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Suppose $M$ is an $R$-module and $N$ is an $S$-module. Then $M\otimes_{\mathbb{Z}} N$ is an $R\otimes_{\mathbb{Z}} S$-module given by $r\otimes s\cdot m\otimes n:= rm\otimes sn$. My question is whether the following claim is valid or not:

$(M\otimes_{\mathbb{Z}} N)\otimes_{R\otimes_\mathbb{Z}S} \mathbb{Z}\cong (M\otimes_{R} \mathbb{Z})\otimes_\mathbb{Z}(\mathbb{Z}\otimes_S N)$ as $R\otimes_\mathbb{Z}S$-modules.

EDIT: $R=\mathbb{Z} G$ and $S=\mathbb{Z}H$ for groups $G$ and $H$, and I am viewing $\mathbb{Z}$ as both an $R$ and an $S$ module by actions induced by $Gx=x$ and $Hx=x$ for all $x\in \mathbb{Z}$.

If so, would one cook up the maps in both directions and show they are inverse to one another, or are there isomorphism theorems that we can take advantage of for tensors over tensors?

Thanks.

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  • $\begingroup$ What do you mean by a trivial action? If the action annihilates everything then the result is true because the tensor products are all trivial. $\endgroup$ – Matt Samuel Mar 6 '15 at 5:23
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    $\begingroup$ Does Proposition 11.13 (or, if the numbering changes, just search for "straightforward (repeated) application of the universal property") in web.mit.edu/~darij/www/algebra/HopfComb-sols.pdf help you? Arguably I leave the proof to the reader, but I think the more symmetric form of it (one of your $\mathbb Z$'s is secretly a $\mathbb Z \otimes_{\mathbb Z} \mathbb Z$) should clear up some things. $\endgroup$ – darij grinberg Mar 6 '15 at 5:38
  • $\begingroup$ @MattSamuel By trivial action I meant $r\cdot x=x$ for all $r\in R$ and $z\in \mathbb{Z}$. I've edited this for clarity. $\endgroup$ – mwmjp Mar 6 '15 at 16:03
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    $\begingroup$ That doesn't define a module structure. Try it with $\mathbb Z$. $1\cdot 1+1\cdot 1=2\neq (1+1)\cdot 1=1$. $\endgroup$ – Matt Samuel Mar 6 '15 at 16:07
  • $\begingroup$ @MattSamuel I see, I guess what I mean is if $R=\mathbb{Z} G$, the group ring, and $G$ acts trivially on $\mathbb{Z}$. Does that remedy the situation? $\endgroup$ – mwmjp Mar 6 '15 at 16:13

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