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If $R$ is a local ring and $M$ and $N$ are finitely generated R-modules such that $M\otimes N = 0$ then it follows from Nakayama's lemma that either $M=0$ or $N=0$. This I know. But now I am looking for an example of an local ring $R$ and 2 not finitely generated modules such that $M\otimes N = 0$ but neither $M=0$ nor $N=0$.


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1 Answer 1

Let $R$ be a domain which is not a field. If $K$ is the field of fractions of $R$. Then $K,K/R$ are non-zero $R$-modules satisfying $K\otimes_R(K/R)=0$. In particular, you could take $R$ local and have an example of the type sought.

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@Kennan Kidwell: is $\mathbb{Q} \mathop{\otimes_{\mathbb{Z}} (\mathbb{Q}/\mathbb{Z})}$ really $0$? – Rob Arthan Dec 8 '12 at 1:58
Take $r\in\mathbf{Q}$ and $s\in\mathbf{Q}/\mathbf{Z}$. Since $\mathbf{Q}/\mathbf{Z}$ is torsion, there is $n\geq 1$ such that $ns=0$. Since $\mathbf{Q}$ is divisible, we can write $r=nr^\prime$ for some $r^\prime\in\mathbf{Q}$. Now $r\otimes s=nr^\prime\otimes s=r^\prime\otimes ns=0$. This idea works generally for a domain $R$ and its quotient field $K$. A divisible $R$-module tensored with a torsion $R$-module is zero. – Keenan Kidwell Dec 8 '12 at 2:46
@Kennan Kidwell: of course! Thanks. – Rob Arthan Dec 8 '12 at 2:52

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