Let $f\rightarrow S$ be an open morphism of reduced finite dimensional complex spaces (or a universally open morphism of locally noetherian excellents without embedded components or reduced schemes) with fibers of dimension $n$.

If $f^{*}G$ is torsion free for all torsion free coherent sheaves $G$ on $S$, then is it true that $f$ is flat ?

Thank you.

  • $\begingroup$ Kaddar, just to clarify, does "torsion-free" mean each stalk has positive depth, or the (seemingly stronger, if I'm not mistaken) condition that there is a dense Zariski-open $j:U \rightarrow S$ on which $G$ is a vector bundle s.t. $G \rightarrow j_{\ast}j^{\ast}(G)$ has trivial kernel (or equiv., $G$ is subsheaf of $j_{\ast}(E)$ for such a $U$ and some vector bundle $E$ on $U$)? Also, do you insist on the hypothesis only for $G$ given over all of $S$, or are you willing to make the hypothesis over all opens of $S$ too (as otherwise hard to localize the problem)? $\endgroup$
    – BCnrd
    Jun 20 '10 at 1:53

I think so. It seems to me that the two definitions proposed by Brian are equivalent. Let us use the following: a coherent sheaf $F$ on a reduced space $X$ is torsion-free if whenever $a$ is a non-zero divisor in the local ring ${\cal O}_{X,p}$, multiplication by $a$ is injective on the stalk $F_p$. Or, for every $p\in X$ the only associated primes of $F_p$ as an ${\cal O}_{X,p}$-module are the minimal primes of ${\cal O}_{X,p}$.

Now, let $f\colon X \to S$ be a morphism, $p \in X$, $q := f(p)$, $A := {\cal O}_{S,q}$, $B := {\cal O}_{X,p}$. Let $M$ be the maximal ideal of $A$; this is the stalk of the torsion free sheaf $I_{S,q}$ (the sheaf of ideals of $q$ in $S$); set $k := A/M$. The fact that $f$ is open implies that every minimal prime of $B$ maps to a minimal prime in $A$.

By Grothendieck's local criterion of flatness, it is enough to show that $\mathop{\rm Tor}_1^A(k, B) = 0$, which is equivalent to saying that the natural homomorphism $M \otimes_A B \to B$ is injective. However, this homomorphism is injective outside of the inverse image of the fiber on the maximal ideal of $A$, which does not contain any of the minimal primes of $B$, so is nowhere dense. Since $M \otimes_A B$ is a torsion-free $B$-module, it can not contain a non-zero submodule supported on a nowhere dense closed subset; so the kernel has to be trivial, and this proves the result.

  • $\begingroup$ Thank you Angelo. Of course, the two definition proposed by Brian are equivalent in this setting. $\endgroup$
    – kaddar
    Jun 21 '10 at 7:50

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