Supports of flat modules, part B

Part A is this post. Let me prove the opposite of what the exercise in part A wrongly claimed (sigh).

Lemma: Let Z —> Y be a finite morphism of affine schemes. Then there exists a closed immersion Z —> Z’ of schemes over Y such that Z’ is finite syntomic over Y.

Remark: If we embed Z’ into a smooth scheme X over Y, then F = O_{Z’} is a coherent O_X-module flat over Y such that the generic points of Z are associated points of the restriction of F to their fibres.

Proof: Write Y = Spec(A) and Z = Spec(B). Choose generators b_1, …, b_r of B as an algebra over A. As B is finite over A, each b_i is the root of a monic polynomial P_i with coefficients in A. Then B’ = A[x_1, …, x_r]/(P_1(x_1), …, P_r(x_r)) is finite syntomic over A and Z’ = Spec(B’) works. EndProof.

The point I want to make in this post is that we have some equidimensionality result for associated points of flat modules, namely EGA IV, Proposition (see Tag 0GSF). It implies the following: suppose that f : X —> Y is smooth with Y Noetherian and irreducible. Suppose that F is coherent on X and flat over Y. Let x be a point of the generic fibre of f which is an associated point of F. Then the zariski closure Z ⊂ X of the singleton {x} has the property that Z —> Y is equidimensional!

So for example, there is no finite module M over k[x, y, z] which is flat over k[x, y] such that (x – yt) is an associated prime of M. Presumably, you can see this directly? Is it easy? I didn’t try.


Supports of flat modules

Let Z —> Y be the normalization of an affine cuspidal curve over an algebraically closed field k. Let i : Z —> X be a closed immersion over Y with X smooth over Y.

Question: Does there exist a coherent module F on X, flat over Y, whose support is equal to Z set theoretically?

Answer: No in characteristic 0 and yes in characteristic p > 0.

To see that the answer is no in characteristic 0 you show that the map O_Y —> O_Z has an O_Y-linear section if you have F (and of course this isn’t possible for the normalization of the cuspidal curve). Namely, consider the map tau : O_Z —> O_Y which sends an element f of O_Z to the trace over O_Y of multiplication by f’ on F where f’ is any lift of f to O_X. You show that the choice of f’ doesn’t matter by checking at the generic point; the key fact is that the support condition tells us that f’ which vanish on Z give nilpotent operators on F. Finally, this gets us a section as tau(g) = rg for g in O_Y. Here r = rank_Y(F) > 0 which is invertible as we have char 0.

Remark: I think there doesn’t even exist a coherent F on X, flat over Y, such that the generic point of Z is an associated point of F. Exercise! [Edit on 2/12/22: Jason did the exercise and, uh, it isn’t true!]

To see that the answer is yes in characteristic p > 2, say Y is the spectrum of A = k[a, b]/(a^3 – b^2). Let X be the spectrum of A[t] and consider the closed subscheme, finite flat over Y, cut out by t^p – a^{(p – 3)/2}b. The reduction of this subscheme is isomorphic to Z. For p = 2 use t^2 – a. Cheers!