ag.algebraic geometry - Is the following a sufficient condition for flatness?

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 $pin X$ 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 $fcolon X to S$ 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.