Math 256B. Answers to Problem Set #8

1. (20 points) Do III Ex. 5.7.

(a). Let ℒ be ample on X, and let ℱ be any coherent sheaf on Y . Then i_{∗}ℱ is coherent (II Ex. 5.5), and by Lemma 2.10, (II Ex. 6.8a), (II Ex. 5.1d), and Prop. 5.3, we have

H^{i}(Y, ℱ ⊗ (i^{∗}ℒ)^{⊗n}) = H^{i}(X, i_{∗}(ℱ ⊗ (i^{∗}ℒ)^{⊗n})) = H^{i}(X, i_{∗}ℱ ⊗ ℒ^{⊗n}) = 0

for all n ≫ 0 . Thus i^{∗}ℒ is ample on Y , by Prop. 5.3 again.

(b). Since i: X_{red} → X is a closed immersion and ℒ_{red} = i^{∗}ℒ (II Prop. 5.2), one implication is immediate from part (a).

Conversely, suppose that ℒ_{red} is ample, and let ℱ be a coherent sheaf on X. Let 𝒩 be the sheaf of nilpotent elements on X, so that 𝒪_{X_{red}} ≅ 𝒪_{X}/𝒩 , and recall from the proof of Ex. 3.1 (given in class on 3 February) that 𝒩 ^{r} = 0 for some r ∈ ℕ. Then, for 0 ≤ j < r, the sheaf 𝒩 ^{j}ℱ /𝒩 ^{j+1}ℱ is a coherent sheaf on X_{red} , so H^{i}(X_{red}, (𝒩 ^{j}ℱ /𝒩 ^{j+1}ℱ) ⊗_{𝒪_{X_{red}}} ℒ_{red}^{⊗n}) = 0 for all i > 0 and all n ≥ n_{0}(j) for some n_{0}(j) ∈ ℤ. By elementary properties of tensor, we have that

(𝒩 ^{j}ℱ /𝒩 ^{j+1}ℱ) ⊗_{𝒪_{X_{red}}} ℒ_{red}^{⊗n} ≅ (𝒩 ^{j}ℱ /𝒩 ^{j+1}ℱ) ⊗_{𝒪_{X}} ℒ^{⊗n}

for all n, and the latter can be regarded as a sheaf on X. Thus, by Lemma 2.10,

H^{i}(X, (𝒩 ^{j}ℱ /𝒩 ^{j+1}ℱ) ⊗ ℒ^{⊗n}) = 0

for all i > 0 , all n ≥ n_{0}(j) , and all 0 ≤ j < r. Let n_{0} = max{n_{0}(0), …, n_{0}(r − 1)}. From the short exact sequence

0 ⟶ 𝒩 ^{j+1}ℱ ⊗ ℒ^{⊗n} ⟶ 𝒩 ^{j}ℱ ⊗ ℒ^{⊗n} ⟶ (𝒩 ^{j}ℱ /𝒩 ^{j+1}ℱ) ⊗ ℒ^{⊗n} ⟶ 0

we get an exact sequence

H^{i}(X, 𝒩 ^{j+1}ℱ ⊗ℒ^{⊗n}) ⟶ H^{i}(X, 𝒩 ^{j}ℱ ⊗ℒ^{⊗n}) ⟶ H^{i}(X, (𝒩 ^{j}ℱ /𝒩 ^{j+1}ℱ)⊗ℒ^{⊗n}) .

By descending induction on j, and noting that 𝒩 ^{r}ℱ = 0, so H^{i}(X, 𝒩 ^{r}ℱ ⊗ℒ^{⊗n}) = 0 for all i and all n, it follows from the above sequence that H^{i}(X, ℱ ⊗ ℒ^{⊗n}) = 0 for all i > 0 and all n ≥ n_{0} , so ℒ is ample on X by Prop. 5.3.

(c). Again, since j_{i}: X_{i} → X is a closed immersion and ℒ ⊗ 𝒪_{X_{i}} = j_{i}^{∗}ℒ for all i, one implication is immediate from part (a) (and Lemma 2.10).

Conversely, suppose that ℒ ⊗ 𝒪_{X_{i}} is ample on X_{i} for all i. It will suffice to prove the following lemma.

Lemma. Let X be a reduced scheme, proper over A, let Y and Z be reduced closed subschemes of X such that X = Y ∪ Z, and let ℒ be a line sheaf on X such that 𝒪 ⊗ 𝒪_{Y} and ℒ ⊗ 𝒪_{Z} are ample on Y and Z, respectively. Then ℒ is ample (on X).

Proof. Let ℐ and 𝒥 be the sheaves of ideals corresponding to Y and Z, respectively. Then ℐ · 𝒥 ⊆ ℐ ∩ 𝒥 = 0 , since ℐ ∩ 𝒥 is associated to a closed subscheme of X whose underlying topological space is all of X, and since X is reduced.

Let ℱ be a coherent sheaf on X. Consider the filtration

ℱ ⊇ ℱ · ℐ ⊇ ℱ · ℐ · 𝒥 = 0 .

The sheaf ℱ · ℐ is killed by 𝒥 , so as in the first lemma in the solution of III Ex. 3.2, ℱ · ℐ is a coherent sheaf on Z, so by Prop. 5.3, H^{i}(Z, (ℱ · ℐ) ⊗ ℒ^{⊗n}) = 0 for all i > 0 and all n ≫ 0 (depending only on ℱ and ℐ ). Next consider the short exact sequence

0 ⟶ ℱ · ℐ ⟶ ℱ ⟶ 𝒢 ⟶ 0 ,  (*)

where 𝒢 = ℱ /(ℱ · ℐ) . This quotient is killed by ℐ , so again it is a coherent sheaf on Y , and H^{i}(Y, 𝒢 ⊗ ℒ^{⊗n}) = 0 for all i > 0 and all n ≫ 0 depending only on 𝒢 (and Y ).

Twist the short exact sequence (*) by ℒ^{⊗n} and take the long exact sequence in cohomology. We get (in part)

H^{i}(Z, (ℱ · ℐ) ⊗ ℒ^{⊗n}) ⟶ H^{i}(X, ℱ ⊗ ℒ^{⊗n}) ⟶ H^{i}(Y, 𝒢 ⊗ ℒ^{⊗n})

for all i > 0 . For all n ≫ 0 the two end terms are zero, so the middle term must also be zero. By Prop. 5.3 it then follows that ℒ is ample. □

(d). Suppose first that ℒ is ample on Y , and let ℱ be a coherent sheaf on X. Then f_{∗}ℱ is coherent on Y (II Ex. 5.5c), so by (II Ex. 5.1d) and Prop. 5.3,

H^{i}(Y, f_{∗}(ℱ ⊗ (f^{∗}ℒ)^{⊗n})) = H^{i}(Y, f_{∗}ℱ ⊗ ℒ^{⊗n}) = 0

for all i > 0 and all n ≫ 0 depending only on ℱ and on Y . But by Ex. 4.1, this gives H^{i}(X, ℱ ⊗ (f^{∗}ℒ)^{⊗n}) = 0 for all i > 0 and all n ≫ 0 ; hence f^{∗}ℒ is ample by Prop. 5.3.

We prove the converse by contradiction. Suppose the converse is false. Let f: X → Y be a finite surjective morphism of proper schemes over A, and let ℒ be a line sheaf on Y which is not ample but such that f^{∗}ℒ is ample on X.

Consider a closed subscheme i: Y′ → Y , and let X′ = X ×_{Y} Y′. The projection j: X′ → X is a base change of the closed immersion i, so it is a closed immersion, so by part (a), j^{∗}f^{∗}ℒ is ample. The projection f′: X′ → Y′ is a base change of f , so it is finite and surjective. Thus (noting that (f′)^{∗}i^{∗}ℒ ≅ j^{∗}f^{∗}ℒ ), one may ask whether the result is true for f′ and i^{∗}ℒ . If it is, then it would imply that i^{∗}ℒ is ample on Y′.

We therefore prove the result by noetherian induction on Y ; specifically, the collection of closed subschemes i: Y′ → Y for which the assertion is false is nonempty (since it contains Y itself), therefore it contains a minimal element. After replacing Y with this minimal element and pulling back ℒ , we may assume that Y is minimal in the sense that i^{∗}ℒ is ample on Y′ for all proper closed immersions i: Y′ → Y .

By part (b), we then have that Y is reduced, since otherwise Y_{red} → Y would be a proper closed immersion and therefore ℒ_{red} would be ample on Y_{red} by minimality. Similarly, by part (c), Y must be irreducible, again by minimality. Note also that Y ≠ ∅, since in that case there is nothing to prove.

Thus, Y is integral. By surjectivity of f , there is some point in X mapping to the generic point η of Y ; if we replace X with the irreducible component of X containing that point (with reduced induced subscheme structure), then f is still finite and surjective (since it is a closed morphism and its image contains η), and f^{∗}ℒ is still ample by part (a). Thus we may assume that X is also integral.

Since ℒ is not ample, by Prop. 5.3 there is a coherent sheaf ℱ on Y such that H^{i}(Y, ℱ ⊗ ℒ^{⊗n}) ≠ 0 for some i > 0 and arbitrarily large n. Fix such an ℱ and such an i.

By Ex. 4.2 and its published solution, there are: (i) a proper closed subscheme j: Z → X, (ii) coherent sheaves 𝒦 and 𝒦 ′ on Z, (iii) a coherent sheaf 𝒢 on X, (iv) an integer r > 0 , and (v) and exact sequence

0 ⟶ j_{∗}𝒦 ⟶ f_{∗}𝒢 ⟶^{β} ℱ^{r} ⟶ j_{∗}𝒦 ′ ⟶ 0

on Y . By minimality of Y , j^{∗}ℒ is ample on Z, so by II Ex. 5.1d, Lemma 2.10, and Prop. 5.3, H^{i+1}(X, j_{∗}𝒦 ⊗ ℒ^{⊗n}) = H^{i+1}(X, j_{∗}(𝒦 ⊗ (j^{∗}ℒ)^{⊗n})) = H^{i+1}(Z, 𝒦 ⊗ (j^{∗}ℒ)^{⊗n}) = 0 for all n ≫ 0 . Similarly, H^{i}(X, j_{∗}𝒦 ′ ⊗ ℒ^{⊗n}) = 0 for all n ≫ 0 .

Now twist the short exact sequence

0 ⟶ j_{∗}𝒦 ⟶ f_{∗}𝒢 ⟶ Im β ⟶ 0

by ℒ^{⊗n} and take the long exact sequence in cohomology. This gives (in part) an exact sequence

H^{i}(Y, f_{∗}𝒢 ⊗ ℒ^{⊗n}) ⟶ H^{i}(Y, (Im β) ⊗ ℒ^{⊗n}) ⟶ H^{i+1}(Y, j_{∗}𝒦 ⊗ ℒ^{⊗n}) .

The first term vanishes for n ≫ 0 , by the projection formula, by Ex. 4.1, and by Prop. 5.3. We have seen already that the last term is also zero for n ≫ 0 . Thus H^{i}(Y, (Im β) ⊗ ℒ^{⊗n}) = 0 for all n ≫ 0 .

Next twist the short exact sequence

0 ⟶ Im β ⟶ ℱ^{r} ⟶ j_{∗}𝒦 ′ ⟶ 0

by ℒ^{⊗n} and take the long exact sequence in cohomology. This gives an exact sequence

H^{i}(Y, (Im β) ⊗ ℒ^{⊗n}) ⟶ H^{i}(Y, ℱ^{r} ⊗ ℒ^{⊗n}) ⟶ H^{i}(Y, j_{∗}𝒦 ′ ⊗ ℒ^{⊗n}) . The two end terms have been shown to be zero for all n ≫ 0 , so the middle term is zero; thus H^{i}(Y, ℱ ⊗ ℒ^{⊗n})^{r} = 0 and therefore H^{i}(Y, ℱ ⊗ ℒ^{⊗n}) = 0 for all n ≫ 0 . This achieves the required contradiction.

2. (10 points) Do III Ex. 6.7.

For the first part, first note that, since A is noetherian, the category of finitelygenerated A-modules (call it 𝔑𝔬𝔢𝔱𝔥(A) ) is an abelian category.

By II Cor. 5.5 and III Prop. 6.4, the contravariant functors 𝔑𝔬𝔢𝔱𝔥(A) → 𝔄𝔟 given by

M ↦ Ext_{X}^{i}(M̃, Ñ)  and  M ↦ Ext_{A}^{i}(M, N)

are δ-functors, and coincide when i = 0 by II Cor. 5.5.

For any n ∈ ℕ and i > 0 , we have

Ext_{X}^{i}(Ã^{n}, Ñ) = Ext_{X}^{i}(Ã, Ñ)^{n} = H^{i}(X, Ñ) = 0

by Prop. 6.3c and Theorem 3.5, and we also have

Ext_{A}^{i}(A^{n}, N) = Ext_{A}^{i}(A, N)^{n} = 0

because the (covariant) functor Hom(A, ·) is the identity functor, hence is exact.

Since every finitely-generated A-module M admits a surjection from A^{n} for some n, these δ-functors are coeffaceable. By Theorem 1.3A they therefore coincide for all i. This shows the first half of the problem.

For the second half, we first show:

Lemma. If M is a finitely-generated A-module and N is any A-module, then

ℋ om _{X}(M̃, Ñ) ≅ Hom_{A}(M, N) .

Proof. Since Γ(X, ℋ om _{X}(M̃, Ñ)) = Hom_{X}(M̃, Ñ) = Hom_{A}(M, N) by II Cor. 5.5, the identity map Hom_{A}(M, N) → Hom_{A}(M, N) gives a map of sheaves,

Hom_{A}(M, N) ⟶ ℋ om _{X}(M̃, Ñ)

by II Ex. 5.3. By Prop. 6.8, we have for all 𝔭 ∈ Spec A = X that

(Hom_{A}(M, N) )_{𝔭} = (Hom_{A}(M, N))_{𝔭} = Hom_{A_{𝔭}}(M_{𝔭}, N_{𝔭}) = ℋ om _{X}(M̃, Ñ)_{𝔭} ,

so the above map is an isomorphism. □

Now let

… ⟶ L_{1} ⟶ L_{0} ⟶ M ⟶ 0

be an exact sequence with L_{i} ≅ A^{n_{i}} for all i ∈ ℕ. This gives a locally free resolution of M (viewing this sequence as consisting of modules over the single-point ringed space associated to A). Thus, by Prop. 6.5,

Ext_{A}^{i}(M, N) = h^{i}(Hom_{A}(L_{·}, N)) .

Similarly, L̃_{·} → M̃ → 0 is a locally free resolution of M̃ on X, so

ℰxt_{X}^{i}(M̃, Ñ) = h^{i}(ℋ om _{X}(L̃_{·}, Ñ)) = h^{i}(Hom_{A}(L_{·}, N) )
= h^{i}(Hom_{A}(L_{·}, N)) = Ext_{A}^{i}(M, N)

by Prop. 6.5, by the lemma, and by exactness of .

3. (10 points) Do III Ex. 6.3.

For both parts, the question is local, so we may assume by Prop. 6.2 that X = Spec A, ℱ = M̃, and 𝒢 = Ñ , where A is a noetherian ring, M is a finitely-generated A-module, and N is an A-module. (a). In this part, N is also finitely generated.

Let L_{·} → M → 0 be a free resolution of M (which is easy to construct since M is finitely generated and A is noetherian), and let L̃_{·} → M̃ → 0 be the corresponding free resolution of M̃. Then by Prop. 6.5, the lemma in the solution of Ex. 6.7, and II Cor. 5.5,

ℰxt_{X}^{i}(M̃, Ñ) = h^{i}(ℋ om _{X}(L̃_{·}, Ñ)) = h^{i}(Hom_{A}(L_{·}, N) ) = h^{i}(Hom_{A}(L_{·}, N)) .

So it will suffice to show that Hom_{A}(L_{i}, N) is finitely generated. But this is true because L_{i} ≅ A^{n_{i}} for some n_{i} ∈ ℕ, and therefore

Hom_{A}(L_{i}, N) ≅ Hom_{A}(A^{n_{i}}, N) ≅ Hom_{A}(A, N)^{n_{i}} ≅ N^{n_{i}} .

(b). This part is immediate from the second part of Ex. 6.7 (note that that exercise did not use anything from Exercises 6.3–6.6 in its proof) and II Prop. 5.4.