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Let $R$ be an integral domain and let $M$ be a flat $R$-module. Then $M$ is torsion-free, so $am \neq 0$ for all $0 \neq a \in A$ and $0 \neq m \in M$. In particular, if $0 \neq a \in A$, then $aM \neq 0$. So, $Ann_R(M) = 0$.

But then this question implies that all finitely generated flat modules are already faithfully flat, so this seems to be wrong.

So, how can a torsion-free module have non-trivial annihilator? I must be misunderstanding something!

I also thought that when $M$ is torsion-free, then for any prime ideal $P$ of $R$ the localization map $M \rightarrow M_P$ is injective, so if $M$ is non-zero so is $M_P$ and therefore $Supp(M) = Spec(R)$, implying $Ann(M) = 0$ for $M$ finitely generated.

What is wrong here? This must be absolutely stupid.

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llja
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1 Answers1

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The result is true.

If you want to take another route, then use a result of Cartier (see Lemme 5 on page 249) which says that over an integral domain the finitely generated flat modules are projective. Then check that a finitely generated projective module $P\ne0$ has the property that $mP\ne P$ for any maximal ideal $m$, so $P$ is faithfully flat. Suppose the contrary, localize at $m$, use Nakayama and find an $s\in R-m$ such that $sP=0$. But this is not possible since $P$ is a direct summand of a free module, and $R$ is an integral domain.

user26857
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    This is amazing, I am totally surprised. I haven't seen the equivalence flat <=> faithfully flat for finitely generated modules nowhere written down! I really believed this is wrong. – llja Oct 13 '15 at 10:55
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    @llja Geometrically it makes sense. Suppose that $A$ and $B$ are Noetherian domains (for convenience) and that $A\to B$ is both finite and flat. Then, $\mathrm{Spec}(B)\to\mathrm{Spec}(A)$ is both open and closed. Thus, it must surject which implies that $A\to B$ is faithfully flat. – Alex Youcis Oct 13 '15 at 11:21