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If $R$ is a ring with unit and $M$ a left R-module, in which way becomes $R\underset{R}{\otimes}M$ a left R-module? Thanks!

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    Apologies for the other negative solution. I can't believe what I wrote in the first pass, or that it got votes. – 2017-01-06

2 Answers 2

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Whenever you have an $S,R$ bimodule $N$, the abelian group $N\otimes_RM$ has a natural left $S$ module structure given by

$s(n\otimes m):=sn\otimes m$.

The ring is an $R,R$ bimodule over itself, so this is possible.

And the other obvious thing holds, that if $M$ is additionally an $R,T$ bimodule, then $N\otimes_R M$ has an $S, T$ bimodule structure given in the obvious way.

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    Is the first statement also true if $N$ is only a left $S$ module and a right $R$ module but not a $S$,$R$ bimodule? – 2017-01-06
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    @aii I can't verify why in my head, but I don't think that is enough. In all books I have seen anyway, they stick to bimodules – 2017-01-06
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We have a theorem: if $R,S$ rings,

$1)$ if $_SM_R $ a left $S$- module , a right $R$- module and $_RN$ a left $R$- module then $M \otimes_R N$ a left $S$- module.

$2)$ if $M_R $ a right $R$- module and $_RN_S$ a left $R$- module , right $S$- module then $M \otimes_R N$ a right $S$- module.

in your question: $R\otimes_R M $ a left $R$- module if $M$ a left $R$- module. and in a partical we have $R\otimes_R M \cong M$