =

12i(A�B). In (A,B)-cordinates, (A,B) is real if A 2 C

lies on the unit circle. The point (A,B) = (z, 1/z) is, in (X1, X2) coordinates, given by

(X1, X2) =

✓

1

2i

✓

z � 1

z

◆

,1

2

✓

z +1

z

◆◆

with complex conjugate:✓

1

2i

✓

1

z� z

◆

,1

2

✓

z +1

z

◆◆


givingA = 1/z.

Another way to see this: When we conjugate X1 and X2, (A,B) is replaced with

(A0, B0) = (X2 + iX1, X2 � iX1) = (B,A).

In other we conjugate A,B and switch them. So, the new A is A0 = B = 1/z. The mappingz 7! 1/z interchanges the outside of the circle and the inside minus the origin. (In polarcoordinates z = re�i✓, 1/z = 1

r

e�✓.) These two points are identified in Spec(R). By dimen-sion theory (to be discussed later) there is only one more point in Spec(R) corresponding tothe unique minimal prime ideal 0. Thus Spec(R) is the punctured disk union an open point.As I said in class, the unique generic point [0] 2 Spec(R) is usually drawn as a “cloud”which is smeared over the whole space since the closure of the point [0] is the whole space.

Spec(R) = [ (generic point)

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