ALGEBRAIC GEOMETRY COURSE NOTES, LECTURE 9: SCHEMES AND THEIR MODULES.


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1 ALGEBRAIC GEOMETRY COURSE NOTES, LECTURE 9: SCHEMES AND THEIR MODULES. ANDREW SALCH 1. Affine schemes. About notation: I am in the habit of writing f (U) instead of f 1 (U) for the preimage of a subset U under a function f, to avoid any confusion with the other meanings of the symbol f 1. DefinitionProposition 1.1. (Direct image.) Let C be a category with equalizers and all small products, and let X Y be a continuous map of topological spaces, and let F be a Cvalued sheaf on X. Then by f F we mean the sheaf on Y defined by letting ( f F )(U) = F ( f (U)) for all open subsets U of Y. We call f F the direct image sheaf of F along f or the pushforward of F along f. Proof. We need to check that f F really is a sheaf! Let U be an open subset of Y and let {U i } i I be an open cover of U. Then it is a trivially easy exercise to check that { f (U i )} i I is an open cover of f (U). Hence F satisfies the sheaf axiom with respect to the open cover { f (U)} i I of f (U) is equivalent to f F satisfying the sheaf axiom with respect to the open cover {U i } i I of U. So F being a sheaf implies that f F is a sheaf. That f is functorial is easy; so f is a functor f : Sheaves C (X) Sheaves C (Y). Later we will see that, under mild assumptions on C, the functor f has a left adjoint, which we will wind up calling f, inverse image instead of direct image. Definition 1.2. By a locally ringed space we mean a topological space X equipped with a sheaf of commutative rings O X such that, for every x X, the stalk (O X ) x is a local ring. A morphism of locally ringed spaces (X, O X ) (Y, O Y ) is a continuous map f : X Y together with a homomorphism of sheaves of commutative rings f : O Y f O X such that, for each point x X, the induced map of stalks f x : (O Y ) f (x) (O X ) x is a local homomorphism, that is, ( f x ) (m x ) = m f (x), where m x is the maximal ideal in (O X ) x and m f (x) is the maximal ideal in (O Y ) f (x). Definition 1.3. Let R be a commutative ring. By Spec R, the prime spectrum of R, we mean the locally ringed space defined as follows: The set of points of Spec R is the set of prime ideals of R. Date: February
2 2 ANDREW SALCH The topology on Spec R is given by letting the closed subsets of Spec R be all intersections of subsets of the form V( f ) = {p Spec R : f p} Spec R for the various elements f R. This topology on Spec R is called the Zariski topology. The sheaf of commutative rings O on Spec R is defined as follows: for any element f R, the complement Spec R\V( f ) of V( f ) in Spec R is open, and these opens form a basis for the Zariski topology, so if we define O(Spec R\V( f )) for all elements f R, then the sheaf axiom extends O uniquely to being defined on all open subsets of Spec R. We define O(Spec R\V( f )) to be the localization R[ f 1 ] of R inverting the ideal ( f ) generated by f. This sheaf O is called the structure sheaf of Spec R. This is indeed a locally ringed space: for any point p Spec R, the stalk O p of O is the localization R p of R at the ideal p, i.e., the localization which inverts all elements of R not contained in the ideal p. The construction of Spec R is functorial in R, as follows: if f : R S is a homomorphism of commutative rings, then we get an induced map Spec S Spec R by sending each point p Spec S, i.e., each prime ideal p of S, to the preimage f R. In Definition 1.3 you use the fact, from commutative algebra, that the preimage of a prime ideal under a ring homomorphism is a prime ideal. This fails for maximal ideals, and this is one big reason why we define Spec R as the prime ideals and not just the maximal ideals: even though the maximal ideals are more geometrically meaningful (in the case that R is a Noetherian integral domain over an algebraically closed field, the maximal ideals of R correspond to points on the affine variety given by R!), we have to have the prime ideals in there, so that a homomorphism of commutative rings will actually give us a function between ther prime spectra! Here is an example: Example 1.4. Let A be a discrete valuation ring with maximal ideal m. (If you are not totally comfortable with discrete valuation rings, keep in mind the following two examples: the case where A = Z (p), the plocal integers, and m = (p); and the case where A = k[x] (x), the polynomial ring k[x] localized at (x), and m = (x). These (and also their adic completions!) are probably the most important examples of discrete valuation rings. More generally: for any finite field extension K/Q and any prime ideal p in the ring of integers O K, the localization (O K ) p and also the completion (O K )ˆp of O K at p are both discrete valuation rings, so discrete valuation rings are the central object of study in local number theory; meanwhile, if X is an algebraic curve over a field k and x X is a nonsingular point, then the local ring (O X ) x and also its completion (O X )ˆx are both discrete valuation rings. So discrete valuation rings are the central object of study in the (local, at least) study of nonsingular algebraic curves.) Discrete valuation rings are, in particular, local principal ideal domains, and every ideal in a discrete valuation ring is a power of the maximal ideal. Hence the only prime ideals in A are m and (0). Hence Spec A has just two elements, m and (0). Let f be any generator of m, i.e., m = ( f ). Then V( f ) = m, so {m} Spec A is a closed subset, but there isn t any element of A which is contained in 0 but not in m, so {(0)} is not a closed subset of Spec A. So the point m is a closed point in Spec A, and sometimes this point is called the special point of Spec A. The other point (0) in Spec A isn t closed, and in fact its closure is the whole space; this point is called the generic point of Spec A.
3 ALGEBRAIC GEOMETRY COURSE NOTES, LECTURE 9: SCHEMES AND THEIR MODULES. 3 So there are three open sets in Spec A, namely, and {(0)} and {{(0), m} = Spec A. Of course O({(0), m}) is A localized at the ideal m, i.e., A itself, since A is already local with maximal ideal m. Similarly, O( ) is the zero ring, since every sheaf on a topological space takes value the terminal object when evaluated on the empty set (this is a trivial case of the sheaf axiom, which we did in class). So the interesting open set is {(0)}. Since {(0)} = Spec A\V(π) for any generator π of m, we have O({(0)}) A[π 1 ], i.e., O({(0)}) is the field of fractions of A. I will write K for this field of fractions. There are a couple of natural maps of commutative rings out of A, namely, the quotient map q : A A/m, and the localization map l : A K. Since A/m and K are both fields (namely, the residue field and the fraction field of A, respectively), each of Spec A/m and Spec K are a single point (but whose sheaves of commutative rings are different). The map Spec A/m Spec A induced by q is the inclusion of the special point, while the map Spec K Spec A induced by i is the inclusion of the generic point. Example 1.5. (Why it matters that a map of locally ringed spaces has to be local homomorphism on the stalks.) Let A be a discrete valuation ring with fraction field K and maximal ideal m. Here is something you are really not supposed to do: there is a morphism of ringed spaces, that is, spaces equipped with sheaves of commutative rings, f : Spec K Spec A, given by sending the unique point of Spec K to the special point, i.e., the closed point m; then ( f O Spec K )({0}) = 0 and ( f O Spec K )(Spec A) = K, and the map O Spec A f O Spec K is given on the open sets {(0)} and Spec A by letting (O Spec A )(Spec A) ( f O Spec K )(Spec A) is the standard embedding A K of A into its field of fractions, and (O Spec A )({(0)}) ( f O Spec K )({(0)}) is (of course) the zero map K 0. This map f of ringed spaces isn t a map of locally ringed spaces, even though both Spec K and Spec A are locally ringed spaces: f isn t a local homomorphism on the stalk at m! Since the only open set containing m is Spec A itself, the stalk F m of any sheaf F at m is just F (Spec A), and the morphism (O Spec A )(Spec A) ( f O Spec K )(Spec A) has the property that the preimage of the maximal ideal 0 in the codomain is much smaller than the maximal ideal in the domain. The above example matters the definition of a locally ringed space is designed to exclude examples like it because of the following result: Proposition 1.6. Let R, S be commutative rings. Let α be the map α : hom Rings (R, S ) hom LRS (Spec S, Spec R), from the set of ring homomorphisms R S to the set of morphisms of locally ringed spaces Spec S Spec R, which sends a ring homomorphism f to its induced map Spec f : Spec S Spec R. Then α is a bijection. In categorytheoretic terms: the functor Spec is full and faithful. Proof. I claim that α has an inverse function, β : hom LRS (Spec S, Spec R) hom Rings (R, S ), given by letting β( f ) be the map f (Spec R) : O Spec R (Spec R) ( f O Spec S )(Spec R). Since f (Spec R) = Spec S no matter which locally ringed space morphism f : Spec S Spec R is, we have that ( f O Spec S )(Spec R) S and of course O Spec R (Spec R), and now we need to check that the ring homomorphism f (Spec R) : R S has the property that
4 4 ANDREW SALCH Spec ( f (Spec R) ) : Spec S Spec R is exactly the locally ringed space morphism f ; see Hartshorne s Proposition II.2.3 for proof of this part (but the argument is just this: because f is a local homomorphism, for each point p of Spec S, we have a commutative diagram of commutative rings R f (Spec R) S f (Spec R) f (p) R f (p) and hence the preimage of p under the ring map f (Spec R) is just f (p), so the map of schemes Spec( f (Spec R)) : Spec S Spec R agrees with f on the underlying topological spaces. Then, for each open subset U Spec R, the value of the structure sheaf O Spec R (U) on U is determined (via localizations inverting principal ideals, together with the sheaf axiom) by the value of the structure sheaf O Spec R (Spec R) on all of Spec R, i.e., R itself; so Spec( f (Spec R)) coinciding with f on the underlying topological spaces, together with the two maps of locally ringed spaces coinciding when evaluated on all of Spec R, implies that the two maps of locally ringed spaces Spec( f (Spec R)) and f are equal.) Hence α β is the identity function. The other direction is much easier: (β α)( f ) is just f, by the definition of the map induced on the structure sheaf of Spec by a morphism of commutative rings. By now you are noticing that, given a sheaf F on a topological space X, we often find ourselves having to evaluate the sheaf on the entire space, i.e., we often have to consider F (X). This happens so often that it gets a special name: Definition 1.7. Let C be a category, let X be a topological space, and let F be a Cvalued sheaf on X. We write Γ(F ) for F (X), and we call Γ(F ) the global sections of F. If U X is an open subset, we sometimes refer to F (U) as the sections of F in U. Definition 1.8. An affine scheme is a locally ringed space which is isomorphic to Spec R for some commutative ring R. A scheme is a locally ringed space which admits an open cover {U i } i I such that each U i is an affine scheme. A morphism of affine schemes or of schemes is simply a morphism of the underlying locally ringed spaces. We sometimes write Aff for the category of affine schemes and Sch for the category of schemes. Theorem 1.9. The functors S p Spec : (Comm Rings) op Aff Γ : (Aff) op Comm Rings are mutually inverse up to isomorphism. Hence the category of affine schemes is equivalent to the opposite category of the category of commutative rings. Proof. This is really just a restatement of Proposition 1.6! Lemma Let X be a scheme and let a Γ(X). Let V(a) be the subset of X consisting of the points p X such that the map h p : Γ(X) (O X ) p
5 ALGEBRAIC GEOMETRY COURSE NOTES, LECTURE 9: SCHEMES AND THEIR MODULES. 5 from Γ(X) to the stalk of O X at p sends a into the maximal ideal of (O X ) p. Then V(a) is closed in X. Proof. Choose an open cover {U i } i I of X by affine schemes, i.e., U i Spec R i for each i I. For each U i, the subset V(a) U i of U i is exactly the set of primes p of R i such that a p, since it is for those primes that a will map to the maximal ideal (that is, p) in the stalk (O X ) p. Hence V(a) U i = V(r(a)), where r is the restriction map r : O X (X) O X (U i ) = R i. Hence V(a) U i is a closed set (in fact, the complement of a basic open set) in U i. Since V(a) U i is closed in U i for each U i in an open cover {U i } i I of X, V(a) is closed in X. Theorem There exists a natural bijection i.e., the functor is right adjoint to the functor hom Comm Rings (R, Γ(X)) hom Sch (X, Spec R), Spec : (Comm Rings) op Sch Γ : Sch (Comm Rings) op. Proof. First, for each scheme X, we define a function γ X : X Spec(Γ(X)) as follows: for each point p X, we have the restriction map Γ(X) O X (U) for each open subset U of X containing p. Passing to the colimit over all such U, we have a map of commutative rings h p : Γ(X) (O X ) p. Let m p denote the maximal ideal of the stalk (O X ) p, and let γ X (p) = h p (m p ), the preimage of m p under the map h p. I claim that this function γ X is continuous. Suppose that U Spec(Γ(X)) is a basic open set U = Spec(Γ(X))\V(a), for some element a Γ(X). Then U is the set all prime ideals in Γ(X) which do not contain a. A point p X is in γ X (U) if and only if γ X(p) does not contain a, i.e., if and only if the map h p : Γ(X) (O X ) p sends a to an element of (O X ) p not in the maximal ideal m p (O X ) p. But now by Lemma 1.10, the complement X\γ X (U) is closed, so γ X (U) is open. So the preimage under γ X of any basic open set is open. So γ X is continuous. It is also the case that γ X is a morphism of schemes; the check is routine and I ll leave the details to you if you re feeling like checking such things. Now we proceed to the proof. Clearly we have a function α : hom Sch (X, Spec R) hom Comm Rings (R, Γ(X)) given as follows: for any scheme morphism f : X Spec R, we let α( f ) : R Γ(X) be the composite of the isomorphism R Γ(Spec R) with the map Γ( f ) : Γ(Spec R) Γ(X) (recall that Γ is contravariant!). We also have a function β : hom Comm Rings (R, Γ(X)) hom Sch (X, Spec R) given by sending a morphism of rings g : R Γ(X) to the composite of the map γ X : X Spec(Γ(X)) with Spec g : Spec(Γ(X)) Spec R. Since Spec g and γ X are morphisms of schemes, so is the composite Spec g γ X. There s our function β. Now all that s left is to check that α β and β α are each the identity function, which again, I ll leave to you (just follow the definitions backward). Corollary The functor Spec sends colimits of commutative rings to limits in the category of schemes. In particular, if A, C are commutative Balgebras, then Spec(A B C) Spec A Spec B Spec C.
6 6 ANDREW SALCH Corollary The functor Γ sends colimits in the category of schemes to limits in commutative rings. In particular, if X, Y are schemes and X Y their disjoint union, then Γ(X Y) Γ(X) Γ(Y).
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