Student Q&A

Professor Kedlaya provided answers to some of the questions commonly asked by students in the class. The questions, as well as their answers, are listed below.


Are the hypotheses of Hartshorne exercise II.2.18(d) (PS 3) all necessary?

No, it is enough to assume that phi is surjective; the hypothesis that f is a homeomorphism onto a closed subset will follow as a consequnece.

^ Back to top

Help! I'm stuck on the Stone-Cech compactification (PS 3, problem 2).

A: Look up the definition of an ultrafilter, and relate Spec F_2^S to those. Then read a proof of the theorem written in the language of ultrafilters.

^ Back to top

For f: X -> Y a morphism and y in Y a point, what is the definition of the scheme-theoretic fibre X_y?

Let k(y) be the residue field of the local ring O_{Y,y}. Then there is a canonical map Spec k(y) -> Y given by choosing an open affine neighborhood U = Spec(A) of y in Y, then noticing that A maps to O_{Y,y} and hence to k(y). Now let X_y be the fibre product of Spec k(y) with Y over X.

^ Back to top

What does the notation M(n), for M a graded module over a graded ring, mean? I couldn't find the definition in Hartshorn

It means that the i-th graded piece of M(n) is the (i+n)-th graded piece of M.

^ Back to top

Does the proof of Theorem II.5.4 in Hartshorne work without the noetherian hypothesis?

Yes and no. The proof does work as written. However, the definition of coherent sheaf given in Hartshorne is not the standard one in the nonnoetherian case; he is defining what I would call a finitely generated quasicoherent sheaf. The correct (as per EGA) definition of coherent sheaf will be given in class later.

^ Back to top

Does the definition of "locally compact" used in this course include the Hausdorff condition?

Yes. According to Bourbaki, "locally compact" means Hausdorff and having a neighborhood basis consisting of open sets with compact closure.

^ Back to top

Let S be a graded ring generated in degree 1. Do the D_+(f) (called D(f) in class) for f in S_1 form a basis of Proj S?

No, they only form a cover of Proj S. For instance, the intersection of D_+(f) with D_+(g) is D_+(fg), but fg is then in degree 2 rather than 1.

^ Back to top

Suppose X is a topological space covered by the open sets U_i. Let F and G be sheaves on X, and suppose F -> G is a morphism which induces bijections F(U_i) -> G(U_i). Is F a bijection?

In general, no. But there is an important special case where this is true: if X is a scheme, each U_i is affine, and F and G are quasicoherent O_X-modules, then this follows from the third fundamental theorem of affine schemes (proof left to the reader).

^ Back to top

In Hartshorne exercise II.4.8, what exactly is meant by the assertion that "a product of morphisms having property P has property P"?

If X,Y,X',Y' are schemes over a common base S, and X -> Y and X' -> Y' are two morphisms (commuting with the maps to S) having P, then X x_S X' -> Y x_S Y' also has P.

^ Back to top

I'm having trouble dealing with the topological notion of properness (PS 4, exercise 9).

The most difficult part is probably (c). Everything can be found in Bourbaki's Topologie Generale, section 1, or the English translation of same, but you might find that difficult to read because it uses the language of filters and ultrafilters.

Here is a translation of a key lemma: if f: X -> {point} is proper, then X is quasicompact. Proof sketch: suppose that {U_i} is an open cover of X. Topologize Y = X union {y} with a basis consisting of all subsets of X, plus sets of the form (X - U_{i_1} - ... - U_{i_n}) union {y}. Show that if the closure of {(x,x): x in X} in X x Y has closed image under f x id_Y, then the original cover of X must admit a finite subcover.

^ Back to top

Does "stable under base change", for a property of morphisms of schemes, imply "local on the base"?

It implies only part of that. For a property P of morphisms of schemes, we say P is stable under base change if for any morphism f: Y -> X with P, and any morphism Z -> X at all, the morphism Y x_X Z -> Z has P also. If we take Z to be an open subscheme of X, then Y x_X Z is just the inverse image f^{-1}(U) viewed as an open subscheme of Y. So stability under base change implies that if f has P, then so do the maps f^{-1}(U_i) -> U_i for any open cover U_i of X. We say f is local on the base if this is true and you can go the other way: for any open cover {U_i} of X, if f^{-1}(U_i) -> U_i has P for each i, then f itself has P.

^ Back to top

Does the category of schemes admit arbitrary (small) colimits?

No, it does not even admit finite colimits. It does admit (small) coproducts via the disjoint union, but in general one cannot construct coequalizers, i.e., colimits of diagrams consisting of two morphisms from X to Y. In the category of sets, the coequalizer of f_1, f_2: X -> Y is defined by taking the equivalence relation generated by the pairs (y_1, y_2) in Y x Y with the property that for some x in X, f_1(y_1)= f_2(y_2); the coequalizer is then the quotient by this equivalence relation. Unfortunately, in the category of schemes, it is possible to have equivalence relations (defined in the appropriate categorical fashion) for which the quotient does not exist. Any Moishezon manifold which is not a scheme gives an example; see the appendices to Hartshorne. (More precisely, an equivalence relation on Y is a scheme R carrying two maps R -> Y satisfying the usual three axioms for an equivalence relation, and such that the induced map R -> Y x Y is an immersion. The quotient by R if it exists is the coequalizer of the two maps R -> Y.) By contrast, the category of affine schemes does admit equalizers (and in fact arbitrary small colimits), because the category of rings has arbitrary small limits (reflected from the category of sets).

^ Back to top

Is the statement of Hartshorne exercise II.5.13 still true without the hypothesis that S is generated by S_1 as an S_0-algebra?

I believe so, but I didn't check all the details. I think the assumption is included so that the explanation in terms of the d-uple embedding is valid.

^ Back to top

How can one approach Hartshorne exercise II.5.14(a)?

One approach (possibly not what Hartshorne intended) is to view Spec S as a G_m-bundle over Proj S (i.e., locally a product of Proj S with Spec k[t,t^{-1}] over Spec k). You can then use the fact from class that normality of a ring is a local condition to reduce to checking that if R is a normal domain, then so is R[t,t^{-1}].

^ Back to top

Can you explain in more detail what the map Spec (B tensor_A B) -> Spec B is in PS 6, problem 4(a)?

Recall that we were given initially a homomorphism f: A -> B of rings. View that as a map of A-algebras, and then tensor over A with B using f. Let's say we're doing this on the right side; you now have a map A tensor_A B -> B tensor_A B computed by a tensor b --> f(a) tensor b. Then identify A tensor_A B with B by identifying 1 tensor b with b.

^ Back to top

I'm still having trouble with PS 6, problem 4(a) (faithful flat descent along B -> B tensor_A B).

In fact, the argument is the same for any faithfully flat morphism B -> C which admits a splitting C -> B in the category of rings; this might simplify your notation. To wit, let M be a C-module equipped with a descent datum. We claim that the underlying B-module is M tensor_C B; to establish this, we must (among other things) exhibit an isomorphism M -> (M tensor_C B) tensor_B C. We get it from the descent isomorphism M tensor_C (C tensor_B C) -> M tensor_C (C tensor_B C), in which on the left side C operates on the first copy of C in C tensor_B C, and on the right side C acts on the second copy of C in C tensor_B C.

The two sides of the descent isomorphism are modules over C tensor_B C; using the right action of C on C tensor_B C, we view them both as C-modules. What happens if we now tensor with B? The left side of the isomorphism simply becomes (M tensor_C C) tensor_B (C tensor_B B) = M tensor_B B = M. The right side is a bit trickier: we can pull the first C of C tensor_B C all the way to the outside to get C tensor_B (M tensor_C C tensor_C B) = C tensor_B (M tensor_C B). In other words, the descent isomorphism gives us the map M -> (M tensor_C B) tensor_B C that we were looking for.

There is more to do, but I'll leave it to you. You must use the cocycle condition to check that the operation M -> (M tensor_C B) tensor_B C is functorial in M. If you get confused, you might want to keep in mind the example k -> k[x] -> k.

^ Back to top

How do I show that a finite surjective morphism between nonsingular algebraic varieties over an algebraically closed field is finite (Hartshorne exercise III.9.3)?

Note that if x is a nonsingular point on the k-variety X, then the map from the local ring O_{X,x} to its completion is faithfully flat, and the latter is isomorphic to a power series ring k[[x_1, ..., x_n]] by the Cohen structure theorem. Using this, one can reduce to checking: if f: R -> R is a finite morphism for R = k[[x_1, ..., x_n]], then f is flat. To prove this, you may want to use the Auslander-Buchsbaum formula (Hartshorne Proposition III.6.11A); you will need to know that if A -> B is a finite injective morphism of local rings, the depth of a B-module M is the same as the depth of M considered as an A-module.

^ Back to top

In the proof of Hartshorne Proposition II.7.3, it is shown that assuming conditions (1) and (2), then the map phi is injective on closed points. Why does it follow that phi is injective on all points?

Suppose the point z in P^n_k has more than one point in its inverse image. Let Z be the closure of z in P^n_k, let y be any point in the inverse image of z, and let Y be the closure of y in X. Since phi is proper, it is a closed map, so phi(Y) is a closed subset of P^n_k contained in Z (by continuity) but containing z (since phi(y) = z). We must thus have phi(Y) = Z. That is, for each point y in phi^{-1}(z), the closure of y in X surjects onto Z. So now we hit trouble: if y_1, y_2 are distinct preimages of Z with closures Y_1, Y_2, then any closed point of Z has one preimage in Y_1 and one in Y_2, which must then coincide. Since closed points are dense in both Y_1 and Y_2 (closed points are dense in any quasiprojective scheme over a field; I think that was an exercise a while back), this forces Y_1 = Y_2. But then y_1 = y_2 since they are the generic points of Y_1 and Y_2, contradiction.

^ Back to top

Again in the proof of Hartshorne Proposition II.7.3, while we are trying to prove that phi is projective, Corollary II.5.20 is invoked. But doesn't that statement require phi to be projective already, not just proper?

Yes it does! This is a real problem with the argument. Corollary II.5.20 turns out to be true for a proper morphism, but some proof is required. It can be deduced from Chow's lemma (exercise II.4.10) as follows. (I think this is the approach used in EGA III but I haven't checked yet.)

Namely, if f: X -> Y is proper, then we can find g: X' -> X a birational regular morphism such that X' -> Y is projective. Here birational means that there is a dense open subset U of X over which g is an isomorphism. If F is a coherent sheaf on X, then g^* F is a coherent sheaf on X', so Corollary II.5.20 implies that f_* g_* g^* F is a coherent sheaf on S. There is an adjunction map F -> g_* g^* F which is an isomorphism over U, so the kernel H is supported on X-U. If I can show that f_* H is coherent, then I'll be done because f_* F will then be trapped in an exact sequence between the coherent sheaves f_* H and f_* g_* g^* F. I do this by finding a closed subscheme Z supported on X-U such that H is the pushforward of a coherent sheaf on Z. (What this is saying on affines is that if you have a finitely generated module M over a noetherian ring R, and it is supported entirely on V(I) for some ideal I, then there is another ideal I' with the same support as I such that I'M = 0.) Now I can replace X -> Y with Z -> Y and argue by induction on the dimension of Z.

^ Back to top

What is the correct statement of Hartshorne Proposition 8.12? The exact sequence given seems to mix sheaves on X and Z.

One way you can say it is to write it using sheaves on X. Letting j denote the closed immersion Z -> X, we have J/J^2 --> Omega_{X/Y} tensor_{O_X} j_* O_Z --> j_* Omega_{Z/Y} --> 0.

Or you can note that from Proposition 8.4A, the sheaf on the left is the pushforward of a sheaf F on Z, which coincides with j^* (J/J^2). So we also have

j^* (J/J^2) --> j^* Omega_{X/Y} --> Omega_{Z/Y} --> 0.

^ Back to top

The example of a flat family in Hartshorne Example III.9.8.4 seems a bit mysterious. Where does the ideal I come from?

It is obtained as follows. First, working in k[x,y,z,a,a^{-1}], find the defining ideal for the scheme $X_a$ by eliminating t from the parametric equations. Then take generators for this ideal and multiply by a suitable power of a to obtain generators of some ideal in k[x,y,z,a]. That only gives you the desired I if k[x,y,z,a]/I is flat over k[a], or equivalently torsion-free. That will fail precisely if there is a-torsion in the quotient, i.e., if you can form a combination of generators of the ideal which is divisible by a, but the result is no longer in the ideal. Once you "saturate" to eliminate the a-torsion in the quotient, you have a flat family.

^ Back to top

In Hartshorne exercise II.6.1, once I have an exact sequence 0 -> Z -> Cl(X x P^n) -> Cl(X) -> 0, how do I show that it splits?

Use the map from Cl(X) to Cl(X x P^n) taking a divisor D to its inverse image in X x P^n.

^ Back to top

I got 2g-2 instead of g+1 in Hartshorne exercise IV.1.6. What did I do wrong?

Instead of using the canonical divisor, pick a divisor D of degree as small as possible such that l(D) > = 2.

^ Back to top

On PS 7, problem 8, to exhibit a group of automorphisms of order 168, is it sufficient to exhibit subgroups of orders 2^3, 3, 7?

In principle, no, because they might generate a larger group. (You would then have to write down relations between the elements to show that this does not happen.) But if you assume characteristic 0, then the Hurwitz bound rules this out. I decided to allow this assumption retroactively.

^ Back to top

I need a hint on how to write down a nonzero section of the canonical sheaf on a hyperelliptic curve (over a field of characteristic different from 2).

Suppose your curve has an affine part of the form Spec k[x,y]/(y^2 - P(x)) with P of odd degree 2g+1 having no repeated roots. Then dx/y = 2 dy/P'(x); these two representations together give a section of omega defined everywhere in the affine part where either y or P'(x) are nonzero. But since P has no repeated roots, this is all of the affine part. Moreover, you can compute the order of vanishing of x, y at the unique point at infinity (by rewriting the defining equation in terms of 1/x and 1/y, which are both regular there) and see that dx/y is holomorphic also at infinity.

^ Back to top

What is the length of a scheme (PS 7, problem 11)?

It is the length of the longest chain of (nonisomorphic) closed immersions ending at your scheme (with starting index 0). For instance, the scheme Spec Z/p^n Z has length n because Spec Z/Z -> Spec Z/p Z -> Spec Z/p^2 Z -> ... -> Spec Z/p^n Z is a chain of length n. It can be shown that if (and only if) k is an algebraically closed field, the length of Spec A for A an artinian k-algebra is equal to dim_k A.

^ Back to top

So then what do I need to do on PS 7, problem 11?

Relate the length of the intersection to the vanishing of some section of the canonical divisor on C itself. (This really has nothing specific to do with the canonical divisor; any divisor giving an embedding works the same way.)

^ Back to top

I'm confused about PS 8, problem 7(b). Am I supposed to construct a commuting diagram?

No, that may not be possible in general. All you have to do is find a quasi-isomorphism D^. -> J^. with J^. a complex of injectives, and a morphism I^. -> J^. which on cohomology induces the map that you get from f by using the quasi-isomorphisms C^. -> I^. and D^. -> J^. to identify cohomology. You need not ensure that C^. -> D^. -> J^. and C^. -> I^. -> J^. are the same maps!

^ Back to top

What does the suggestion about the pushout mean in PS 8, problem 3?

It means that instead of directly comparing what happens when you consider monomorphisms u: A -> B and u': A -> B' for which T^i(u) = T^i(u'), you should reduce to the case when these are related by a map B -> B'. You do that by forming the pushout of u and u'.

^ Back to top

What is the basic idea behind PS 8, problem 7(a)?

Say C^. is your original complex, and you have defined a complex of injectives up to I^i and comparison maps up to C^i -> I^i. The kernel of the arrow I^i -> I^{i+1} must then have kernel equal to the image of ker(C^i -> C^{i+1}) under C^i -> I^i, plus the image of I^{i-1} -> I^i. Let J be the quotient of I^i by that stuff. Now form the pushout of C^i -> C^{i+1} and C^i -> J and stick that into an injective.

^ Back to top

Isn't the corollary of PS 8, problem 13 already a corollary of the simpler fact that M is flat if and only if Tor_1(R/I,M) = 0 for any ideal I of R?

Yes, it is.

^ Back to top

How do I compute the cohomology in Hartshorne exercise III.2.7(b)?

You may want to use the fact that any open cover can be refined to a finite cover by open intervals, and then further refined by replacing each interval with a subinterval whose closure is contained in the original interval. Then use partitions of unity.

^ Back to top

Why doesn't Hartshorne exercise III.2.1 contradict the acyclicity theorem for affine schemes?

Because the sheaf Z_U is not quasicoherent!

^ Back to top

Should I be using cohomology to solve Hartshorne III.3.2?

You can do it either with or without cohomology. In both cases, what you should be trying to prove is: if X is reduced and noetherian, and is the union of two disjoint closed subschemes X_1, X_2 which are both affine, then X itself is affine (then argue by induction on the number of components).

For a cohomological argument: one must show that the higher cohomology of a coherent sheaf F on X is zero. Use the ideal sheaves defining X_1 and X_2 to put F in the middle of a short exact sequence.

For a direct argument: put X_1 = Spec R, X_2 = Spec S, X_1 cap X_2 = Spec T; we then have surjective ring homomorphisms R -> T and S -> T. Let U be the kernel of R oplus S -> T; then there is a map X -> Spec U by adjunction, which we want to be an isomorphism. But now we have a claim that we can check locally on Spec U.

^ Back to top

I gather that Hartshorne exercise II.1.16(b) requires Zorn's Lemma, but I don't quite understand how to apply it here.

Recall that the statement is: if F is a flasque sheaf on a topological space X, and 0 -> F -> G -> H -> 0 is an exact sequence of sheaves (of abelian groups, say), then Gamma(X, G) -> Gamma(X, H) is surjective. To prove this, choose a section s in Gamma(X, H). We know that X is covered by open sets U such that s lifts to Gamma(U, G). Consider the collection of such open sets U; it has the property that every chain of such open sets has an upper bound (namely its union, by the sheaf axiom). So Zorn's lemma implies that there is a maximal element U, i.e., an open set U such that s lifts to Gamma(U, G) but not to Gamma(V, G) for any open set V properly containing U.

We claim this forces U = X. Suppose the contrary; pick a point x in X not in U. We can then find an open subset V of X containing x such that s lifts to a section t_2 in Gamma(V, G). We already know that s lifts to a section t_1 in Gamma(U, G). After restricting to Gamma(U \cap V, G), we have t_1 - t_2 = 0; so by exactness, this difference lifts to a section u in Gamma(U \cap V, F). By flasqueness of F, this section extends to a section u in Gamma(X, F). So now the restrictions of t_1 and (t_2 + u) to Gamma(U \cap V, G) agree, which means we can glue them to get a section of Gamma(U \cup V, G). This maps to s in Gamma(U \cup V, H) because u maps to 0 as a section of H by exactness of the sequence. This contradicts the choice of U, so we must indeed have U = X.

If you know about ordinals, you might prefer to argue using transfinite induction instead. The idea is the same: if U_i is an increasing sequence of open subsets over which s lifts to a section of G, we can always extend it unless it has a last element equal to X. At nonlimit stages we do this as in the previous argument; at limit stages we do it by taking the union.

^ Back to top

Is there a simple solution of PS 9, exercise 1 not using spectral sequences?

Yes; here is a solution suggested by Fucheng Tan. Let 0 -> F -> G -> H -> 0 be a short exact sequence of sheaves on U with F as given and G flasque. Claim: the sequence 0 -> F(U) -> G(U) -> H(U) -> 0 is exact, or equivalently, any section s in H(U) lifts to G(U). Proof: we know that U admits a cover by basic open subsets V_i such that the restriction of s to V_i lifts to a section s_i in G(V_i). For each pair i, j, the difference s_i - s_j is the zero section of G(V_i cap V_j), so comes from a unique element f_{ij} of F(V_i cap V_j). Moreover, the f_{ij} form a Cech 1-cocycle for the covering by the V_i, so by our hypothesis on F, they also form a 1-coboundary. That is, there exist g_i in F(V_i) such that g_i - g_j = f_{ij}. Now the sections s_i - g_i in G(V_i) glue to a section in G(U) lifting s.

It follows that 0 -> C-cech(U, F) -> C-cech(U, G) -> C-cech(U, H) -> 0 is exact (since we need only work with basic open coverings). If we take the long exact sequence in cohomology, the first connecting homomorphism is zero (because H-cech^0 = H^0 always and we just checked that the H^0 give an exact sequence) and the higher G terms vanish because G is flasque. Hence H also satisfies the input hypothesis (i.e., it is Cech-acyclic on any basic open). We can now perform a dimension shifting argument: we get the isomorphism H-check^1(U, F) -> H^1(U, F) = 0 by comparing long exact sequences, and the higher ones by comparing to H.

We now know that F is sheaf-acyclic on each basic open U. By the Leray theorem (and the niceness of the basis), it follows that the Cech cohomology of F for any basic open covering computes sheaf cohomology.

^ Back to top

Is a coherent sheaf necessarily (locally) finitely generated?

Yes, that is part of the definition.

^ Back to top

How do I get started on PS 11, problem 5(a)?

Remember that the property of a sheaf being finitely generated is by definition a local property. So we can check the finite generation of ker(phi) locally. In particular, we need only check it on an open set U for which F itself is generated by finitely many sections over U. We can then add phi to a surjection O_X^m -> F to get another surjection, and proceed from there.

^ Back to top

If A is a coherent ring, is the structure sheaf on Spec A necessarily coherent?

Yes. This follows from the fact that if A is a coherent ring, then so is the localization A_f for any f in A. That can be checked directly from my definition, or it can be deduced by establishing an alternate criterion: a ring is coherent if and only if any finitely generated ideal is finitely presented.

^ Back to top

I'm confused about the Hilbert function (PS 10, problem 11(b)).

That's because I was confused about it too. The classical definition is not the one I gave: if X is a closed subscheme of P^r_k = Proj S for S = k[x_0, ..., x_n], and I is the saturated ideal defining S, then the Hilbert function is n -> dim_k (S/I)_n. For n large, this agrees with the Hilbert polynomial and hence with the number I wrote down, but not for small n; e.g., if X is zero-dimensional, then my Hilbert function is constant but the classical one is not.

^ Back to top