LECTURES ON ALGEBRAIC GEOMETRY MATH 202A 41

5. Affine schemes

The definition of an a�ne scheme is very abstract. We will bring it down to Earth.However, we will concentrate on the definitions. Properties of schemes will be discussedlater.

5.1. Definitions. First recall the definition of the topological space Spec(R), then con-struct the structure sheaf.

5.1.1. The space X = Spec(R). Let R be any commutative ring. Let X = Spec(R), thespace of prime ideals in R (where we sometimes use the notation [p] 2 Spec(R) for p ⇢ R)with the Zariski topology: closed subsets are V (f) = {[p] : f 2 p}. (Recall: when R = k[X]and p = ⇡(x) then f 2 p , f(x) = 0.) The complement of V (f) is

Xf

= {[p] : f /2 p}.These are the basic open sets in X. We also recall:

Xf

⇠= Spec(Rf

)

where Rf

is R with f inverted.

5.1.2. Structure sheaf oX

. To makeX = Spec(R) into an a�ne scheme, we need to constructits structure sheaf o

X

. This will be a sheaf of local rings. The stalk o[p] will be the localring Rp, R localized at the prime p. Recall that this is a ring of fractions:

Rp =na

bwhere a, b 2 R, b /2 p

o

The stalk bundle of the structure sheaf oX

is the set of all pairs ([p], a) where [p] 2 X, a 2 Rp.Projection to the first coordinate gives the bundle map:

a

[p]2X

Rp �! X

Note that sections of the stalk bundle over any U ⇢ X form a ring (but not a local ring)by pointwise addition and multiplication: s1 + s2 and s1s2 are given by (s1 + s2)(x) =s1(x) + s2(x) and (s1s2)(x) = s1(x)s2(x).

For every basic open set Xf

, let �(Xf

, oX

) = Rf

. Each element a = r

f

m 2 Rf

gives a

section �a

of the stalk bundle over Xf

as follows. For any [p] 2 Xf

(f /2 p), let �a

([p]) =([p], a) where a = r

f

m considered as an element of Rp. (Since fm /2 p, this fraction defines

an element of Rp.)The construction of the structure sheaf o

X

follows the following pattern which may beeasier to understand as a separate concept using standard examples.

Definition 5.1.1. Suppose that X is any topological space and {V↵

} is a basis for thetopology on X. Recall that this means:

Every open set in X is a union of basic open sets V↵

.

Equivalently, for every element x of every open set U , there is a basic open set V↵

so that

x 2 V↵

✓ U

Define a basic presheaf on X to be a functor F defined only on the basic open sets V↵

:

(1) For every basic open set V↵

, we are given FV↵

in some category C which has directlimits. We assume that FV

↵

is a set with additional structure, e.g., a ring.

42 KIYOSHI IGUSA BRANDEIS UNIVERSITY

(2) For every pair of basic open sets V↵

✓ V�

, we have the restriction map:

res�↵

: FV�

! FV↵

(3) F is a functor: res↵↵

is the identity map on FV↵

and res�↵

� res��

= res�↵

.

Note that the basic presheaf is given by two things (1) FV↵

, (2) res�↵

.

Theorem 5.1.2. A basic presheaf F defines a stalk bundle and sheaf F 0in the following

way.

(1) The stalk Fx

of F at x 2 X is defined to be the direct limit of FV↵

for all basic open

nbhs V↵

of x:

Fx

= dir. limx2V↵ FV

↵

(2) For all open U ✓ X, F 0U is the set of all sections � of the stalk bundle over U so

that, for every x 2 U there is a basic open nbh x 2 V↵

✓ U and s↵

2 FV↵

so that

�(y) is the germ of s↵

at y for all y 2 V↵

(the germ represented by (V↵

, s↵

).)

Example 5.1.3. If Mn is a di↵erentiable n-manifold then M has an atlas of pairs (U,�)where U is an open set in M and � : U ⇠= B

✏

(0) is a C1 di↵eomorphism of U with the ✏ ballaround 0 in Rn. Then the basic presheaf of C1 embeddings of M into Rm is given by lettingF (U,�) be the set of all mappings U ! Rm so that the composition B

✏

(0) ! U ! Rm is aC1 embedding.

Example 5.1.4. Suppose that (⌃, o⌃) is a prevariety over (k,⌦). By definition, ⌃ is a finiteunion of a�ne open sets U

i

so that (Ui

, o⌃|Ui

) is an a�ne variety with ring of global sectionsR

i

= �(Ui

, o⌃). Basic open sets in ⌃ are given by (Ui

)f

where f 2 Ri

is not nilpotent. Thestructure sheaf o⌃ is given by the basis presheaf:

F ((Ui

)f

) = (Ri

)f

=

⇢

a

fm

: a 2 Ri

�

If (Ui

)f

✓ (Uj

)g

then what is the restriction map

res : (Rj

)g

! (Ri

)f

?

From the point of view of the rings, this is di�cult to describe. But, with a�ne varieties,we have a elementary description of the restriction map: F ((U

j

)g

) = (Rj

)g

is a subring ofthe ring of all functions (U

j

)g

! ⌦. The restriction map F ((Uj

)g

) = F ((Ui

)f

) is given byrestriction of these mappings to the subset (U

i

)f

✓ (Uj

)g

.

In this example, we started with a sheaf o⌃ and we defined the basic presheaf in terms ofthe sheaf. In such a case, the sheafification will be the sheaf we started with. This followsfrom the definition of a sheaf. We need to make this easy concept precise.

Lemma 5.1.5. Suppose that G is a sheaf on X. Suppose that F is a basic presheaf with the

property that FV↵

⇠= GV↵

for every basic open set V↵

and furthermore, that this isomorphism

is natural: I.e., the following diagram commutes.

FV�

res

�↵✏✏

⇠=// GV

�

res

�↵

✏✏

FV↵

⇠=// GV

↵

Then G is isomorphic to the sheafification F 0of F . In particular, F 0V

↵

⇠= GV↵

⇠= FV↵

.

LECTURES ON ALGEBRAIC GEOMETRY MATH 202A 43

Proof. Since FV↵

⇠= GV↵

, we get an isomorphism of stalks:

Fx

= dir. limx2V↵ FV

↵

⇠= dir. limx2V↵ GV

↵

= Gx

So, F,G have the same (isomorphic) stalk bundles. The sheafification F 0, G0 are thus equalby definition. (They are both sections of the same stalk bundle which agree with FV

↵

= GV↵

in a nbh V↵

of each x 2 U .) Since G is already a sheaf, we have G = G0 = F 0. ⇤Definition 5.1.6. The structure sheaf of X = Spec(R) for any commutative ring R isdefined to be the sheafification o

X

= F 0 of the basic presheaf F on X given by FXf

= Rf

with restriction map FXf

! FXg

defined to be the unique ring homomorphism Rf

! Rg

making the following diagram commute.

R

~~

Rf

// Rg

The pair (X, oX

) is called an a�ne scheme.

One of the basic theorems about a�ne schemes is:

�(Xf

, oX

) ⇠= Rf

In other words, F 0Xf

= FXf

for all basic open sets Xf

in X. We will prove this in thespecial case when R = �(⌃, o⌃) is the coordinate ring of an a�ne variety. By Lemma 5.1.5it su�ces to find a sheaf G and a natural isomorphism GX

f

⇠= Rf

for every basic open Xf

.

5.2. Example: R = Z.

5.2.1. The space SpecZ. X = SpecZ hasclosed points: [p] for p = (2), (3), (5), (7), · · · (all primes)generic point: 0 with 0 = X.The basic open sets are

Xn

= {[p] : p - n} [ {0}For example, X6 is everything except [2], [3]: V (6) = {[2], [3]} is the complement of X6.Note that X

n

✓ Xm

i↵ m|n. For example, X6 ✓ X3.The only other open sets are the empty set and the whole space X.

5.2.2. Structure sheaf. The structure sheaf is given on basic open sets by

�(Xn

) = Rn

=n a

nk

: a 2 Zo

= Z

1

n

�

and on all of X by �(X) = Z.So, the local rings are (avoiding Z

p

which means something else)

o[p] = Rp

=na

b: a, b 2 Z, p - b

o

o[0] = dir. limRn

=[

n

Rn

= Q

In terms of the stalk bundle: an arbitrary section � of the stalk bundle over U ✓ X is afunction which assigns

�(p) =ap

bp

p - bp

bp

/2 (p)

�(0) =a0b0

b0 6= 0 b0 /2 (0)

44 KIYOSHI IGUSA BRANDEIS UNIVERSITY

(The general condition is that the denominator is not in the prime ideal.) In order for � tobe an element of �(U), �(x) must be given by the same fraction for all x 2 U . For example:

�(p) =3

2is a section on X6 since p - 5 for all p 6= 2, 3.

Indeed, this is equal to 96 2 R6 = �(X6).

Recall that this is the same condition which defines sections of o⌃ over U ✓ ⌃: �(x) mustbe given by the same rational function f

g

2 K(⌃) for all x 2 U . In the case X = SpecZ,the field is K(X) = Q. So:

Rn

= �(Xn

) ✓ K(X) = Qwhich we already knew. But now we should realize that it means: For each section � thereis a single element of K(X) which gives the value of � at all points.

5.3. Relation to a�ne varieties. Now consider the special case R = k[Z]/P = �(⌃)where P is a prime ideal and X = Spec(R). The prime ideals in R are p/P where p is aprime in k[Z] containing P . We will abuse notation and use p, p/P interchangeably.

⌃ ✓ An is an a�ne variety with structure sheaf o⌃. Recall that we have an epimorphism⇡ : ⌃ ⇣ X given by ⇡(x) = p = {f 2 k[Z] : f(x) = 0}.

Theorem 5.3.1. The structure sheaf of X = Spec(R) is the push-forward (direct image)

of the structure sheaf of ⌃, i.e.,

�(U, oX

) = �(⇡�1U, o⌃)

for every open subset U of X.

It is easy to see that the push-forward of any sheaf along any continuous epimorphismis a sheaf.

Here is the lemma we used in the theorem.

Lemma 5.3.2. Let � : X ⇣ Y be a continuous epimorphism and let G be a sheaf on X.

Then the push-forward presheaf

G(U) = G(��1U)

is a sheaf on Y .

Proof. This follows from the definition of a sheaf. Suppose that {V↵

} is a collection of opensets in Y with union U . Then we want to show that the following diagram is exact:

0 ! F (U) !Y

F (V↵

) ◆Y

F (V↵

\ V�

)

Given any collection of sections s↵

2 F (V↵

) which agree on intersections:

s↵

|V↵

\ V�

= s�

|V↵

\ V�

we need to show that there exists a unique s 2 F (U) so that s|V↵

= s↵

for all ↵. But weknow that the same statement is true for G(��1V

↵

) = F (V↵

). Note that {��1V↵

} is anopen covering of ��1U in X. The collection of sections �⇤s

↵

2 G(��1V↵

) are compatibleon overlaps since

��1V↵

\ ��1V�

= ��1(V↵

\ V�

)

So, the sections �⇤s↵

extend uniquely to a section s 2 G(��1U) = F (V↵

). Therefore, F isa sheaf. ⇤

LECTURES ON ALGEBRAIC GEOMETRY MATH 202A 45

Proof of Theorem 5.3.1. We use Lemma 5.1.5 in the case where G is the push-forward ofthe structure sheaf of ⌃ along ⇡ : ⌃ ⇣ X:

G(U) := �(⇡�1U, o⌃)

and F is the basic presheaf FXf

= Rf

used to define oX

. In other words, we need to showthat

�(⇡�1Xf

, o⌃) = Rf

First, ⇡�1Xf

= ⌃f

since

x 2 ⌃f

, f(x) 6= 0 , f /2 ⇡(x) , ⇡(x) 2 Xf

So, by Theorem 3.3.3, we have:

�(⇡�1Xf

, o⌃) = �(⌃f

) = Rf

as required.We also need to show naturality. IfX

g

✓ Xf

then the restriction maps make the followingdiagram commute.

FXf

= Rf

✏✏

⇠=// �(⌃

f

, o⌃)

✏✏

FXg

= Rg

⇠=// �(⌃

g

, o⌃)

�(⌃f

, o⌃)

✏✏

=// �(X

f

, oX

)

✏✏

�(⌃g

, o⌃)=// �(X

g

, oX

)

The vertical arrow on the left is the one making the left hand triangle in the followingdiagram commute. The vertical arrow on the right is the one making the right hand trianglein the following diagram commute.

Rf

##

R

??

��

K(⌃)

Rg

;;

But the whole diagram below commutes. So the diagram above commutes. ⇤

Corollary 5.3.3. �(Xf

, oX

) = Rf

in the special case at hand (when R is a finite generated

k-algebra without zero divisors). ⇤

This is true in general but not that easy to prove. I will give the proof later.

5.4. Basic properties of a�ne schemes.

46 KIYOSHI IGUSA BRANDEIS UNIVERSITY

5.4.1. X is quasi-compact. First, we need the general statement of the Nullstellensatz. LetX = Spec(R). For any subset S ✓ R, let

V (S) :=\

f2SV (f) = {[p] : S ✓ p}.

This is closed since each V (f) is closed by definition. Then we clearly have:

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

For any subset Z ✓ X, let

I(Z) :=\

[p]2Z

p = {f 2 R : f 2 p for all [p] 2 Z}.

Since I(Z) is an intersection of prime ideals we have: r(I(Z)) = I(Z).

Theorem 5.4.1 (Nullstellensatz). For any ideal I in R, I(V (I)) = r(I).

Proof. By definition, I(V (I)) is the intersection of all prime ideals which contain I. This iseasily seen to be the radical of I: By passing to R/I, we may assume that I = 0. Then thestatement is that the intersection of all prime ideals is equal to the set of nilpotent elements.To see this take any f which is not nilpotent. Take any maximal ideal in R

f

. The inverseimage in R is a prime ideal which does not contain f . Conversely, nilpotent elements lie inevery prime ideal. ⇤

Corollary 5.4.2. V (I(Z)) = Z.

Proof. Any closed subset of X is equal to V (S) for some S ✓ R. So, Z = V (S) and

Z ✓ V (I(Z)) ✓ V (I(Z)) = V (I(V (S))) = V (r(S)) = Z.

Since V (I(Z)) is closed, it is equal to Z. ⇤

Proposition 5.4.3. Let X = Spec(R). Then X =S

Xf↵ if and only if 1 2 (· · · f

↵

· · · ),where I = (· · · f

↵

· · · ) is the ideal generated by the f↵

.

Corollary 5.4.4. X = Spec(R) is quasi-compact.

Proof. Given any open covering of X, we can refine it to an open covering by basic opensets X

f↵ . Then, 1 =P

g↵if↵i . So, the X

f↵icover X. (Each of these basic open sets is

contained in one of the original open sets of the covering and those form a finite subcoverof the original covering.) ⇤

The proposition follows from the following lemma by setting g = 1.

Lemma 5.4.5. Xg

=S

Xf↵ if and only if r(g) = r(· · · f

↵

· · · ).

Proof. The complements of the sets Xg

, Xf↵ are V (g) and V (f

↵

). By the Nullstellensatz,

V (g) =\

V (f↵

) = V ({f↵

}) = V (I)

) r(g) = I(V (g)) = I(V ({f↵

})) = r(I) ) V (r(g)) = V (r(I))

But V (r(g)) = V (g) and V (r(I)) = V (I). So, these conditions are equivalent. ⇤

LECTURES ON ALGEBRAIC GEOMETRY MATH 202A 47

5.4.2. Local rings.

Proposition 5.4.6. Suppose that p ✓ q ⇢ R are prime ideals. Let x = [p], y = [q]. Then

y 2 x. In other words, every open nbh of y contains x.

Proof. For any basic open set Xf

in X we have:

y 2 Xf

, f /2 q ) f /2 p , x 2 Xf

So, every basic open nbh of y contains x. ⇤This implies that we have a ring homomorphism:

oy

! ox

This is easy to see knowing that oy

= Rq and ox

= Rp. After inverting the elements in thecomplement of q we can then invert the remaining elements in the complement of p. Thisalso follows from the definition of these stalks:

oy

= dir. limy2V �(V, o

X

) ! dir. limx2W �(W, o

X

) = ox

Since the set of all V ’s is contained in the set of all W ’s, the direct system on the left isa subsystem of the one on the right. So, the left hand direct limit maps to the right handdirect limit.

Example 5.4.7. Let R = Z, x = (0), y = (3). Then the map oy

! ox

is the inclusion map:

oy

= R(3) =na

b: 3 - b

o

,! ox

= Q.

This is a non-example of the following definition.

Definition 5.4.8. Let R1, R2 be local rings with maximal ideals M1,M2. Then a ringhomomorphism � : R1 ! R2 is called local if ��1(M2) = M1.

In the example, R1 = R(3),M1 = (3), R2 = Q, M2 = 0. The inverse image of M2 = 0 inR1 is 0 which is not the maximal ideal in R(3). So, � is not local.

If � : R1 ! R2 is a local morphism then, we get an induced morphism of quotient rings:

� : R1/M1 ! R2/M2

In the example, R1 = Z(3), R2 = Q, there is no mapping

Z(3)/m = Z/3Z ! Q.

5.4.3. The function a(x). If x = [p] 2 X = Spec(R) then ox

is the local ring Rp withmaximal ideal m

x

. For any a 2 ox

let

a(x) := the image of a in k(x) = ox

/mx

If a 2 �(V, oX

) then, for each x 2 V ,

a(x) := ax

(x) 2 k(x)

where ax

2 ox

is the germ of a at x. (Thus ax

is the equivalence class of the pair (V, a)where (V, a) ⇠ (U, b) if 9x 2 W ✓ U \ V so that a|W = b|W .)

The idea is that the “function” a(x) generalizes the idea that, for a�ne varieties, sectionsof the structure sheaf are ⌦-valued functions. Suppose that g 2 �(V, o⌃) where ⌃ is an a�nevariety. Then g is a function

g : V ! ⌦

48 KIYOSHI IGUSA BRANDEIS UNIVERSITY

I explained in class that the definition of a(x) for the a�ne scheme corresponding to ⌃is the same as the definition of g(x):

a(x) = g(x).

To show this, we went through the definitions. If g 2 R and x 2 ⌃ then

g(x) = evaluation of g on x = evx

(g)

The corresponding element of X = Spec(R), R = k[Z]/I(⌃), is

⇡(x) = ker(evx

: R ! ⌦)

The local ring at x is

ox

= R⇡(x) =

⇢

f

h: h(x) 6= 0

�

with unique maximal ideal

mx

=

⇢

f

h: f(x) = 0, h(x) 6= 0

�

So,ox

mx

⇠= im(evx

) ✓ ⌦

Recall that the function g is given by the same fraction f/h at all points in ⌃. The sectiona 2 �(X, o

X

) corresponding to the function g is this fraction considered as an element ofR

h

on the set Xh

. Then

a(x) = image of a = f/h in k(x) =ox

mx

= evx

(a) = f(x)/h(x) = g(x).

When X = Spec(R) is not an a�ne variety then there is no ⌦ and the “function”x 7! a(x) 2 k(x) has target sets k(x) which vary as x varies. For varieties, k(x) is naturallya subset of ⌦ making a an ⌦-valued function.

5.5. Morphisms of a�ne schemes. We need morphisms of a�ne schemes to have acategory which we will show is isomorphic to the category of rings.

5.5.1. Definition. Let (X, oX

), (Y, oY

) be a�ne schemes. Then a morphism:

(f, f⇤) : (X, oX

) ! (Y, oY

)

consists of (1), (2) satisfying (3), (4).

(1) f : X ! Y is continuous.(2) For all open subsets V ✓ Y we have a ring homomorphism

f⇤V

: �(V, oY

) ! �(f�1V, oX

)

(3) The homomorphisms f⇤V

are natural. I.e., for open subsets U ✓ V , we have acommuting diagram:

�(V, oY

)

res

VU

✏✏

f

⇤V// �(f�1V, o

X

)

res

f�1V

f�1U✏✏

�(U, oY

)f

⇤U// �(f�1U, o

X

)

LECTURES ON ALGEBRAIC GEOMETRY MATH 202A 49

(4) For every x 2 X, every open nbh V of f(x) in Y , and every a 2 �(V, oY

) we have:

a(f(x)) = 0 , f⇤V

(a)(x) = 0

In other words, the germ f⇤x

: of(x) ! o

x

of f at x is a local morphism. So, itinduces a homomorphism

k(f(x)) ! k(x).

Claim: This homomorphism sends a(f(x)) 2 k(f(x)) to f⇤V

(a)(x) 2 k(x).Examples ...

50 KIYOSHI IGUSA BRANDEIS UNIVERSITY

5.6. Proof that �(Xf

, oX

) = Rf

. Recall that

(5.1) Xf

=[

Xf↵ () r(f) = r(· · · f

↵

· · · )

Lemma 5.6.1. f, g 2 R, X = Spec(R). Then

Xg

✓ Xf

() g 2 r(f)

Proof. By (5.1) we have:

Xg

✓ Xf

() Xf

= Xf

[Xg

() r(f) = r(f, g) () g 2 r(f).

⇤So, gm = fh for some h. So, we have ring homomorphism

Rf

! Rg

a

fn

7! ahn

fnhn=

ahn

gnm

This is 0 in Rg

i↵ agk = 0 for some k. (ahngs = 0 ) agn+s = 0). I rephrased this asfollows. Define the annihilator of a 2 R to be

Ann(a) = {b 2 R : ab = 0}Then a/fn is zero in R

g

i↵ g 2 Ann(a). Similarly, a/fn = 0 in Rf

is f 2 Ann(a).

Lemma 5.6.2. Suppose that Xf

=S

Xf↵ and

a

f

n 2 Rf

. Then

a

f

n goes to zero in Rf↵ for

all ↵ i↵

a

fn= 0 in R

f

.

Proof. Certainly, a/fn = 0 in Rf

implies it is 0 in each Rf↵ . Conversely, suppose a/fn = 0

in each Rf↵ . Then f

↵

2 r(Ann(a)). So,

f 2 r(f) = r(· · · f↵

· · · ) ⇢ r(Ann(a))

So, a/fn = 0 in Rf

. ⇤Exercise 5.6.3. Show that X

f

is quasi-compact.

Basically, this follows from the fact that, if f 2 r(· · · f↵

· · · ), then some power of f is afinite sum: fn =

P

hi

f↵i . So, Xf

is a union of the corresponding Xf↵i

.

Lemma 5.6.4. Suppose that Xf

= Xf1 [ · · ·[X

fs and gi

2 Rfi so that the images of g

i

, gj

in Rfifj agree for all i, j. Then there is some g 2 R

f

which maps to gi

for all i.

Proof. gi

= ai

/fnii

. Since there are only finitely many, assume ni

are all equal to n. Sincegi

= gj

in Rfifj we have:

ai

fn

j

� aj

fn

i

fn

i

fn

j

= 0

Since there are only finitely many pairs i, j, there is an m so that

(ai

fn

j

� aj

fn

i

)(fi

fj

)m = 0

Let bi

= ai

fm

i

. Then:

bi

fN

j

= bi

fn+m

j

= bj

fn+m

i

= fN

i

bj

where N = n+m. Then gi

= bi

/fN

i

for all i. But

f 2 r(· · · fi

· · · ) = r(· · · fN

i

· · · )

LECTURES ON ALGEBRAIC GEOMETRY MATH 202A 51

So, f s =P

hi

fN

i

for some hi

2 R. Let g = 1f

s

P

hi

bi

2 Rf

. Then, in Rf

we have:

gfN

j

=

P

hi

bi

fN

j

f s

=

P

hi

fN

i

bj

f s

= bj

So, in Rfi we have g = b

j

/fN

j

= gj

. ⇤Theorem 5.6.5. If X = Spec(R) and f 2 R is not nilpotent then

�(Xf

, oX

) = Rf

In other words, the sheafification F 0of the basic presheaf F (X

f

) = Rf

has the same value

on the basic open sets: F 0(Xf

) = F (Xf

).

Proof. By definition of F 0, any element of �(Xf

, oX

) is locally given by g↵

2 F (Xf↵) = R

f↵ .And these are compatible on overlaps X

f↵ \Xf�. Since X

f

=S

Xf↵ , there is an element

g 2 Rf

which maps to each of these. By the other lemma, g is unique. ⇤5.7. Preschemes.

Definition 5.7.1. A ringed space is a pair (X, oX

) where X is a topological space and oX

is a sheaf of (commutative) rings on X. A prescheme is a ringed space which is locallyisomorphic to an a�ne scheme: X =

S

U↵

so that (U↵

, oX

|U↵

) ⇠= Spec(R↵

) with itsstandard structure sheaf.

Definition 5.7.2. A morphism (f, f⇤) : (X, oX

) ! (Y, oY

) of preschemes is

(1) f : X ! Y a continuous map(2) A ring homomorphism f⇤

V

: �(V, oY

) ! �(f�1V, oX

) for every open subset V ✓ Yso that

(3) f⇤ is natural with respect to restriction and(4) f⇤

x

: of(x) ! o

x

is a local homomorphism for all x 2 X.

f⇤x

is the germ of f⇤ at f(x). So, it might be more accurate to write it as: (fx

)⇤ = (f⇤)f(x).

Proposition 5.7.3. Given X,Y preschemes with a�ne open sets X =S

U↵

, Y =S

V�

.

Then, a morphism X ! Y can be given as follows. Suppose that for all ↵ there is a � and

a morphism of a�ne schemes f↵�

: (U↵

, oX

|U↵

) ! (V�

, oY

|V�

) which agree on overlaps:

(a) The composition f↵�

: U↵1 ! V

�1 ,! Y is equal to U↵2 ! V

�2 ,! Y on U↵1 \ U

↵2

giving a continous map f =S

f↵�

: X ! Y .

(b) For every open B ✓ V�1 \ V

�2 and A ✓ U↵1 \ U

↵2 \ f�1(B), the two induced ring

homomorphisms:

(f↵i�i)

⇤B

: �(B, oY

) ! �(A, oX

)

agree.

Proof. This is straightforward. The continuous mappings paste together to give f . Themap on sheaves is given first on each of open set W ✓ Y by: W =

S

�

V�

\W and

0 // �(W ) //

9!f⇤W

✏✏

Q

�

�(V�

\W ) +3

resf

⇤↵�

✏✏

Q

�(V�

\ V�

0 \W )

res

✏✏

0 // �(f�1W ) //

Q

�

�(U↵

\ f�1W ) +3

Q

�(U↵

\ U↵

0 \ f�1W )

The key point is that condition (b) implies that the right hand square commutes andtherefore induces a unique map on equalizers. ⇤

52 KIYOSHI IGUSA BRANDEIS UNIVERSITY

Example 5.7.4. Suppose that R =L

d�0Rd

is a graded ring and X = Proj(R) is thespace of homogeneous prime ideals which do not contain

L

d>0Rd

. Recall that the basicopen sets are, for any homogeneous f 2 R

d

, d > 0,

Xf

:= {[P ] 2 Proj(R) : f /2 P}Note that P does not contain

L

d>0Rd

. So, there exists an f 2L

d>0Rd

not contained inP . (So, one of the homogeneous components of f will not be in P .) Therefore,

X =[

Xf

.

To make this into a prescheme, we define a basic presheaf by

F (Xf

) = R(f) :=

⇢

a

fn

: a 2 Rnd

�

This is the subring of Rf

of all elements of degree 0. (Recall that on Pn(⌦), functions arelocally defined by g

h

where g, h are homogeneous polynomials of the same degree.) Then:X

f

is homeomorphic to the a�ne scheme Spec(R(f)) and the basic presheaf defined above isthe same as the basic presheaf defining the structure sheaf of Spec(R(f)). Furthermore, thesheafification of the basic presheaf is the same by what we just proved: F 0(X

f

) = F (Xf

).So, the structure sheaves of the various a�ne schemes agrees on overlaps. So, they define aprescheme structure on all of Proj(R).

Lemma 5.7.5. Xf

is homeomorphic to the a�ne scheme Spec(R(f)) with “the same struc-

ture sheaf.”

Proof. The points in Xf

are homogeneous prime ideals P which do not contain f . Beinghomogeneous means P =

L

Ps

. If f 2 Rd

, the corresponding prime in R(f) is:M

f�nPnd

Conversely, given any prime ideal Q in R(f), let I be the direct sum of all numerators of allelements of Q. Then I is a homogeneous ideal in R. Let P = r(I). Then P is a homogeneousprime ideal not containing f . To see that P is prime, suppose that a 2 R

j

, b 2 Rk

andab 2 P . Then, for some n, s, anbn/f s 2 Q. So nj + nk = ds and

adnbdn

fds

=

✓

adn

fnj

◆✓

bdn

fnk

◆

2 Q

Since Q is prime, either adn or bdn is in I. So, either a or b is in P = r(I). This gives a 1-1correspondence P $ Q.

Suppose that g 2 Rs

is not nilpotent. Then [P ] 2 Xfg

= Xf

\Xg

i↵ gd/f s is not in Q.Conversely, a/fn is not in Q i↵ [P ] 2 X

af

= Xf

\Xa

. So, the basic open sets of Xf

andSpec(R(f)) correspond.

The structure sheaves are equal by definition! ⇤Example 5.7.6. Suppose that (⌃, o⌃) is a prevariety. So, ⌃ =

S

U↵

so that each U↵

isan a�ne variety. Let R

↵

= �(U↵

) be the ring of regular function U↵

! ⌦. Define anequivalence relation on the set ⌃ by x ⇠ y i↵ x = y. Note that, if x 2 U

↵

, then, also,y 2 U

↵

since, otherwise, y would be disjoint from the open set U↵

and could not contain x.Therefore, the prime ideals are equal: ⇡(x) = ⇡(y).

Let X = ⌃/⇠ be the quotient space with the quotient topology and let oX

be thepush-forward sheaf of o⌃.

LECTURES ON ALGEBRAIC GEOMETRY MATH 202A 53

Proposition 5.7.7. (X, oX

) is a prescheme.

Proof. The projection map f : ⌃ ⇣ X = ⌃/⇠ is an open map since f(U) is open i↵f�1f(U) is open. But:

f�1f(U) = U

⌃ =S

U↵

, U↵

a�ne variety. Then

f(U↵

) = U↵

/⇠ = (Spec(R↵

), oX↵)

with the push-forward structure sheaf.But: push-forward commutes with restriction by definition. So, the restriction of o

X

to X↵

is the push-forward of oU↵ which we already showed to be the structure sheaf of

Spec(R↵

). ⇤