= Spec(Rf ) R p. 2 R m f gives a section a of the stalk bundle over X f as follows. For any [p] 2 X f (f /2 p), let a([p]) = ([p], a) wherea =


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1 LECTURES ON ALGEBRAIC GEOMETRY MATH 202A 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 discussed later Definitions. First recall the definition of the topological space Spec(R), then construct the structure sheaf 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) thenf 2 p, f(x) = 0.) The complement of V (f) is X f = {[p] :f/2 p}. These are the basic open sets in X. We also recall: where R f is R with f inverted. X f = Spec(Rf ) Structure sheaf o X. To make X = Spec(R) into an a ne scheme, we need to construct its structure sheaf o X. This will be a sheaf of local rings. The stalk o [p] will be the local ring R p, R localized at the prime p. Recall that this is a ring of fractions: n a o R p = b where a, b 2 R, b /2 p The stalk bundle of the structure sheaf o X is the set of all pairs ([p],a)where[p] 2 X, a 2 R p. Projection to the first coordinate gives the bundle map: a! X [p]2x R p Note that sections of the stalk bundle over any U X form a ring (but not a local ring) by pointwise addition and multiplication: s 1 + s 2 and s 1 s 2 are given by (s 1 + s 2 )(x) = s 1 (x)+s 2 (x) and (s 1 s 2 )(x) =s 1 (x)s 2 (x). For every basic open set X f, let (X f,o X )=R f. Each element a = r f 2 R m f gives a section a of the stalk bundle over X f as follows. For any [p] 2 X f (f /2 p), let a([p]) = ([p], a) wherea = r f considered as an element of R m p. (Since f m /2 p, this fraction defines an element of R p.) The construction of the structure sheaf o X follows the following pattern which may be easier to understand as a separate concept using standard examples. Definition 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 FV in some category C which has direct limits. We assume that FV is a set with additional structure, e.g., a ring.
2 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 A basic presheaf F defines a stalk bundle and sheaf F 0 in the following way. (1) The stalk F x of F at x 2 X is defined to be the direct limit of FV for all basic open nbhs V of x: F x =dir. lim x2v FV (2) For all open U X, F 0 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 FV so that (y) is the germ of s at y for all y 2 V. Example If M n is a di erentiable nmanifold 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 around 0 in R n. Then the basic presheaf of C 1 embeddings of M into R m is given by letting F (U, ) be the set of all mappings U! R m so that the composition B (0)! U! R m is a C 1 embedding. Example Suppose that (,o ) is a prevariety over (k, ). By definition, is a finite union of a ne open sets U i so that (U i,o U i ) is an a ne variety with ring of global sections R i = (U i,o ). Basic open sets in are given by (U i ) f where f 2 R i is not nilpotent. The structure sheaf o is given by the basis presheaf: a F ((U i ) f )=(R i ) f = f m : a 2 R i If (U i ) f (U j ) g then what is the restriction map res :(R j ) g! (R i ) 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 )=(R j ) g is a subring of the ring of all functions (U j ) g!. The restriction map F ((U j ) g )=F ((U i ) f ) is given by restriction of these mappings to the subset (U i ) f (U j ) 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 Suppose that G is a sheaf on X. SupposethatF 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 = / GV res res FV = / GV Then G is isomorphic to the sheafification F 0 of F. In particular, F 0 V = GV = FV.
3 ~ LECTURES ON ALGEBRAIC GEOMETRY MATH 202A 43 Proof. Since FV = GV, we get an isomorphism of stalks: F x =dir. lim x2v FV = dir. limx2v GV = G x So, F, G have the same (isomorphic) stalk bundles. The sheafification F 0,G 0 are thus equal by 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 = G 0 = F 0. Definition The structure sheaf of X = Spec(R) for any commutative ring R is defined to be the sheafification o X = F 0 of the basic presheaf F on X given by FX f = R f with restriction map FX f! FX g defined to be the unique ring homomorphism R f! R g making the following diagram commute. The pair (X, o X ) is called an a One of the basic theorems about a R f ne scheme. R / R g ne schemes is: (X f,o X ) = R f In other words, F 0 X f = FX f for all basic open sets X f in X. We will prove this in the special case when R = (,o ) is the coordinate ring of an a ne variety. By Lemma it su ces to find a sheaf G and a natural isomorphism GX f = Rf for every basic open X f Example: R = Z 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 X n = {[p] :p  n}[{0} For example, X 6 is everything except [2], [3]: V (6) = {[2], [3]} is the complement of X 6. Note that X n X m i m n. For example, X 6 X 3. The only other open sets are the empty set and the whole space X Structure sheaf. The structure sheaf is given on basic open sets by n a o apple 1 (X n )=R n = n k : a 2 Z = Z n and on all of X by (X) =Z. So, the local rings are (avoiding Z p which means something else) n a o o [p] = R p = b : a, b 2 Z,p b o o =dir. lim R n = [ n R n = Q In terms of the stalk bundle: an arbitrary section of the stalk bundle over U X is a function which assigns (p) = a p p  b p b p /2 (p) b p (0) = a 0 b 0 b 0 6=0 b 0 /2 (0)
4 44 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 5 is a section on X 6 since 2, 35 Recall that this is the same condition which defines sections of o over U : (x) must be 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: R n = (X n ) K(X) =Q which we already knew. But now we should realize that it means: The formula for each section at each point is given a single element of K(X) 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 a prime in k[z] containing P. We will abuse notation and use p, p/p interchangeably. A n 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 The structure sheaf of X = Spec(R) is the pushforward (direct image) of the structure sheaf of, i.e., for every open subset U of X. (U, o X ) = ( 1 U, o ) It is easy to see that the pushforward of any sheaf along any continuous epimorphism is a sheaf. (Proved in Lemma below.) Proof. We use Lemma in the case where G is the pushforward of the structure sheaf of along : X: G(U) := ( 1 U, o ) and F is the basic presheaf FX f = R f used to define o X. In other words, we need to show that ( 1 X f,o )=R f First, 1 X f = f since So, by Theorem 3.3.3, we have: x 2 f, f(x) 6= 0, f/2 (x), (x) 2 X f ( 1 X f,o ) = ( f )=R f as required. We also need to show naturality. If X g X f then the restriction maps make the following diagram commute. FX f = R = f / ( f,o ) FX g = R g = / ( g,o )
5 ? ; LECTURES ON ALGEBRAIC GEOMETRY MATH 202A 45 ( f,o ) = / (X f,o X ) ( g,o ) = / (X g,o X ) 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. R f # R K( ) But the whole diagram below commutes. So the diagram above commutes. R g Corollary (X f,o X )=R f in the special case at hand (when R is a finite generated kalgebra without zero divisors). This is true in general but not that easy to prove. (See [4].) Here is the lemma we used in the theorem. Lemma Let : X Y be a continuous epimorphism and let F be a sheaf on X. Then the pushforward presheaf G(U) =F ( 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 { 1 V } is an open covering of 1 U in X. So, any collections of sections s 2 F ( 1 V ) extends uniquely to a section s 2 F ( 1 U). But F ( 1 V )=G(V ) and F ( 1 U)=G(U) by definition of G. Therefore, G is a sheaf. Interpretation of Theorem (I won t repeat this part of the last lecture.) Note that the scheme does not have in its definition. However, the a ne scheme in the Theorem contains the essential information of the points of the variety. For example, take R = k[z]/p. ThenSpec(R) and its structure sheaf contains the information in Y = V (p) n and its sheaf o Y which is a sheaf of functions to. This hidden information comes in the following form. A section s 2 (U, o Y ) is a function which sends each [x] 2 U to s(x) 2 o x = R x where we think of x as a prime ideal. To interpret this as an valued function on U, choose a ring homomorphism (not unique) ' x : R x R x /xr x,! whose kernel is the maximal ideal xr x. Then we take the image s(x) =' x (s(x)) Although this is not well defined, s(x) =0i s(x) 2 xr x, the maximal ideal in o x. The point is that: The function s : U! will be called regular if it comes from an element s 2 R = (U, o Y ). interpretation: x 7! s(x) 2 o x m x
6 46 KIYOSHI IGUSA BRANDEIS UNIVERSITY actual: x 7! s(x) 2 o x In order for this interpretation to be correct, the element s must be uniquely determined by {s(x)}. Buts 0 (x) =s(x) impliess 0 = s 2 o x for all x if and only if \ x =0 but the intersection is over all x 2 U = SpecR. And this intersection is equal to r(0). Definition AringR is called reduced if r(0) = 0. reduced. Thus the interpretation of a section s 2 R = s(x) 2 o x ) is accurate only when R is reduced. References [1] Atiyah, Macdonald, Introduction to commutative algebra [2] J. Harris, Algebraic geometry: A first course [3] I. Macdonald, Algebraic Geometry: Introduction to schemes [4] D. Mumford, The red book of varieties and schemes [5] D. Mumford, Algebraic geometry I: Complex Projective Varieties For example, any domain is (U, o Y ) as a function on U (with values Department of Mathematics, Brandeis University, Waltham, MA address:
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