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Let $M$ be a flat $R$-module. Then the following are equivalent:

1) for every $R$-module $N$ we have $M\otimes_R N\neq0$

2) for every maximal ideal $m$ of $R$ we have $M\neq mM$

I did 1 implies 2. I'm having some problem in 2 implies 1: if $M\otimes_R N=0$ then also $M/mM\otimes_{R/m} N/mN=0$ and so (being vector spaces) or $M=mM$ and we have a contradiction or $N=mN$ but I don't know how to continue, could you help me please?

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Nitpick: In (1) you want to require $N \neq 0$. –  Dylan Moreland May 18 '12 at 3:30
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2 Answers

up vote 3 down vote accepted

Suppose there exists a module $N$ such that $M\otimes N=0$. Let $n\in N$ be non-zero and consider the submodule $N'=Rm\subseteq N$. Since $M$ is flat, the map $M\otimes N'\to M\otimes N$ is injective, and therefore $M\otimes N'=0$. Now $N'$ is isomorphic to $R/I$ where $I$ is the annihilator of $n$ in $R$. If $\newcommand\m{\mathfrak m}$ is a maximal ideal containing $I$, there is a surjection $N'\cong R/I\to R/\m$, and tensoring it with $M$ we get that the map $M\otimes N'\to M\otimes R/\m$ is surjective. Since its domain is zero, $M\otimes R/\m$ is zero.

But $M\otimes R/\m\cong M/\m M$, so this impossible.

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I think your idea for (2) $\Rightarrow$ (1) works after a couple of reductions. [Isn't commutative algebra always this way?] I'm not sure that what follows is essentially different from Mariano's excellent answer, but just in case:

To show that $N = 0$ it's enough to show that $N_\mathfrak m = 0$ for each maximal ideal $\mathfrak m$. So we can assume that $(A, \mathfrak m)$ is local. If $N' \subset N$ is a submodule then $M \otimes_A N' = 0$ by flatness, so assume that $N$ is finitely generated. Now when you run your argument, Nakayama is available.

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