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It is well-know that the tensor product of two finitely generated modules over a given ring $R$ is finitely generated. Now my question is

Let $M$ and $N$ be $R$-modules where $R$ is a commutative ring. If $N$ is a FG module and we know that $M$ is not a FG module. Then what happens for the tensor product of $M$ and $N$ over $R$. Can we say that it is not FG or it may be FG. I am interested in the later case.

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The tensor product may be finitely generated, even if both factors are not: For example consider the $\mathbf Z$-modules $M := \def\Z{\mathbf Z}\bigl(\Z/(3)\bigr)^{(\mathbf R)} = \bigoplus_{x\in \mathbf R} \Z/(3)$ and $N := \bigl(\Z/(2)\bigr)^{(\mathbf R)}$. Then neither $M$ nor $N$ is finitely generated, but $M \otimes N = 0$ is.


I do not exactly know, what you mean by "non-trivial", but the example can be modified, to have non-zero product, let $$ M:= \bigl(\Z/(2)\bigr)^{(\Z)} \oplus \Z/(3), N := \Z(3) $$ then $M$ is not finitely generated, but $N$ is and $$ M \otimes N = \Z/(3) $$ is also.

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  • $\begingroup$ Can you give me a non-trivial example. I mean the tensor product is not zero. One of my factors is FG and another not. $\endgroup$ – user108209 Nov 5 '15 at 10:29
  • $\begingroup$ Modified my answer. $\endgroup$ – martini Nov 5 '15 at 11:35

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