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 12.1.1.5 (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.

Enjoy!

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!