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Suppose $R$ is some commutative ring and $G$ a finite group so that $R[G]$ is the usual group ring. If $M$ is some $R[G]$-module, then we can inject $M\hookrightarrow M\oplus R[G]^n$ for some $n\geq 1$. Is this necessarily an $R[G]$-homomorphism?

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  • $\begingroup$ I think so...why wouldn't it be? You mean the map $m\mapsto (m, 0)$, right? $\endgroup$ – Nishant Aug 28 '15 at 20:40
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    $\begingroup$ Yes, the inclusion of a $R[G]$-submodule in a $R[G]$-module is a homomorphism. Here of course the factor $M$ is a "submodule" by way of being a direct factor as a matter of convention, but mapping $m\in M$ to $(m,0)$ is clearly an injective homomorphism. $\endgroup$ – hardmath Aug 28 '15 at 20:43
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Yes, the inclusion of a factor into a direct product $M \to M \oplus N$ defined by $m \mapsto (m, 0)$ is always an injective homomorphism no matter what modules you choose for $M$ and $N$. In particular, you can inject $M \to M \oplus R[G]^n$ not just for some $n$, but in fact for any $n$.

If $M$ and $N$ are $G$-modules then it would be a good exercise for you to write out the definition of the $G$-module action on $M \oplus N$ and the axioms that a homomorphism must satisfy, and then check that the inclusion map $M \to M \oplus N$ does indeed satisfy these axioms.

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