MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


Let $f:A\rightarrow B$ be a local morphism of locally noetherian (reduced) rings with $B$ $A$-flat. Let $M$ be an $B$-module of finite type.

Question: Which conditions ensure the following:

$N\otimes_{A} M$ is $A$-torsion free for every $A$-torsion free module of finite type $N$ $\Longrightarrow$ $M$ is $A$-flat ?

Thank you very much.

share|cite|improve this question
I'm curious, what conditions are you hoping for? Conditions on $f, A, B, M$? – Karl Schwede Sep 22 '10 at 16:47
Have you done any special cases, $f$ being the identity for example? Also, why do you want this? – Karl Schwede Sep 22 '10 at 16:51

Assuming by "torsion-free" you mean any zero-divisor on $M$ must be zero-divisor on $R$, which is standard for Noetherian rings. Then you don't need any extra conditions. In fact I will claim the following (no need for flatness of $B$):

Let $(A,m,k)$ be a local, Noetherian, reduced ring. Let $B$ be a Noetherian A-algebra such that $mB \subset rad(B)$ and $M$ be a finitely generated $B$-module. Then $M$ is a flat $A$-module if and only if $M\otimes_Am$ is A-torsion-free.

Proof: One direction is clear, as $m$ is always torsion-free. So now assume $M\otimes_Am$ is A-torsion-free.

Tensor $M$ with $0 \to m \to A \to k\to 0$ we get that $\text{Tor}_1^A(M,k)$ embeds into $M\otimes_Am$. The latter is torsion-free, and the former is killed by $m$. But reducedness implies that $m$ is not an associated prime of $A$ (if it is, then $m$ kills some nonzero element $x\in m$, so $x^2=0$). So $\text{Tor}_1^A(M,k)$ must be $0$, and $M$ is flat by Matsumura Theorem 22.3 (all you need is that $M$ is $m$-adically separated).

Note that without reducedness the result is false. Take $A=B=k[[t]]/t^2$. Then any $A$-module is torsion-free, as everything in $m=(t)$ is a zero-divisor.

share|cite|improve this answer
Thank you very much dear Dao. – kaddar Sep 23 '10 at 8:06

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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