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Let $R$ be a ring and $M$ be a $R$-module. In Example 5.10 of Algebra: Chapter 0 that for $r\in R$, \begin{equation} rM=\{r\cdot m:m\in M\} \end{equation} is a submodule of $M$.

I have no problem showing that this is a subgroup of $M$, but why is it closed under the actions of $R$? That is, given $m\in M$ and $s\in R$, how can we find $n\in M$ such that \begin{equation} s\cdot (r\cdot m)=r\cdot n. \end{equation}


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$A$ is not commutative? – Manos Dec 25 '12 at 16:38
up vote 2 down vote accepted

Here is an explicit counterexample when $R$ is not commutative:

Let $R=k\langle x,y\rangle$, the free associative $k$-algebra on the indeterminates $x$ and $y$. We can of course consider the left $R$-module $M=R$. However, the subset $xM$ is not a left submodule of $M$, because $yx$ cannot be written as $xr$ for any $r\in R$.

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Thanks! So this is a typo in the book. I will send the author an email. – Hui Yu Dec 26 '12 at 5:22

Usually while dealing with modules Ring $R$ will be commutative with unity , if that is the case, $ s.(r.m) = r.(s.m)$ and as $M$ is module $ s.m \in M,$ so $s.m = n$ for some $n\in M$ If the ring is not commutative, usually it need not be a sub module.

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