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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.