Download Sec 5: Affine schemes

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41
5. Affine schemes
The definition of an affine scheme is very abstract. We will bring it down to Earth.
However, we will concentrate on the definitions. Properties of schemes will be discussed
later.
5.1. Definitions. First recall the definition of the topological space Spec(R), then construct the structure sheaf.
5.1.1. The space X = Spec(R). Let R be any commutative ring. Let X = Spec(R), the
space 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 make X = Spec(R) into an affine scheme, we need to construct
its structure sheaf oX . This will be a sheaf of local rings. The stalk o[p] will be the local
ring Rp , R localized at the prime p. Recall that this is a ring of fractions:
na
o
Rp =
where a, b 2 R, b 2
/p
b
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
Rp ! X
[p]2X
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 s1 s2 are given by (s1 + s2 )(x) =
s1 (x) + s2 (x) and (s1 s2 )(x) = s1 (x)s2 (x).
For every basic open set Xf , let (Xf , oX ) = Rf . Each element a = frm 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 = frm considered as an element of Rp . (Since f m 2
/ p, this fraction defines
an element of Rp .)
The construction of the structure sheaf oX follows the following pattern which may be
easier 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 the
topology 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 F V↵ in some category C which has direct
limits. We assume that F V↵ is a set with additional structure, e.g., a ring.
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KIYOSHI IGUSA BRANDEIS UNIVERSITY
(2) For every pair of basic open sets V↵ ✓ V , we have the restriction map:
res↵ : F V ! F V↵
(3) F is a functor: res↵↵ is the identity map on F V↵ and res↵ res = res↵ .
Note that the basic presheaf is given by two things (1) F V↵ , (2) res↵ .
Theorem 5.1.2. A basic presheaf F defines a stalk bundle and sheaf F 0 in the following
way.
(1) The stalk Fx of F at x 2 X is defined to be the direct limit of F V↵ for all basic open
nbhs V↵ of x:
Fx = dir. limx2V↵ F V↵
0
(2) For all open U ✓ X, F U 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 F V↵ 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 M n 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 C 1 di↵eomorphism of U with the ✏ ball
n
around 0 in R . Then the basic presheaf of C 1 embeddings of M into Rm is given by letting
F (U, ) be the set of all mappings U ! Rm so that the composition B✏ (0) ! U ! Rm is a
C 1 embedding.
Example 5.1.4. Suppose that (⌃, o⌃ ) is a prevariety over (k, ⌦). By definition, ⌃ is a finite
union of affine open sets Ui so that (Ui , o⌃ |Ui ) is an affine variety with ring of global sections
Ri = (Ui , o⌃ ). Basic open sets in ⌃ are given by (Ui )f where f 2 Ri is not nilpotent. The
structure sheaf o⌃ is given by the basis presheaf:
⇢
a
F ((Ui )f ) = (Ri )f =
: a 2 Ri
fm
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 difficult to describe. But, with affine varieties,
we have a elementary description of the restriction map: F ((Uj )g ) = (Rj )g is a subring of
the ring of all functions (Uj )g ! ⌦. The restriction map F ((Uj )g ) = F ((Ui )f ) is given by
restriction of these mappings to the subset (Ui )f ✓ (Uj )g .
In this example, we started with a sheaf o⌃ and we defined the basic presheaf in terms of
the sheaf. In such a case, the sheafification will be the sheaf we started with. This follows
from 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 F V↵ ⇠
= GV↵ for every basic open set V↵ and furthermore, that this isomorphism
is natural: I.e., the following diagram commutes.
FV
⇠
=
/ GV
res↵
res↵
✏
F V↵
⇠
=
✏
/ GV↵
Then G is isomorphic to the sheafification F 0 of F . In particular, F 0 V↵ ⇠
= GV↵ ⇠
= F V↵ .
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Proof. Since F V↵ ⇠
= GV↵ , we get an isomorphism of stalks:
Fx = dir. limx2V↵ F V↵ ⇠
= dir. limx2V↵ GV↵ = Gx
So, F, G have the same (isomorphic) stalk bundles. The sheafification F 0 , G0 are thus equal
by definition. (They are both sections of the same stalk bundle which agree with F V↵ = 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 is
defined to be the sheafification oX = F 0 of the basic presheaf F on X given by F Xf = Rf
with restriction map F Xf ! F Xg defined to be the unique ring homomorphism Rf ! Rg
making the following diagram commute.
R
Rf
~
/ Rg
The pair (X, oX ) is called an affine scheme.
One of the basic theorems about affine schemes is:
(Xf , oX ) ⇠
= Rf
In other words, F 0 Xf = F Xf for all basic open sets Xf in X. We will prove this in the
special case when R = (⌃, o⌃ ) is the coordinate ring of an affine variety. By Lemma 5.1.5
it suffices to find a sheaf G and a natural isomorphism GXf ⇠
= Rf for every basic open Xf .
5.2. Example: R = Z.
5.2.1. The space SpecZ. X = SpecZ has
closed 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 Xn ✓ 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

na
o
1
(Xn ) = Rn =
: a2Z =Z
n
nk
and on all of X by (X) = Z.
So, the local rings are (avoiding Zp which means something else)
na
o
o[p] = Rp =
: a, b 2 Z, p - b
b
[
o[0] = dir. lim Rn =
Rn = Q
n
In terms of the stalk bundle: an arbitrary section of the stalk bundle over U ✓ X is a
function which assigns
ap
(p) =
p - bp bp 2
/ (p)
bp
a0
(0) =
b0 6= 0 b0 2
/ (0)
b0
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KIYOSHI IGUSA BRANDEIS UNIVERSITY
(The general condition is that the denominator is not in the prime ideal.) In order for to
be an element of (U ), (x) must be given by the same fraction for all x 2 U . For example:
(p) =
3
is a section on X6 since p - 5 for all p 6= 2, 3.
2
Indeed, this is equal to 96 2 R6 = (X6 ).
Recall that this is the same condition which defines sections of o⌃ over U ✓ ⌃: (x) must
be given by the same rational function fg 2 K(⌃) for all x 2 U . In the case X = SpecZ,
the field is K(X) = Q. So:
Rn = (Xn ) ✓ K(X) = Q
which we already knew. But now we should realize that it means: For each section
is a single element of K(X) which gives the value of at all points.
there
5.3. Relation to affine 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 a
prime in k[Z] containing P . We will abuse notation and use p, p/P interchangeably.
⌃ ✓ An is an affine 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 ) = (⇡
1
U, o⌃ )
for every open subset U of X.
It is easy to see that the push-forward of any sheaf along any continuous epimorphism
is 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( 1 U )
is a sheaf on Y .
Proof. This follows from the definition of a sheaf. Suppose that {V↵ } is a collection of open
sets in Y with union U . Then we want to show that the following diagram is exact:
Y
Y
0 ! F (U ) !
F (V↵ ) ◆
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 we
know that the same statement is true for G( 1 V↵ ) = F (V↵ ). Note that { 1 V↵ } is an
1 U in X. The collection of sections ⇤ s 2 G(
1 V ) are compatible
open covering of
↵
↵
on overlaps since
1
V↵ \ 1 V = 1 (V↵ \ V )
So, the sections
a sheaf.
⇤s
↵
extend uniquely to a section s 2 G(
1U )
= F (V↵ ). Therefore, F is
⇤
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Proof of Theorem 5.3.1. We use Lemma 5.1.5 in the case where G is the push-forward of
the structure sheaf of ⌃ along ⇡ : ⌃ ⇣ X:
1
G(U ) := (⇡
U, o⌃ )
and F is the basic presheaf F Xf = Rf used to define oX . In other words, we need to show
that
(⇡
First, ⇡
1X
f
1
X f , o⌃ ) = Rf
= ⌃f since
x 2 ⌃f , f (x) 6= 0 , f 2
/ ⇡(x) , ⇡(x) 2 Xf
So, by Theorem 3.3.3, we have:
(⇡
1
Xf , o⌃ ) = (⌃f ) = Rf
as required.
We also need to show naturality. If Xg ✓ Xf then the restriction maps make the following
diagram commute.
⇠
=
F Xf = Rf
✏
⇠
=
(⌃f , o⌃ )
=
F Xg = Rg
✏
=
(⌃g , o⌃ )
(⌃f , o⌃ )
/
✏
(⌃g , o⌃ )
/
(Xf , oX )
/
✏
(Xg , oX )
/
The vertical arrow on the left is the one making the left hand triangle in the following
diagram commute. The vertical arrow on the right is the one making the right hand triangle
in 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 affine schemes.
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KIYOSHI IGUSA BRANDEIS UNIVERSITY
5.4.1. X is quasi-compact. First, we need the general statement of the Nullstellensatz. Let
X = Spec(R). For any subset S ✓ R, let
\
V (S) :=
V (f ) = {[p] : S ✓ p}.
f 2S
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 is
easily seen to be the radical of I: By passing to R/I, we may assume that I = 0. Then the
statement 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 Rf . The inverse
image in R is a prime ideal which does not contain f . Conversely, nilpotent elements lie in
every 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.
S
⇤
Proposition 5.4.3. Let X = Spec(R). Then X = 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 openPcovering of X, we can refine it to an open covering by basic open
sets Xf↵ . Then, 1 =
g↵i f↵i . So, the Xf↵i cover X. (Each of these basic open sets is
contained in one of the original open sets of the covering and those form a finite subcover
of the original covering.)
⇤
The proposition follows from the following lemma by setting g = 1.
S
Lemma 5.4.5. Xg = 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.
⇤
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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 the
complement of q we can then invert the remaining elements in the complement of p. This
also follows from the definition of these stalks:
oy = dir. limy2V
(V, oX ) ! dir. limx2W (W, oX ) = ox
Since the set of all V ’s is contained in the set of all W ’s, the direct system on the left is
a subsystem of the one on the right. So, the left hand direct limit maps to the right hand
direct limit.
Example 5.4.7. Let R = Z, x = (0), y = (3). Then the map oy ! ox is the inclusion map:
na
o
oy = R(3) =
: 3 - b ,! ox = Q.
b
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 ring
1 (M ) = M .
homomorphism : R1 ! R2 is called local if
2
1
In the example, R1 = R(3) , M1 = (3), R2 = Q, M2 = 0. The inverse image of M2 = 0 in
R1 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 with
maximal ideal mx . 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 affine varieties, sections
of the structure sheaf are ⌦-valued functions. Suppose that g 2 (V, o⌃ ) where ⌃ is an affine
variety. Then g is a function
g:V !⌦
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KIYOSHI IGUSA BRANDEIS UNIVERSITY
I explained in class that the definition of a(x) for the affine 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
The local ring at x is
⇡(x) = ker(evx : R ! ⌦)
ox = R⇡(x) =
with unique maximal ideal
mx =
So,
⇢
⇢
f
: h(x) 6= 0
h
f
: f (x) = 0, h(x) 6= 0
h
ox ⇠
= im(evx ) ✓ ⌦
mx
Recall that the function g is given by the same fraction f /h at all points in ⌃. The section
a 2 (X, oX ) corresponding to the function g is this fraction considered as an element of
Rh on the set Xh . Then
ox
a(x) = image of a = f /h in k(x) =
mx
= evx (a) = f (x)/h(x) = g(x).
When X = Spec(R) is not an affine 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 naturally
a subset of ⌦ making a an ⌦-valued function.
5.5. Morphisms of affine schemes. We need morphisms of affine schemes to have a
category which we will show is isomorphic to the category of rings.
5.5.1. Definition. Let (X, oX ), (Y, oY ) be affine 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
fV⇤ : (V, oY ) ! (f
1
V, oX )
(3) The homomorphisms fV⇤ are natural. I.e., for open subsets U ✓ V , we have a
commuting diagram:
(V, oY )
fV⇤
/
(f
✏
X)
resf
resV
U
(U, oY )
1 V, o
⇤
fU
/
(f
✏
f
1 U, o
1V
1U
X)
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(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 , fV⇤ (a)(x) = 0
In other words, the germ fx⇤ : of (x) ! ox of f at x is a local morphism. So, it
induces a homomorphism
k(f (x)) ! k(x).
Claim: This homomorphism sends a(f (x)) 2 k(f (x)) to fV⇤ (a)(x) 2 k(x).
Examples ...
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KIYOSHI IGUSA BRANDEIS UNIVERSITY
5.6. Proof that
(5.1)
(Xf , oX ) = Rf . Recall that
[
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
⇤
= f h for some h. So, we have ring homomorphism
Rf ! Rg
a
ahn
ahn
!
7
=
fn
f n hn
g nm
This is 0 in Rg i↵ ag k = 0 for some k. (ahn g s = 0 ) ag n+s = 0). I rephrased this as
follows. Define the annihilator of a 2 R to be
Ann(a) = {b 2 R : ab = 0}
is zero in Rg i↵ g 2 Ann(a). Similarly, a/f n = 0 in Rf is f 2 Ann(a).
S
Lemma 5.6.2. Suppose that Xf = Xf↵ and fan 2 Rf . Then fan goes to zero in Rf↵ for
all ↵ i↵ fan = 0 in Rf .
Then
a/f n
Proof. Certainly, a/f n = 0 in Rf implies it is 0 in each Rf↵ . Conversely, suppose a/f n = 0
in each Rf↵ . Then f↵ 2 r(Ann(a)). So,
So,
a/f n
= 0 in Rf .
f 2 r(f ) = r(· · · f↵ · · · ) ⇢ r(Ann(a))
⇤
Exercise 5.6.3. Show that Xf is quasi-compact.
Basically, this P
follows from the fact that, if f 2 r(· · · f↵ · · · ), then some power of f is a
n
finite sum: f =
hi f↵i . So, Xf is a union of the corresponding Xf↵i .
Lemma 5.6.4. Suppose that Xf = Xf1 [ · · · [ Xfs and gi 2 Rfi so that the images of gi , gj
in Rfi fj agree for all i, j. Then there is some g 2 Rf which maps to gi for all i.
Proof. gi = ai /fini . Since there are only finitely many, assume ni are all equal to n. Since
gi = gj in Rfi fj we have:
ai fjn aj fin
=0
fin fjn
Since there are only finitely many pairs i, j, there is an m so that
(ai fjn
Let bi = ai fim . Then:
aj fin )(fi fj )m = 0
bi fjN = bi fjn+m = bj fin+m = fiN bj
where N = n + m. Then gi = bi /fiN for all i. But
f 2 r(· · · fi · · · ) = r(· · · fiN · · · )
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So, f s =
P
MATH 202A
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P
hi fiN for some hi 2 R. Let g = f1s
hi bi 2 Rf . Then, in Rf we have:
P
P
hi bi fjN
hi fiN bj
N
gfj =
=
= bj
fs
fs
So, in Rfi we have g = bj /fjN = gj .
⇤
Theorem 5.6.5. If X = Spec(R) and f 2 R is not nilpotent then
(Xf , oX ) = Rf
F0
In other words, the sheafification
of the basic presheaf F (Xf ) = Rf has the same value
0
on the basic open sets: F (Xf ) = F (Xf ).
Proof. By definition of F 0 , any element of (Xf , oX ) is locally given
S by g↵ 2 F (Xf↵ ) = Rf↵ .
And these are compatible on overlaps Xf↵ \ Xf . Since Xf = 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.SA prescheme is a ringed space which is locally
isomorphic to an affine scheme: X =
U↵ so that (U↵ , oX |U↵ ) ⇠
= Spec(R↵ ) with its
standard 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 fV⇤ : (V, oY ) ! (f 1 V, oX ) for every open subset V ✓ Y
so that
(3) f ⇤ is natural with respect to restriction and
(4) fx⇤ : of (x) ! ox is a local homomorphism for all x 2 X.
fx⇤ is the germ of f ⇤ at f (x). So, it might be more accurate to write it as: (fx )⇤ = (f ⇤ )f (x) .
S
S
Proposition 5.7.3. Given X, Y preschemes with affine open sets X = U↵ , Y = V .
Then, a morphism X ! Y can be given as follows. Suppose that for all ↵ there is a and
a morphism of affine schemes f↵ : (U↵ , oX |U↵ ) ! (V , oY |V ) which agree on overlaps:
(a) The composition f↵ : U↵1 !
S V 1 ,! Y is equal to U↵2 ! V 2 ,! Y on U↵1 \ U↵2
giving a continous map f = 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 . The
S
map on sheaves is given first on each of open set W ✓ Y by: W =
V \ W and
Q
Q
/ (W )
/
+3
(V \ W )
(V \ V 0 \ W )
0
0
/
(f
✏
resf↵⇤
⇤
9!fW
1W )
/
Q
✏
(U↵ \ f
1W )
+3
res
Q
✏
(U↵ \ U↵0 \ f
1W )
The key point is that condition (b) implies that the right hand square commutes and
therefore induces a unique map on equalizers.
⇤
52
KIYOSHI IGUSA BRANDEIS UNIVERSITY
L
Example 5.7.4. Suppose that R =
ring and X = P roj(R) is the
d 0 Rd is a gradedL
space of homogeneous prime ideals which do not contain d>0 Rd . Recall that the basic
open sets are, for any homogeneous f 2 Rd , d > 0,
Xf := {[P ] 2 P roj(R) : f 2
/ P}
L
L
Note that P does not contain d>0 Rd . So, there exists an f 2 d>0 Rd not contained in
P . (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
⇢
a
F (Xf ) = R(f ) :=
: a 2 Rnd
fn
This is the subring of Rf of all elements of degree 0. (Recall that on P n (⌦), functions are
locally defined by hg where g, h are homogeneous polynomials of the same degree.) Then:
Xf is homeomorphic to the affine scheme Spec(R(f ) ) and the basic presheaf defined above is
the same as the basic presheaf defining the structure sheaf of Spec(R(f ) ). Furthermore, the
sheafification of the basic presheaf is the same by what we just proved: F 0 (Xf ) = F (Xf ).
So, the structure sheaves of the various affine schemes agrees on overlaps. So, they define a
prescheme structure on all of P roj(R).
Lemma 5.7.5. Xf is homeomorphic to the affine scheme Spec(R(f ) ) with “the same structure sheaf.”
Proof. The points in Xf are
L homogeneous prime ideals P which do not contain f . Being
homogeneous means P =
Ps . If f 2 Rd , the corresponding prime in R(f ) is:
M
f n Pnd
Conversely, given any prime ideal Q in R(f ) , let I be the direct sum of all numerators of all
elements of Q. Then I is a homogeneous ideal in R. Let P = r(I). Then P is a homogeneous
prime ideal not containing f . To see that P is prime, suppose that a 2 Rj , b 2 Rk and
ab 2 P . Then, for some n, s, an bn /f s 2 Q. So nj + nk = ds and
✓ dn ◆ ✓ dn ◆
adn bdn
a
b
=
2Q
nj
ds
f
f
f nk
Since Q is prime, either adn or bdn is in I. So, either a or b is in P = r(I). This gives a 1-1
correspondence P $ Q.
Suppose that g 2 Rs is not nilpotent. Then [P ] 2 Xf g = Xf \ Xg i↵ g d /f s is not in Q.
Conversely, a/f n is not in Q i↵ [P ] 2 Xaf = Xf \ Xa . So, the basic open sets of Xf and
Spec(R(f ) ) correspond.
The structure sheaves are equal by definition!
⇤
S
Example 5.7.6. Suppose that (⌃, o⌃ ) is a prevariety. So, ⌃ = U↵ so that each U↵ is
an affine variety. Let R↵ = (U↵ ) be the ring of regular function U↵ ! ⌦. Define an
equivalence 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 the
push-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 1 f (U ) is open. But:
f 1 f (U ) = U
S
⌃ = U↵ , U↵ affine 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 oX
to X↵ is the push-forward of oU↵ which we already showed to be the structure sheaf of
Spec(R↵ ).
⇤