MATH 8254 ALGEBRAIC GEOMETRY HOMEWORK 1


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1 MATH 8254 ALGEBRAIC GEOMETRY HOMEWORK 1 CİHAN BAHRAN I discussed several of the problems here with Cheuk Yu Mak and Chen Wan Let X be a normal and proper algebraic variety over a field k. Show that O X (X) = K(X) k. Via Proposition we embed k, every stalk and section of O X inside K(X) in a compatible way. By Corollary , O X (X) is a finite extension of k. Thus O X (X) K(X) k. To get the reverse inclusion, let us fix s K(X) k. Now for every x X, since s is algebraic over k, s is a fortiori integral over O X,x. But O X,x is normal and K(X) is the field of fractions of O X,x, so s O X,x. As x X was arbitrary, by Proposition (c) we get s x X O X,x = O X (X) Let O K be a discrete valuation ring with uniformizing parameter t and field of fractions K. Let n 1. We suppose that char(k) 2, 3. (a) Show that K[x, y]/(y 2 + x 3 + t n ) is regular. (b) Let A = O K [x, y]/(y 2 + x 3 + t n ). Show that the special fiber of Spec A Spec O K is reduced. Deduce from this that A is normal (see Lemma 1.18). (c) Let m be the maximal ideal of A generated by x, y, and t. Show that A m is regular if and only if n = 1. (a) Let B = K[x, y]/(y 2 + x 3 + t n ). Write P (x) = x 3 + t n K[x]. Since the formal derivative P (x) = 3x 2 is relatively prime with P (x) (char(k) 3!), P (x) has no repeated roots in the algebraic closure of K. Hence P (x) does not have a square factor. As char(k) 2, by Exercise (b) Y := Spec B is normal. Furthermore, since dim K[x, y] = 2 and y 2 + P (x) K[x, y] is irreducible, we have dim Y = dim B = 1. Thus Y is a Noetherian Dedekind scheme. So by the remark at the end of Example 4.2.9, Y is regular. (b) Writing k = O K /to K for the residue field, the special fiber of Spec O K Spec A is the spectrum of the ring A OK k = A/tA = k[x, y]/(y 2 + x 3 ) which is a domain because y 2 + x 3 k[x, y] is irreducible. Thus the special fiber is reduced. (c) Let us write R = O K [x, y] and n for the maximal ideal (x, y, t) of R. Note that the residue field of the local ring (R n, nr n ) is k. As nr n can be generated by 3 elements 1
2 MATH 8254 ALGEBRAIC GEOMETRY HOMEWORK 1 2 and n has height 3 in R (0 (x) (x, y) (x, y, t)), we have dim k nr n /(nr n ) 2 3 = dim R n. Thus R n is a regular Noetherian local ring. Now writing f = y 2 + x 3 + t n R, by Corollary A m = Rn /fr n is regular if and only if f / n 2. But y 2, x 3 n 2, so f / n 2 if and only if t n / n 2 if and only if n = Let f : X Y be a morphism of locally Noetherian schemes. Let Z be a closed subscheme of X. We suppose that there exists a point y Y such that Z y = X y as schemes. (a) Show that if Z is flat over Y at z X y, then Z equals to X in a neighborhood of z (use Exercise and Nakayama s lemma) (b) Show that if Z is flat over Y and if f is moreover proper over Y, then there exists an open neighborhood V y such that Z f 1 (V ) f 1 (V ) is an an isomorphism (use Exercise 3.2.5). (a) Let V be an affine neighborhood of y in Y. Then there exists an affine neighborhood U of z in X such that f(u) V. Then f U : U V is a morphism of locally Noetherian schemes. C := Z U is a closed subscheme of U. Since the map Z y X y is an isomorphism, so is C y U y (where the fibers in the latter morphism are taken with respect to f U ). If C equals to U in a neighborhood of z, that means that z has a neighborhood W in U such that W C. Then W is also a neighborhood of z in X such that W Z, that is, Z equals X in a neighborhood of z. Thus we may assume X and Y are affine. So X = Spec B, Y = Spec A where A, B are Noetherian rings, and f = Spec ϕ where ϕ : A B is a ring homomorphism. Moreover Z = Spec(B/J) for some ideal J of B and y = p is a prime ideal of A such that the map B A (A p /pa p ) B/J A (A p /pa p ) induced by B B/J is an isomorphism. So we have an isomorphism B p /pb p = Bp /(pb p + JB p ), which means the ideals pb p and pb p + JB p of B p are equal, hence JB p pb p. Note that since ϕ 1 (q) = p, we have ϕ(a p) B q, so B q is a localization of B p ; therefore JB q pb q. Finally z = q V (J) such that ϕ 1 (q) = p where the local ring homomorphism is flat. Thus by Exercise , we get A p (B/J) q/j = Bq /JB q JB q = pb q JB q = pjb q. Since JB q qjb q pjb q, this yields JB q = qjb q = qb q JB q. So by Nakayama s lemma (B q is Noetherian) we get JB q = 0. So every element of J is annihilated by an element in B q. As J is finitely generated, by taking the products of the annihilators
3 MATH 8254 ALGEBRAIC GEOMETRY HOMEWORK 1 3 of its generators, we get a g B q such that gj = 0. Therefore the principal open set D(g) in X = Spec B contains z and lies in V (J) = Z. (b) Consider the following diagram where every rectangle is a fibersquare since they are formed by fiber squares starting from the bottom right corner: Z y X y k(y) Z Y Spec O Y,y j X Y Spec O Y,y g Spec O Y,y Z i Since f is proper, g is a closed map. So if t is a closed point of X Y Spec O Y,y, then g(t) is a closed point of Spec O Y,y. But the only closed point of Spec O Y,y is m y, thus f(p(t)) = y. Therefore p(t) = z X y. Now by assumption Z is flat over Y at z X y, so by part (a) Z equals to X in a neighborhood of U of z. Thus W t := U Y Spec O Y,y is a neighborhood of t such that the restriction j t : j 1 (W t ) W t of j is an isomorphism. So every closed point of T := X Y Spec O Y,y has a neighborhood such that j restricts to an isomorphism. But T is locally Noetherian (it has an affine chart given by localizations of the affine charts of X) therefore every point in T specializes to a closed point(this is Lemma in Stacks project, this result can be avoided by reducing the statement to the affine case where we know the specialization claim is true). Thus {W t : t is a closed point in T } is an open covering of T and hence j is an isomorphism. Note that Z and X are schemes of finite type over Y, as f is proper and Z is a closed subscheme of X. Therefore by Exercise 3.2.5, y has a neighborhood V such that Z Y V X Y V is an isomorphism. Since V is an open subscheme of Y, this map of fiber products can be identified with the inclusion f 1 (V ) Z f 1(V ) Let k be a field of characteristic char(k) 2. Let Y = Spec k[u, v]/(v 2 u 2 (u + 1)), and let f : X Y be the normalization morphism. Show that f is unramified surjective, but not étale. Let A = k[u, v]/(v 2 u 2 (u + 1)), so Y = Spec A. Since u 2 (u + 1) is not a square in k[u], by Exercise (a), Y is an integral scheme. Let B = k[t] and X = Spec B. Consider the kalgebra map Since ψ induces a kalgebra map X ψ : k[u, v] B p u t 2 1 v t(t 2 1). ψ(v 2 u 2 (u + 1)) = î t(t 2 1) ó 2 (t 2 1) 2 t 2 = 0, ϕ : A B. f Y
4 MATH 8254 ALGEBRAIC GEOMETRY HOMEWORK 1 4 Let us show that ϕ is injective. It suffices to check that ker ψ = (v 2 u 2 (u + 1)). Since k[u, v]/ ker ψ embeds in the domain B, ker ψ is a prime ideal of k[u, v]. Consider the chain 0 (v 2 u 2 (u + 1)) ker ψ of prime ideals in k[u, v]. Suppose, for the sake of contradiction, that the last inclusion above is also strict. Now since dim k[u, v] = 2, ker ψ must be a maximal ideal of k[u, v]. Then im ψ = k[u, v]/ ker ψ is a finite extension of k that is embeds in B. But the only subring of B that is a field is k, so im ψ = k. This is a contradiction. Since ϕ is injective, it induces an injection ϕ : Frac A Frac B. We claim that ϕ is an isomorphism. It suffices to show that B im ϕ. And since ϕ is also a kalgebra map, it suffices to show t im ϕ. Indeed, r 0 in A and ϕ(s/r) = t(t2 1) t 2 1 = t. Next, we show that ϕ is integral. Since ϕ is a kalgebra map and B is generated by t as a kalgebra, it suffices to check that t is integral over A. And indeed, writing r and s for the images of u and v in A, respectively, we have t 2 = ϕ(r + 1). Thus, X is normal and the scheme morphism f = Spec ϕ : X Y is birational and integral. So by Proposition f is the normalization morphism. But f is not an isomorphism: indeed u 2 (u + 1) has a square factor so by Exercise (a) Y is not normal. To see that f is unramified, we need to show that for every prime ideal q of B with p = ϕ 1 (q), we have pb q = qb q. Recall that we are writing r and s for the images of u and v in A, respectively. We consider two cases: r p. Then since s 2 = r 2 (r + 1) p we get s p. So ϕ(s) = t 2 1 pb q. Hence q = (t 1) or q = (t + 1). Suppose q = (t 1). Then t + 1 / q so t 1 1 = t2 1 t + 1 pb q hence qb q = pb q. The case q = (t + 1) is similar. r / p. Then ϕ(r) = t 2 1 / q, hence t 1 = t(t2 1) t 2 1 lies in the image of the kalgebra homomorphism ϕ : A p B q. But t/1 generates B q as a kalgebra; thus ϕ is surjective and we get pb q = qb q Let X = Spec k[t 1,..., T n ]/(F 1,..., F r ) be an affine variety over a field k, and x a closed point of X. Let us consider J x = (( F i / T j )(x)) i,j as a matrix with coefficients in k(x). Show that X is smooth at x if and only if rank J x = n dim O X,x. Hence the Jacobian criterion is a criterion for smoothness. First, note that X k = Spec k[t1,..., T n ]/I
5 MATH 8254 ALGEBRAIC GEOMETRY HOMEWORK 1 5 where I is the ideal of k[t 1,..., T n ] generated by F 1,..., F r. Let y X k be a point lying above x. Since O Y,y is a finitely generated kalgebra, k(y) = O Y,y /m y is a finite extension of k, hence is equal to k as k is algebraically closed. Thus y is also a krational point of X k. As y lies over x, the maps and k[t 1,..., T n ] k F F (x) k[t 1,..., T n ] k G G(y) is compatible with k[t 1,..., T n ] k[t 1,..., T n ]. By the Jacobian criterion, X k is regular at y if and only if the matrix Ç å Ç å Fi Fi (y) = (x) = J x T j 1 i r,1 j n T j 1 i r,1 j n has rank n dim O Xk,y. As X k is integral over X k, we have dim O Xk,y = dim O X,x. Thus X k is regular at y if and only if J x has rank n dim O X,x. Since the point y X k lying above x was arbitrary, by definition of smoothness we get that X is smooth at x if and only if J x has rank n dim O X,x Let Y be a locally Noetherian scheme, and let f : X Y be a quasifinite and quasiprojective morphism. Show that f factors into an open immersion X Z followed by a finite morphism Z Y. Suppose the claim is true when Y is affine. Let {Y i } be an affine open covering of Y. Fix i and let X i = f 1 (Y i ). Then the restriction f i := f Xi : X i Y i is quasifinite and quasiprojective (it can be realized as a base change over the open immersion Y i Y ). Therefore by the affine case, there is a scheme Z i, an open immersion h i : X i Z i and a finite morphism g i : Z i Y i such that g i h i = f i. For each i, j Z ij := h i (X i X j ) is an open subscheme of Z i because h i is an open immersion. Also for each i, j we write p ij : Z ij Z ji for the isomorphism Then clearly p ii = id Zi. Now observe that h j Xi X j (h i Xi X j ) 1. The restriction of the isomorphism (h i Xi X j ) 1 : Z ij X i X j on Z ij Z ik is the isomorphism (h i Xi X j X k ) 1 : Z ij Z ik X i X j X k. The restriction of the isomorphism h j Xi X j : X i X j Z ji on X i X j X k is h j Xi X j X k : X i X j X k Z ji Z jk.
6 MATH 8254 ALGEBRAIC GEOMETRY HOMEWORK 1 6 The restriction of the isomorphism (h j Xj X k ) 1 : Z jk X j X k on Z ji Z jk is the isomorphism (h j Xi X j X k ) 1 : Z ji Z jk X i X j X k. The restriction of the isomorphism h k Xj X k : X j X k Z kj on X i X j X k is the isomorphism Z kj Z ki. Thus the restriction of p jk p ij on Z ij Z jk is equal to On the other hand, (h j Xi X j X k ) 1 h j Xi X j X k (h i Xi X j X k ) 1 = (h i Xi X j X k ) 1 : Z ij Z ik Z kj Z ki The restriction of the isomorphism (h i Xi X k ) 1 : Z ik X i X k on Z ij Z ik is (h i Xi X j X k ) 1 : Z ij Z ik X i X j X k. The restriction of the isomorphism h k Xi X k : X i X k Z ki on X i X j X k is Thus the restriction of p ik on Z ij Z ik is X i X j X k Z ki Z kj. (h i Xi X j X k ) 1 which is exactly the same with the restriction of p jk p ij on Z ij Z jk. Therefore, applying Lemma on the Y schemes {Z i } (the structure morphism is Z i Y i Y for each Z i ), we get a Y scheme Z with open immersions q i : Z i Z of Y schemes such that q i = q j p ij on Z ij, and that Z = i q i (Z i ). Let us write h for the structure morphism Z Y. Now since for each i the morphism g i : Z i Y i is finite, so is h qi (Z i ) : q i (Z i ) Y. As {q i (Z i )} is an open covering of Z, we obtain that h is finite. For each i write β i = q i h i : X i Z. As β i Xi X j = q i Zij h i Xi X j = q j Zji p ij h i Xi X j = q j Zji h j Xi X j = β j Xi X j the β i s glue together to give a morphism β : X Z. Since each h i is an open immersion, so is β. And by checking on the coverings we obtain g β = f. So we reduce to the case where Y = Spec A is affine, hence Noetherian. As f is quasiprojective there exists a scheme T, an open immersion of finite type j : X T and a projective morphism p : T Y such that p j = f. Note that p is proper so it is separated and of finite type. Moreover being projective, it makes O T (T ) a finitely generated Amodule (mentioned in footnote at p. 151). Thus writing S = {t T : t is isolated in T p(t) } by Theorem 4.4 S is open in T and T Spec O T (T ) induces an open immersion from S to Spec O T (T ). And we know Spec O T (T ) Y = Spec A is finite and they compose to f.
7 MATH 8254 ALGEBRAIC GEOMETRY HOMEWORK 1 7 So since j is an open immersion, if we can show that j(x) S we can pick Z = Spec O T (T ) and get the desired decomposition. Because then X S Z is an open immersion and Z Y is finite. And indeed, for every x X, writing y = f(x) the fiber X y can be embedded in T y = T p(j(x)) via j as an open set. But since f is quasifinite, X y has discrete topology so x is isolated in X y. Thus j(x) is isolated in T y, so j(x) S.
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