# 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. (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.)

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Since your question is really about intuition. Let me suggest that another way to visualize $\mathcal{F}$ or $M$ is in terms of its associated primes. These are prime ideals $p_i$ for which $M$ can be built up by a successive extension of $B/p_i$. These define the components of the support plus so called embedded components. It is easy to see, by writing out $Tor$'s, that if the $B/p_i$'s are $A$-flat, then so is $M$. Hope this helps. – Donu Arapura Feb 10 '12 at 21:10
Surely you mean to say "let $M$ be a finitely generated $B$-module" in the algebraic version? – Konstantin Ardakov Feb 10 '12 at 21:15
Konstantin: Yes, of course. It's now corrected--thanks for pointing out the error. – Charles Staats Feb 10 '12 at 21:22
Donu Arapura: Thanks for your comment; it does help me relate my intuition to rigorous math. If the converse to your last sentence (of more than three words) were also true, then it would provide an easy positive answer to my question. Unfortunately, this converse is not true, since--for instance--$\Bbbk[x,y]/(y)$ is not flat over $\Bbbk[x,y]/(xy)$. – Charles Staats Feb 10 '12 at 21:29
By the way, it is not true that a coherent sheaf and its scheme-theoretic support have the same associated points. For example, take $A = k[x]$ and $M = A \oplus k[x]/(x)$. – Angelo Feb 11 '12 at 9:01

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.

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In case anyone is wondering why $V$ is not flat over $S$: If $V$ were flat over $S$, then $V$ would be Cohen-Macaulay, since the flat pullback of a regular sequence is regular. Since Cohen-Macaulay is equivalent to ($S_n$ for all $n$), it would in particular be $S_2$, and thus satisfy the Hartogs condition. But there exist regular functions defined in a punctured neighborhood of the singular point that cannot be extended over the singular point. (These correspond to local regular functions on $T$ that do not agree on the two points lying over the singular point.) – Charles Staats Feb 12 '12 at 15:03

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.

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