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Charles Staats
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Is the support of a flat sheaf flat?

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.

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.)

Charles Staats
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