Algebraic Geometry Homework

William A. Stein

April 4, 1996

1 Solutions

Problem 1 (II.2.14(a)) Let S be a graded ring. Show that P rojS = ∅ iff every element of S_{+} is nilpotent.

Proof. This is equivalent to showing that the nilradical of S is equal to the intersection of all homogenous primes of S. Indeed, if every element of S_{+} is nilpotent then every homogenous prime contains S_{+} so P rojS = ∅. Conversely, if P rojS = ∅, then every homogenous prime contains S_{+} so the nilradical of S contains S_{+} so that every element of S_{+} is nilpotent.

It remains to show that the intersection of all homogenous primes of S is the nilradical of S. Using Zorn’s lemma we can show that if I is a proper homogenous ideal then there is at least one maximal homogenous ideal containing I. (The proof preceeds just as on page 2 of Matsumura except one notes that the union of a chain of homogenous ideals is homogeneous. If I_{1} ⊂ I_{2} ⊂ ⋯ is a chain of homogeneous ideals, then ∪_{n=1}^{∞}I_{n} is an ideal and if x ∈ ∪_{n=1}^{∞}I_{n} then x ∈ I_{n} for some n, so the homogenous components of x are in I_{n}, so they are in ∪_{n=1}^{∞}I_{n}, so ∪_{n=1}^{∞}I_{n} is homogeneous.)

Theorem 1 Let T be a multiplicative set and I a homogeneous ideal disjoint from T; then there exists a homogeneous prime ideal containing I and disjoint from T.

Proof. Using Zorn’s lemma we see that the set of homogeneous ideals disjoint from T and containing I contains a maximal element, say P. Then P is prime. For if x ∉ P, y ∉ P are homogenous, then P + (x) and P + (y) are both homogenous and so they meet T, so their product also meets T. However,

(P + (x))(P + (y)) ⊂ P + (xy)

so xy ∉ P since P does not meet T.

Theorem 2 The nilradical of S is the intersection of the homogeneous primes of S.

Proof. Suppose x is in the nilradical of S so that x is nilpotent, say x^{n} = 0. If I is a homogenous prime then x^{n} = 0 ∈ I so, by induction, x ∈ I. Conversely, suppose x is not nilpotent. Then T = {1, x, x^{2}, …} is a multiplicative set disjoint from (0). So, by the above theorem, there is a homogeneous prime ideal containing (0) disjoint from T. Thus x is not in the intersection of all homogeneous prime ideals of S.

Problem 2 (II.2.14(b)) Let φ : S → T be a graded homomorphism of graded rings. Let U = {p ∈ P rojT : p ⊉ φ(S_{+})}. Show that U is an open subset of P rojT, and show that φ determines a natural morphism f : U → P rojS.

Proof. U is open because P rojT − U = {p ∈ P rojT : p ⊇ φ(S_{+})} = {p ∈ P rojT : p ⊇ T φ(S_{+})} = V (T φ(S_{+})) and the ideal T φ(S_{+}) is homogenous because it is generated by the homogeneous elements {φ(f) : f ∈ S_{+} is homogeneous}.

The natural morphism f : U − P rojS is defined as follows. As a map on topological spaces we specify that, for x ∈ U, f(x) = φ^{−1}(x). Because of the way U was chosen and since φ is homogeneous f maps U into P rojS, so f is well-defined. If V (a) is a closed subset of P rojS with a homogeneous, then f^{−1}(V (a)) = {p ∈ U : φ^{−1}(p) ⊇ a} = {p ∈ U : p ⊇ φ(a)} = V (T φ(a)). As above, T φ(a) is homogeneous so f^{−1}(V (a)) is closed so f is continuous.

To define the associated map f^{#} : O_{P roj S} → f_{∗}O_{U} of sheaves let V ⊂ P rojS be open. Then an element of O_{P roj S}(V) is a map s : V → ⨆_{p∈V} S_{(p)} such that s is locally a quotient of elements of S. We specify that f^{#}(s) = φ ∘ s ∘ f : f^{−1}(V) → ⨆_{p∈f^{−1}(V)} T_{(p)}. To see that f^{#}(s) ∈ f_{∗}O(V) we must check that it is locally a quotient. So suppose p ∈ f^{−1}(V). Let W ⊂ V be an open neighborhood of f(p) on which s is represented as a quotient. Then f^{#}(s) is represented as a quotient on the open set f^{−1}(W), as required. Since f^{#} respects the restriction maps we see that f is a morphism.

Problem 3 (II.2.14(c)) f can be an isomorphism even when φ is not. For example, suppose that φ_{d} : S_{d} → T_{d} is an isomorphism for all d ≥ d_{0}. Show that U = P rojT and the morphism f : P rojT → P rojS is an isomorphism.

Proof. To see that U = P rojT note that if p ∈ P rojT but p ∉ U then p ⊇ φ(S_{+}). In particular, p ⊇ ⨁_{d≥d_{0}} T_{d}, so p ⊇ (T_{+})^{d} so, since p is prime, p ⊇ T_{+}, a contradiction.

Let {g_{α}} be a set of generators of T_{+}. Then ∪_{α}D_{T} (g_{α}) = ∪_{α}{x ∈ P rojT : g_{α} ∉ x} = P rojT since every prime in P rojT must omit some g_{α}. Since g_{α} ∉ x iff g_{α}^{d_{0}} ∉ x for x prime, we may replace the g_{α} by elements of T_{≥d_{0}} and still have a cover of P rojT by distinguished open sets.

Our strategy is as follows. We first show that f|_{D_{T} (g_{α})} : D_{T} (g_{α}) → D_{S}(φ^{−1}(g_{α}) is an isomorphism for each α and then show that the open sets D_{S}(φ^{−1}(g_{α})) cover P rojS. Then showing that f is injective completes the proof.

Let g = g_{α} be one of our g_{α}. By Proposition 2.5, D_{T} (g) ≅ SpecT_{(g)} and so f′ = f|_{D_{T} (g)} is a morphism of affine schemes f′ : SpecT_{g} → SpecS_{(φ^{−1}(g))}. This map is induced by \overline{φ} : S_{(φ^{−1}(g))} → T_{(g)} where \overline{φ} is the localization of the ring homomorphism φ : S → T. So we just need to verify that \overline{φ} is an isomorphism. Suppose \overline{φ}(a/b) = 0. Then \overline{φ}(aφ^{−1}(g)/bφ^{−1}(g)) = \overline{φ}(a/b) = 0 so φ(aφ^{−1}(g))/φ(bφ^{−1}(g)) = 0 in T_{(g)} so there is n such that g^{n}φ(aφ^{−1}(g)) = 0 in T so φ(aφ^{−1}(g^{n+1}) = 0. Thus aφ^{−1}(g)^{n+1} = 0 since φ is an isomorphism in high enough degree. Thus a = 0 in S_{(φ^{−1}(g))}, so a/b = 0 in S_{(φ^{−1}(g))}. This shows that \overline{φ} is injective. To see that \overline{φ} is surjective let a/g^{n} ∈ T_{(g)}. Then φ^{−1}(ag)/φ^{−1}(g^{n+1}) is a well-defined element of S_{(φ^{−1}(g))} and \overline{φ}(φ^{−1}(ag)/φ^{−1}(g^{n+1})) = ag/g^{n+1} = a/g^{n}, which shows that \overline{φ} is surjective.

Next we verify that ∪_{g_{α}}D_{S}(φ^{−1}(g_{α}) = P rojS. Suppose x ∉ D_{S}(φ^{−1}(g_{α}) for all α. Then φ^{−1}(g_{α}) ∈ x for each α, so, since we may assume that the g_{α} generate T_{≥d_{0}}, and x is prime, φ^{−1}(T_{≥d_{0}}) ⊆ x, a contradiction since S_{+} ⊈ x.

Next we show that the induced map f : P rojT → P rojS is injective. Let p, q ∈ P rojT and suppose f(p) = f(q). Then φ^{−1}(p) = φ^{−1}(q) so, since φ is an isomorphism, for d ≥ d_{0} we see that p ∩ T_{d} = q ∩ T_{d}. So if a ∈ p is homogeneous then a^{n} ∈ p ∩ T_{d} for some n and d ≥ d_{0}. So a^{n} ∈ q ∩ T_{d} so a^{n} ∈ q so a ∈ q. Likewise a ∈ q implies a ∈ p. Thus p = q so f is injective.

Finally, no solution is complete without an actual example of a map φ : S → T which satisfies the hypothesis of the theorem. Let T = k[x, y] and let S = T_{0} + T_{2} + ⋯ and let φ : S ↪ T be the inclusion map. Then φ is graded and an isomorphism for d ≥ 2. But φ is not an isomorphism.

Problem 4 (II.2.14(d)) Let V be a projective variety with homogeneous coordinate ring S. Show that t(V) ≅ P rojS.

Proof. Define f : t(V) → P rojS by f takes a point x ∈ t(V) to its homogeneous ideal I(x) ∈ P rojS. By exercise I.2.4 f is injective and surjective hence a bijection. Furthermore, x ⊇ y iff f(x) ⊆ f(y) so f and f^{−1} send closed sets to closed sets hence f is a homeomorphism.

To define f^{#} let U be an open subset of P rojS. Then we must define f^{#} so that (notation as in the proof of proposition 2.6) f^{#} : O_{P roj S}(U) → f_{∗}O_{t(V)}(U) = O_{t(V)}(f^{−1}(U)) = O_{V} (α^{−1}(f^{−1}(U))). Let s ∈ P roj_{S}(U). Then s can locally be represented in the form g/h where g, h ∈ S have the same degree and h is nonzero on the appropriate subset of V. Thus s naturally defines an element of O_{V} (α^{−1}(f^{−1}(U))) and any element of O_{V} (α^{−1}(f^{−1}(U))) defines an element of O_{P roj S}(U). Thus f^{#} is an isomorphism. [I’m glossing over a lot of details!]

Problem 5 (II.2.16(a)) Let X be a scheme, let f ∈ Γ(X, O_{X}), and define X_{f} to be the subset of points x ∈ X such that the stalk f_{x} of f at x is not contained in the maximal ideal m_{x} of the local ring O_{x}. (a) If U = SpecB and \overline{f} ∈ B = Γ(U, O_{X|U}) is the restriction of f, show that U ∩ X_{f} = D(\overline{f}) so X_{f} is an open subset of X.

Proof. Note that D(\overline{f}) = \{x ∈ U : \overline{f} ∉ x\} = \{x ∈ U : \overline{f}_{x} ∉ m_{x}\} = \{x ∈ U : f_{x} ∉ m_{x}} = U ∩ X_{f}. Thus U ∩ X_{f} is an open subset of U. Now let {U_{α}} be an affine open over of X. Then X_{f} = ∪_{α}(U_{α} ∩ X_{f}) is the union of open sets, hence open.

Problem 6 (II.2.16(b)) Assume that X is quasi-compact. Let A = Γ(X, O_{X}), and let a ∈ A be an element whose restriction to X_{f} is 0. Show that for some n > 0, f^{n}a = 0.

Proof. Using the fact that X is quasi-compact we can find a finite cover {U_{i} = SpecB_{i}}_{i=1}^{m} of X by affine open sets. Since a|_{X_{f}} is 0, the image of a in O_{X_{f} ∩U_{i}} = (B_{i})\overline{f} is 0. Thus there exists n_{i} such that \overline{f}^{n_{i}} \overline{a} = 0 in B_{i}. That is, f^{n_{i}}a|_{X_{f} ∩U_{i}} is 0. Letting n = max{n_{1}, …, n_{m}} we see that f^{n}a|_{X_{f} ∩U_{i}} is 0 for each i whence f^{n}a = 0 in A = Γ(X, O_{X}).

Problem 7 (II.2.16(c)) Assume X has a finite cover by open affines U_{i} such that each intersection U_{i} ∩ U_{j} is quasi-compact. Let b ∈ Γ(X_{f}, O_{X_{f}}). Show that for some n > 0, f^{n}b is the restriction of an element of A.

Proof. Write U_{i} = SpecB_{i}. Then b|_{U_{i}} ∈ (B_{i})_{f} so, from the definition of (B_{i})_{f}, there exists n_{i} such that f^{n_{i}}b|_{U_{i}} ∈ B_{i} = O_{X}(U_{i}). Let N = max{n_{i}} and let g_{i} = f^{N}b|_{U_{i}} ∈ O_{X}(U_{i}). Then g_{i}|_{X_{f} ∩U_{i}∩U_{j}} = f^{N}b|_{U_{i}∩U_{j}∩X_{f}} = g_{j}|_{X_{f} ∩U_{i}∩U_{j}} so (g_{i} − g_{j})|_{(U_{i}∩U_{j})_{f}} = 0. By part (b) since U_{i} ∩ U_{j} is quasi-compact there is an integer n_{ij} such that f^{n_{ij}}(g_{i} − g_{j}) = 0 in O_{U_{i}∩U_{j}}. Let M = max{n_{ij}}, and let h_{i} = f^{M}g_{i}. Then h_{i} ∈ O(U_{i}) for each i and h_{i}|_{U_{i}∩U_{j}} = h_{j}|_{U_{i}∩U_{j}} so we can find h ∈ Γ(X, O_{X}) such that h|_{U_{i}} = h_{i}. But, for each i, h|_{X_{f} ∩U_{i}} = f^{M}g_{i}|_{X_{f} ∩U_{i}} = f^{M}f^{N}b|_{X_{f} ∩U_{i}} = f^{M+N}b|_{X_{f} ∩U_{i}} so by uniqueness (sheaf axiom iii), h|_{X_{f}} = f^{M+N}b, as desired.

Problem 8 (II.2.16(d)) With the hypothesis of (c), conclude that Γ(X_{f}, O_{X_{f}}) ≅ A_{f}.

Proof. Define a homomorphism φ : A_{f} → Γ(X_{f}, O_{X_{f}}) by φ(a/f^{n}) = a|_{X_{f}} /f^{n}|_{X_{f}}. This is well-defined because the stalk f_{x} is invertible in each local ring O_{x} for each x ∈ X_{f}. φ is a homomorphism since restriction is a homomorphism. Suppose φ(a/f^{n}) = 0, then a|_{X_{f}} /f|_{X_{f}}^{n} = 0 so a|_{X_{f}} = 0. By part (b) there exists m such that f^{m}a = 0, so a is 0 in A_{f}, whence φ is injective. Suppose b ∈ Γ(X_{f}, O_{X_{f}}), then by part (c) there exists n such that f^{n}b is the restriction of some a ∈ Γ(X, O_{X}) to X_{f}. Then φ(a/f^{n}) = a|_{X_{f}} /f|_{X_{f}}^{n} = f^{n}b/f^{n}|_{X_{f}} = b so φ is surjective, which establishes the desired isomorphism.

Problem 9 (II.2.17(a)) A Criterion for Affineness. Let f : X → Y be a morphism of schemes, and assume that Y can be covered by open sets U_{i}, such that for each i, the induced map f^{−1}(U_{i}) → U_{i} is an isomorphism. Then f is an isomorphism.

Proof. Define a morphism g : Y → X as follows. On each open set U_{i} let g_{i} be the morphism inverse to f|_{f^{−1}(U_{i})} : f^{−1}(U_{i}) → U_{i}. Then g_{i}|_{U_{i}∩U_{j}} = f|_{f^{−1}(U_{i})∩f^{−1}(U_{j})} = g_{j}|_{U_{i}∩U_{j}} so, as in step 3 of the proof of Theorem 3.3, we can glue the morphisms g_{i} to obtain a morphism g : Y → X such that g|_{U_{i}} = g_{i}. As a map of spaces g is clearly inverse to f. Since each g_{i} is an isomorphism, the induced maps g^{#} on stalks are isomorphisms so the induced map on sheaves is an isomorphism. Thus g is an isomorphism.