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Let $R$ be a unit ring (not necessarily commutative). Then it is clear that for a right $R$-module $M$ we have:

$M$ is flat $R$-module $\Rightarrow$ for any left $R$-module $E$ with $E\otimes_{R}M=0$ and any submodule $E'\leq E$, we have $E'\otimes_{R}M=0$.

Can anyone give me a counterexample to show the inverse implication is not correct?

Thanks in advance!

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$M$ seems to be a left module and $E$ a right module. – egreg Jul 22 '13 at 23:39
up vote 2 down vote accepted

Let $M=R\oplus T$ where $T$ is not flat. Then $M\otimes E=E\oplus (T\otimes E)$ which is 0, if and only if $E=0$.

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$M$ is not a flat $R$-module, this is clear, but you have to verify for all possible modules $E$, the right hand side condition is satisfied for $M$, not a special $E$. – ougao Jul 22 '13 at 23:04
Impressing! Very clever idea! You can put it as an answer so that I could accept it. Thanks! – ougao Jul 22 '13 at 23:31

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