Is the support of a flat sheaf flat?

Question

Note: in the following, all scheme/algebra morphisms should be assumed essentially of finite type.

Geometric version: Let $X$ be a scheme flat over $S$ (both noetherian), and let $\mathscr{F}$ be a coherent sheaf on $X$, also flat over $S$. The scheme-theoretic support $\mathfrak{X}$ for $\mathscr{F}$ is a closed subscheme of $X$. Is it necessarily true that $\mathfrak{X}$ is flat over $S$?

Algebraic version: Let $B$ be a flat $A$-algebra (both noetherian), and let $M$ be a finitely generated $B$-module, also flat over $A$. Is it necessarily true that $B/\operatorname{Ann}(M)$ is flat over $A$?

Motivation: the only way I know how to visualize a coherent sheaf is to visualize its support, which is a closed subscheme. I justify this by the fact that many of the properties of a coherent sheaf are shared by (the structure sheaf of) its scheme-theoretic support. For instance, they have the same associated points. In case $A$ is a DVR, this even provides a proof for the algebraic version above, since a module is flat over a DVR iff all its associated points map to the generic point. (see Angelo's comment below)

This general "visual intuition" tells me that the two (equivalent) statements above should be true. However, I cannot think of a good argument for this. Although it is not really essential to anything I am doing, it is bothering the heck out of me not to know whether this actually works, and distracting me from my other, more "essential" work. Thus, I would appreciate some help here. A positive answer will help me sleep at night (figuratively speaking); a negative answer will, hopefully, give me a useful counterexample against which to test my intuition in the future.

Second motivation: If the statement is true, then it provides evidence for a morphism from the Quot scheme to the Hilbert scheme, that--loosely speaking--takes a coherent sheaf to its support. (Thinking about it in these terms may also suggest solutions to mathematicians who--unlike me--have a great deal of experience with Quot and Hilbert schemes.)

Answer 1

There are many counterexamples to this. Suppose that $S$ is a smooth surface over $\mathbb C$. Let $T \to S$ be a finite morphism from another smooth surface $T$, and consider a factorization $T \to V \to S$, where $V$ is obtained by gluing two points on a fiber. Then $T \to S$ is flat, $V \to S$ is not. Embed $V$ into the product $X$ of a projective space with $S$; as a sheaf $F$ on $X$ take the direct image of the structure sheaf of $T$. Then $F$ and $X$ are flat over $S$, but the scheme-theoretic support is $V$, which is not flat.

Answer 2

Here is an algebraic construction. The way I think about it is based on these two facts: 1) when $A$ is regular domain, a module-finite A-algebra is flat iff it is Cohen-Macaulay (CM) of same dimension and 2) there are non-CM domains which admit a CM module of same dimension.

Now the concrete construction. Let $B = k[x,y,u,v]$, and we map $B$ onto $C = k[a^4, a^3b, ab^3, b^4]$ which is not CM (so $x$ maps to $a^4$ and $v$ to $b^4$). Let $P \subset B$ be the kernel.

Let $A = k[x,v]$ and $M = \bar C$, the integral closure of $C$. $M$ is actually $k[a,b]$ which is flat over $A=k[a^4, b^4]$.

However, the annihilator of $M$ over $C$ is zero, so the annihilator of $M$ over $B$ is $P$. But if $B/P \cong C$ is flat over $A$, it would be Cohen-Macaulay, contradicting our choice.