Algebraic Geometry Homework II.5, 1,2,3,4,5,7,8

William A. Stein

April 4, 1996

Problem 1 (II.5.1) Let (X, O_{X}) be a ringed space, and let E be a locally free O_{X}-module of finite rank. We define the dual of E, denoted by Ě, to be the sheaf Hom_{O_{X}}(E, O_{X}).

(a) Show that (Ě)ˇ ≅ E.

(b) For any O_{X}-module F, Hom_{O_{X}}(E, F) ≅ Ě ⊗_{O_{X}} F.

(c) For any O_{X}-modules F, G, Hom_{O_{X}}(E⊗F, G) ≅ Hom_{O_{X}}(F, Hom_{O_{X}}(E, G)).

(d) (Projection Formula). If f : (X, O_{X}) → (Y, O_{Y}) is a morphism of ringed spaces, if F is an O_{X}-module, and if E is a locally free O_{Y} -module of finite rank, then there is a natural isomorphism f_{∗}(F ⊗_{O_{X}} f^{∗}E) ≅ f_{∗}(F) ⊗_{O_{Y}} E.

Proof. (a) A free module of finite rank is canonically isomorphic to its double-dual via m̌(λ) = λ(m) where m ∈ M, λ ∈ M̌, and m̌ ∈ (M̌)ˇ . Let U be an open set on which E|U is a free O_{X}-module of finite rank. Define a map φ : E|_{U} → (Ě)ˇ |_{U} by, for all V ⊆ U, E(V) → (Ě)ˇ (V) is the isomorphism described above. Since the isomorphisms are canonical, we can patch on intersections and define a global isomorphism.

(b) As above we may assume E is a free O_{X}-module. Let e_{1}, …, e_{n} be a basis for F and let e_{1}^{∗}, …, e_{n}^{∗} be the corresponding dual basis. Let U be an open subset of X. Define φ_{U} : Hom_{O_{X}}(E, F)|_{U} → (Hom(E, O_{X})⊗_{O_{X}} F)|_{U} by f ↦ ∑_{i=1}^{n} e_{i}^{∗} ⊗ f(e_{i}). Define ψ_{U} : (Hom(E, O_{X}) ⊗_{O_{X}} F)|_{U} → Hom_{O_{X}}(E, F)|_{U} by f ⊗ a ↦ (x ↦ f(x)a). For convenience of notation write φ = φ_{U} and ψ = ψ_{U}. Let f ⊗ a ∈ (Hom(E, O_{X}) ⊗_{O_{X}} F)|_{U}. Then φ ∘ ψ(f ⊗ a) = φ(x ↦ f(x)a) = ∑_{i=1}^{n} e_{i}^{∗}⊗f(e_{i})a = ∑_{i=1}^{n} e_{i}^{∗}f(e_{i})⊗a = f ⊗a. Let f ∈ Hom_{O_{X}}(E, F), then ψ ∘ φ(f) = ψ(∑_{i=1}^{n} e_{i}^{∗} ⊗ f(e_{i})) = (x ↦ ∑_{i=1}^{n} e_{i}^{∗}(x)f(e_{i})) = (x ↦ f(∑_{i=1}^{n} e_{i}^{∗}(x)e_{i})) = (x ↦ f(x)). Thus φ and ψ are inverse bijective homomorphisms, hence ring isomorphisms, and since they respect the restriction maps we see that the corresponding sheaves are isomorphic.

(c) On each open set define φ|_{U} : Hom_{O_{X}}(E⊗F, G) → Hom_{O_{X}}(F, Hom_{O_{X}}(E, sg)) by f ↦ (a ↦ (e ↦ f(e ⊗ a))). (For notational convenience we omit the sheaf restrictions.) If φ(f) = 0 then the map (a ↦ (e ↦ f(e ⊗ a))) is 0 so f is the zero map, hence φ is injective. Let f ∈ Hom_{O_{X}}(F, Hom_{O_{X}}(E, sg)). Define g ∈ Hom_{O_{X}}(E ⊗ F, G) by g(a ⊗ b) = (f(b))(a). Then φ(g) = (a ↦ (e ↦ g(e ⊗ a))) = (a ↦ (e ↦ (f(a))(e)) = (a ↦ f(a)) = f so φ is surjective. Thus φ is the desired isomorphism which, since φ evidently commutes with the restriction maps, induces an isomorphism of sheaves.

(d) First we consider the case when E ≅ O_{Y}^{n}. One one hand,

f_{∗}(F) ⊗_{O_{Y}} E ≅ f_{∗}(F) ⊗_{O_{Y}} O_{Y}^{n}
≅ ⊕_{i=1}^{n}(f_{∗}(F) ⊗_{O_{Y}} O_{Y})
≅ ⊕_{i=1}^{n}f_{∗}(F).

On the other hand,

F ⊗_{O_{X}} f^{∗}E = F ⊗_{O_{X}} f^{∗}(O_{Y}^{n})
≅ F ⊗_{O_{X}} (f^{−1}(O_{Y}^{n}) ⊗_{f^{−1}O_{Y}} O_{X})
≅ F ⊗_{O_{X}} ((f^{−1}(O_{Y})^{n} ⊗_{f^{−1}O_{Y}} O_{X})
≅ F ⊗_{O_{X}} (∑_{i=1}^{n} f^{−1}O_{Y} ⊗_{f^{−1}O_{Y}} O_{X})
≅ F ⊗_{O_{X}} (O_{X}^{n})
≅ (F ⊗_{O_{X}} O_{X})^{n} = F^{n}

where f^{−1}(O_{Y}^{n}) ≅ f^{−1}(O_{Y})^{n} since f^{−1} is a left adjoint functor hence commutes with direct sums (which are a right universal construction).

Putting this together we have that

f_{∗}(F ⊗_{O_{Y}} f^{∗}(E)) = f_{∗}(F^{n}) = f_{∗}(⊕_{i=1}^{n}F) = ⊕_{i=1}^{n}f_{∗}(F).

In general, we construct isomorphisms as above on an open cover then, since all of the isomorphisms are canonical, the isomorphisms match up on the intersections so we can glue to obtain an isomorphism.

Problem 2 (II.5.2) Let R be a discrete valuation ring with quotient field K, and let X = SpecR.

(a) To give an O_{X}-module F is equivalent to giving an R-module M, a K-vector space L, and a homomorphism ρ : M ⊗_{R} K → L.

(b) That O_{X}-module is quasi-coherent if and only if ρ is an isomorphism.

Proof. (a) First suppose we are given an O_{X}-module F. Since R is a DVR, X has exactly two nonempty open sets, X and the set consisting of the generic point, {ξ}. Let M = Γ(F, X) and let L = Γ(F, {ξ}). Since Γ(O_{X}, X) = R and Γ(O_{X}, {ξ}) = K, M is an R-module and L is a K-vector space. Let g : M → L be the restriction map. Define ρ by ρ(m ⊗ α) = α · g(m). Since g is a homomorphism, so is ρ.

Now suppose we are given an R-module M, a K-vector space L, and a homomorphism ρ : M ⊗_{R} K → L. Define an O_{X}-module F as follows. Let Γ(F, X) = M and Γ(F, {ξ}) = L. Define the restriction map g : M → L by g(m) = ρ(m ⊗ 1). We just need to check that g is a valid restriction map. Let r ∈ R, m ∈ M, then g is a valid restriction map iff r · g(m) = g(rm), that is, when r · ρ(m ⊗ 1) = ρ(rm ⊗ 1) = ρ(m ⊗ r). So we must verify that the given homomorphism ρ is R-linear. But there is absolutely no reason why this should be the case! (For example, let M = L = K, then ρ : K → K and it is easy to construct nontrivial homomorphisms of the additive group of a field). I think the problem is imprecisely stated. It should be assumed throughout that ρ is K-linear.

(b) First suppose F is quasi-coherent. Then F = M̃ so Proposition 5.1 implies that L = Γ(M̃, {ξ}) = (R − 0)^{−1}M ≅ M ⊗_{R} K. Thus ρ must be an isomorphism. Conversely, if ρ is an isomorphism, we see that F ≅ M̃ since they are the same on each open set and the restriction map is the same.

Problem 3 (II.5.3) Let X = SpecA be an affine scheme. Show that the functors ˜ and Γ are adjoint, in the following sense: for any A-module M, and for any sheaf of O_{X}-modules F, there is a natural isomorphism

Hom_{A}(M, Γ(X, F)) ≅ Hom_{O_{X}}(M̃, F).

Proof.

Define a homomorphism F : Hom_{A}(M, Γ(X, F)) → Hom_{O_{X}}(M̃, F) as follows. Send a ring homomorphism φ : M → Γ(X, F) to the morphism of sheaves F (φ) : M̃ → F. It suffices to define F (φ) on distinguished open sets (Eisenbud & Harris, page 13). For f ∈ A let F (φ)_{D(f)} be the map

\frac{m}{f^{n}} ↦ \frac{1}{f^{n}} · res_{X,D(f)}(φ(m))

where res_{X,D(f)} : F(X) → F(D(f)) is the restriction map of F. F (φ)_{D(f)} is a well-defined homomorphism since both φ and res_{X,D(f)} are homomorphisms and φ is an A-module homomorphism. Next note that F (φ) commutes with the restriction maps since each res_{X,D(f)} does. To see that F is injective suppose φ and ψ are two homomorphisms M → Γ(X, F). If F (φ) = F (ψ) then, in particular, φ = F (φ)_{X} = F (ψ)_{X} = ψ. To see that F is surjective, let φ ∈ Hom_{O_{X}}(M̃, F). Define ψ : M → Γ(X, F) by letting ψ = φ_{X}, that is, by taking the induced map on global sections. Then, for f ∈ A, F (ψ)_{D(f)} : M̃(D(f)) → F(D(f)) is the map (\frac{m}{f^{n}} ↦ \frac{1}{f^{n}} res_{X,D(f)} ∘ φ_{X}(m)) which, since φ is a morphism of sheaves, equals = (\frac{m}{f^{n}} ↦ \frac{1}{f^{n}} φ_{D(f)}(m) = φ_{D(f)}(\frac{m}{f^{n}}) = φ_{D(f)}. (We are just using the fact that φ commutes with the appropriate restriction maps.) Thus F (ψ) agrees with φ on a basis for X hence F (ψ) = φ. This shows that F is surjective. So F is an isomorphism, as required.

Problem 4 (II.5.4) Show that a sheaf of O_{X}-modules F on a scheme X is quasi-coherent if and only if every point of X has a neighborhood U, such that F|_{U} is isomorphic to a cokernel of a morphism of free sheaves on U. If X is noetherian, then F is coherent iff it is locally a cokernel of a morphism of free sheaves of finite rank.

Proof. Suppose first that F is quasi-coherent. Let x ∈ X. Then there is an affine open neighborhood U = SpecA of x such that F|_{U} ≅ M̃, M an A-module. It suffices to show that M is isomorphic to a cokernel of a morphism of finitely generated A-algebras. Indeed, if φ : A^{(I)} → A^{(J)} then Coker(φ̃) = Ã^{(J)}/φ(A^{(I)})˜ = (A^{(J)}/φ(A^{(I)}))˜ since, for all f ∈ A, (A^{(J)})_{f}/φ(A^{(I)})_{f} = (A^{(J)}/φ(A^{(I)}))_{f} so that they agree on a basis.

Let A^{|M|} be the free A-module on the elements of M. Let φ : A^{|M|} → M be the natural map. Similiary, let ψ : A^{| ker(φ)|} → ker(φ) ⊆ A^{|M|} be the natural map. Then Coker(ψ) ≅ A^{|M|}/ ker(φ) ≅ M, as required.

Now assume the F is coherent. We proceed as above but now M is a finitely generated A-module, generated by e_{1}, …, e_{n}, say, and we must show that M is the cokernel of a morphism of free modules of finite rank. Let φ : A^{n} → M be the map which takes the ith generator (0, …, 1, …, 0) of A^{n} to e_{i} ∈ M. Then ker(φ) is a submodule of M so, since M is neotherian (any finitely generated module over a noetherian ring is noetherian), ker(φ) is finitely generated. Let f_{1}, …, f_{m} be a generating set. Let ψ : A^{m} → ker(φ) be the surjection defined by sending the ith basis element of A^{m} to f_{i}. Then Coker(ψ) ≅ A^{n}/ψ(A^{m}) ≅ A^{n}/ ker(φ) ≅ M. Thus M is isomorphic to a cokernel of a morphism of free sheaves of finite rank.

Problem 5 (II.5.5) Let f : X → Y be a morphism of schemes.

(a) Show by example that if F is coherent on X, then f_{∗}F need not be coherent on Y, even if X and Y are varieties over a field k.

(b) Show that a closed immersion is a finite morphism.

(c) If f is a finite morphism of neotherian schemes, and if F is coherent on X, then f_{∗}F is coherent on Y.

Proof. (a) Lex k be a field, let X = Spec(k[x]_{x}), Y = Spec(k[x]) and F = O_{X}. Then f_{∗}(F)(U) = F(f^{−1}(U)), so f_{∗}(F) = (k[x]_{x})_{˜ Y}. But (k[x]_{x})_{˜ Y} is not a coherent sheaf of O_{Y} -modules. Indeed, if it were, there would be a distinguished neighborhood D(f) of 0 so that (k[x]_{x})_{˜ Y} |_{D(f)} is a finitely generated module over k[x] (here I’m using the fact that over open set is distinguished in the topology of Spec(k[x])). But, k[x]_{f} is never a finitely generated module over k[x] for any f of degree at least 1. For k[x]_{f} contains elements of arbitrarily small degree whereas the degrees of elements of k[x]{α_{1}, …, α_{n}} are bounded below. (k[x]{α_{1}, …, α_{n}} is the k[x]-module generated by α_{1}, …, α_{n}.)

(b) Let f : Y → X be a closed immersion. Let U = Spec(A) ⊆ X and let W = f^{−1}(U) ⊆ Y. Then W is a a scheme (give it the induced scheme structure as an open subset of Y). Furthermore, f(W) = U ∩ f(Y) is a relatively closed subset of U, that is, a closed subset of Spec(A). The map f^{#} : O_{U} → O_{W} is surjective since f^{#} : O_{X} → O_{Y} is surjective and surjectivety is a local property. Thus W is a closed subscheme of Spec(A) so, by Corollary 5.10, W ≅ Spec(A/I) for some ideal I of A. Since A/I is a finitely generated A module (generated by 1 + I), f is a finite morphism.

(c) Let U = Spec(A) ⊆ Y be an affine open subset of Y. Then, since f is finite, f^{−1}(U) = Spec(B) is affine with B is a finitely generated A-module. By Proposition 5.4 it suffices to show that f_{∗}(F)|_{U} is M̃ for some finitely generated A-module M. Now by Proposition 5.4, since f^{−1}(U) is affine and B is noetherian, F|_{f^{−1}(U)} = M̃ for some finitely generated B module M.

But f_{∗}(F)|_{U} = (f|_{f^{−1}(U)})_{∗}F|_{f^{−1}(U)} = (f|_{f^{−1}(U)})_{∗}(M̃) = ˜ (A^{M}) where the last equality follows from Proposition 5.2(d). Since B is a finite module over A and M is a finite module over B, it follows that M is a finite module over A which completes the proof.

Problem 6 (II.5.7) Let X be a noetherian scheme, and let sf be a coherent sheaf.

(a) If the stalk F_{x} is a free O_{X}-module for some point x ∈ X, then there is a neighborhood U of x such that F|_{U} is free.

(b) F is locally free iff its stalks F_{x} are free O_{X}-modules for all x ∈ X.

(c) F is invertible iff there is a coherent sheaf G such that F ⊗ G ≅ O_{X}.

Proof. (a) Let U = Spec(A) be a neighborhood of x so that F|_{U} = M̃, M a finitely generated A-module. Then F_{x} = M_{x} so we have reduced the problem to the following purely algebraic result.

Proposition 1 If M is a finitely generated free A module and there is a prime ℘ of A such that M_{℘} is a free A_{℘}-module, then there exists f ∈ A such that f ∉ ℘ and M_{f} is a free A_{f}-module.

Once we have proven this we will know that F|_{D_{U} (f)} is free since Proposition 5.1 asserts that M̃(D(f)) ≅ M_{f}.

Proof. (of Proposition) Let a_{1}, …, a_{n} ∈ M be a free basis of M_{℘} over A_{℘} (we can clear denominators so that we may assume all a_{i} lie in M.) Let b_{1}, …, b_{m} ∈ M be a generating set for M over A. For each i we can write b_{i} as an A_{℘}-linear combination of the a_{i}. Clearing denominators we see that d_{i}b_{i} ∈ A{a_{1}, …, a_{n}} for some d_{i} ∉ ℘. Let f = ∏ d_{i}, then f ∉ ℘ and a_{1}, …, a_{n} have A_{f}-span including all of the b_{i}, and thus including M, and thus including M_{f}. But a_{1}, …, a_{n} is free over A_{℘} hence over A_{f} since A_{f} ⊆ A_{℘}. [This proposition is in Bourbaki, Commutative Algebra, II.5.1, although the proof is more abstract than mine.]

(b) (⟹) Let x ∈ X and let U = Spec(A) be an open neighborhood of x such that F|_{U} = M̃ with M a free A module. Suppose ℘ is the prime of A corresponding to x. Then F_{x} = M_{℘} which is a free A_{℘}-module. Indeed, if e_{1}, …, e_{n} is a free A-basis for M, then it is also a free A_{℘}-basis for M_{℘}. For if \frac{a_{1}}{b_{1}} e_{1} + ⋯ + \frac{a_{n}}{b_{n}} e_{n} = 0, b_{i} ∉ ℘, then a_{1} \frac{b}{b_{1}} e_{1} + ⋯ + a_{n} \frac{b}{b_{n}} e_{n} = 0, b = ∏ b_{i} so b\frac{a_{i}}{b_{i}} = 0 for each i, so \frac{a_{i}}{b_{i}} = 0 in the localization A_{℘} since b ∉ ℘.

(⟸) By part (a) every point has a neighborhood on which F is free. Therefore X can be covered by open affines on which F is free so F is locally free.

(c) (⟹) Let G = F̌ = Hom(F, O_{X}), we will show that F ⊗ G ≅ O_{X}. To define a morphism φ : F ⊗ Hom(F, O_{X}) → O_{X} it is enough to define φ on the presheaf (U ↦ F(U) ⊗_{O_{X}(U)} Hom(F(U), O_{X}(U)). Define φ by a ⊗ f ↦ f(a). Thus φ commutes with the restrictions so φ defines a valid morphism of sheaves.

Let U be an open affine subset of X such that F(U) is a free O_{X}(U)- module of rank 1 with basis e_{0}. Then Hom(F(U), O_{X}(U)) has basis e_{0}^{∗} where e_{0}^{∗}(e_{0}) = 1. Thus Hom(F(U), O_{X}(U)) is a free O_{X}(U)-module of rank 1. Thus every element of F(U) ⊗ Hom(F(U), O_{X}(U)) can be written in the form a ⊗ f (as opposed to as a sum of such products). Now φ_{U}(a ⊗ f) = 0 implies f(a) = 0 which implies a = 0 or f = 0 so a⊗f = 0, so φ_{U} is injective. Since φ_{U}(ae_{0} ⊗ e_{0}^{∗}) = a, φ_{U} is surjective. Thus φ_{U} is an isomorphism.

Use the definition of locally free of rank 1 to cover X by affine open sets U such that F|_{U} is a free O_{X}|_{U} module of rank 1. Then, by Proposition 5.1 (c) and an argument like that for (b) above, any distinguished open subset of U has a F-sections free of rank 1. Since φ is an isomorphism on each of these distinguished open sets (use the argument in the above paragraph) and these distinguished open sets form a basis for the topology on X it follows that φ must be an isomorphism.

(⟸) Because of part (b) above it suffices to show that F_{x} is free of rank one for each x ∈ X. Since F_{x} ⊗_{O_{x}} G_{x} = (F ⊗ G)_{x} ≅ O_{X,x} the problem is reduced to the following purely algebraic statement.

Proposition 2 Let M and N be finitely generated modules over a local ring (A, m) and suppose that M ⊗_{A} N ≅ A. Then M is free of rank 1.

Proof. Let k = A/m. Let ν : M ⊗_{A} N → A be the given isomorphism. Then, taking the product with the identity, we get an isomorphism ν ⊗ 1_{k} : (M ⊗_{A} N) ⊗_{A} k → A ⊗_{A} k ≅ k (it is obvious that ν ⊗ 1_{k} is surjective, but it is not at all obvious that it is injective, for this see the Bourbaki reference below.) Thus k ≅ M ⊗_{A} N ⊗_{A} k = M ⊗_{A} (k ⊗_{A} k) ⊗_{A} N = (M ⊗_{A} k) ⊗_{A} (N ⊗_{A} k) = (M ⊗_{A} k) ⊗_{k} (N ⊗_{A} k) = (M/mM) ⊗_{k} (N/mN) so, since the k-rank of (M/mM) ⊗_{k} (N/mN) is 1 and is the product of the ranks of M/mM and N/mN, each has rank 1. In particular, M/mM is monogeneous (generated by one element) as an A/m-module and hence as an A-module so, by Nakayama’s lemma, M is monogeneous as an A-module. Since Ann_{A}(M) anihilates M ⊗_{A} N = A as well, it is 0 (any element of Ann_{A}(M) would have to annihilate the identity of A and hence be 0). Thus M is a free module of rank one over A. [See Bourbaki, Commutative Algebra, II.5.4, for a more general theorem.]

Problem 7 (II.5.8) Again let X be a noetherian scheme, and F a coherent sheaf on X. We will consider the function

φ(x) = dim_{k(x)}F_{x} ⊗_{O_{X}} k(x)

where k(x) = O_{X}/m_{x} is the residue field at the point x. Use Nakayama’s lemma to prove the following results.

(a) The function φ is upper semi-continuous, i.e., for any n ∈ Z, the set {x ∈ X : φ(x) ≥ n} is closed.

(b) If F is locally free, and X is connected, then φ is a constant function.

(c) Conversely, if X is reduced, and φ is constant, then F is locally free.

Proof.

(a) We must show that {x : φ(x) ≥ n} is closed. A good way to do this is by showing that {x : φ(x) < n} is open. To do this we show that if φ(x) = m then there is an open neighborhood U of x so that, for all y ∈ U, φ(y) ≤ m. Since we need only look locally, we can assume that X = SpecA, F = M̃, M a finitely generated A-module. Note that F_{x} ⊗_{O_{X}} k(x) = M_{℘} ⊗_{A_{℘}} A_{℘}/℘A_{℘} = M_{℘}/℘M_{℘}. Let s_{1}, …, s_{m} ∈ M be elements whose images form a basis for the vector space M_{℘}/℘M_{℘} over A_{℘}/℘A_{℘} (to do this choose a basis for M_{℘}/℘M_{℘} then clear fractions). Note that the images of the s_{i} in fact generate M_{℘}/℘M_{℘} as an A_{℘}-module. By Nakayama’s lemma the s_{i} generate M_{℘} as an A_{℘}- module. Let m_{1}, …, m_{k} be a generating set for M over A. Write m_{j} = ∑ \frac{a_{i}}{b_{i}} s_{i}, b_{i} ∉ ℘, then, if c_{j} = ∏ b_{i}, c_{j}m_{j} is in the A-span of the s_{i}. Let f = ∏ c_{j}. Then ℘ ∈ D(f) and if q ∈ D(f), then m_{1}, …, m_{k} all lie in the A_{q}-span of s_{1}, …, s_{m} (since c_{j}m_{j} is in the A-span of the s_{i} and c_{j} is inverted in A_{q}. Thus M is spanned by the s_{i} over A_{q}, so M_{q} is spanned by the s_{i} over A_{q}. It follows that φ(q) = dimM_{q}/qM_{q} ≤ m since the images of the s_{i} generate M_{q}/qM_{q} as a vector space over A_{q}/qM_{q} = k(q). Taking D(f) as our open neighborhood completes the proof.

(b) Choose n so that some section of F has rank n. Let U be the union of all open sets W such that F|_{W} ≅ O_{X}^{n}|_{W}. Then U is nonempty. Let V be the union of all open sets W such that F|_{W} ≅ O_{X}^{m}|_{W}, m ≠ n. Since F is locally free, U ∪ V = X. Suppose x ∈ U ∩ V, then F_{x} has rank n and rank m ≠ n (since rank is preserved under localization), a contradiction. Thus U ∩V = ∅. Since U is nonempty and open, X − U = V is open and X is connected, thus we conclude that V = X − U = ∅. Thus every point is contained in an open set W such that F|_{W} ≅ O_{X}^{n}|_{W}.

Let x ∈ X and let U = Spec(A) be an affine open set containing x such that F|_{U} ≅ M̃. By the above argument, M is a free A-module of rank n. Thus φ(x) = dim_{k}M_{x} ⊗_{A_{x}} A_{x}/m_{x} = dim_{k}A_{x}^{n} ⊗_{k} A_{x}/m_{x} = dim_{k}(A_{x}/m_{x})^{n} = dim_{k}k^{n} = n, as desired.

(c) Let x ∈ X. By exercise 5.7b it suffices to show that the stalk F_{x} is free. Since F is coherent we can find an affine open set U = Spec(A) such that F|_{U} = M̃ for some finitely generated A-module, x ∈ U, and A_{f} is reduced for each f ∈ A. Let ℘ be the prime of A corresponding to x. We must show that M_{℘} is free over A_{℘}. Let s_{1}, …, s_{n} ∈ M be preimages of a basis of M_{℘}/℘M_{℘} over k(x) = A_{℘}/℘A_{℘} (find these as in part (a)). Then, by Nakayama’s lemma, the s_{i} generate M_{℘} over A_{℘}.

We must show that the s_{i} are linearly independent over A_{℘}. It will then follow that M_{℘} is free of rank n over A_{℘}. So suppose

\frac{a_{1}}{b_{1}} s_{1} + ⋯ + \frac{a_{n}}{b_{n}} s_{n} = 0

in M_{℘} with \frac{a_{i}}{b_{i}} ∈ A_{℘}. Then for each i, b_{i} ∉ ℘ and a_{i} ∈ A. Since the s_{i} are linearly independent over A_{℘}/℘A_{℘}, for each i there exists c_{i} ∉ ℘ such that \frac{c_{i}a_{i}}{b_{i}} ∈ ℘. Thus c_{i}a_{i} ∈ ℘ so a_{i} ∈ ℘. Let r be as in part (a) so that q ∈ D(r) implies the s_{i} generate at least M over A_{q}. By definition there exists c ∈ A such that

c(b_{2} ⋯ b_{n}a_{1}s_{1} + ⋯ + b_{1} ⋯ b_{n−1}a_{n}s_{n}) = 0

in A. Let f = rc ∏ b_{i}, then if q ∈ D(f), then s_{1}, …, s_{n} generate M_{q}/qM_{q} over A_{q} and, since M_{q}/qM_{q} has dimension n (since φ is constant), the s_{i} are actually a basis for M_{q}/qM_{q} over A_{q}/qA_{q}.

Since c|f, c ∉ q so, as above, \frac{a_{1}}{b_{1}} s_{1} + ⋯ + \frac{a_{n}}{b_{n}} s_{n} = 0 in M_{q}, so, as above, a_{i} ∈ q for each i. Thus, for all q ∈ D(f), a_{i} ∈ q, so a_{i} lies in the nilradical of A_{f} which, since A_{f} is reduced, means that a_{i} = 0 in A_{f}. So a_{i} maps to 0 under the map A_{f} → A_{℘}. Thus s_{1}, …, s_{n} are linearly independent over A_{℘} so M_{℘} is free of rank n over A_{℘}. Applying exercise (5.7b) then completes the proof.