If $M$ is an $R$ module, then why is the set $rM = \{rm : m\in M\}$ not an $R$ module if $R$ is not commutative?

Question

I am able to prove that $rM$ is a module over $R$ if $R$ is commutative. Which property of modules fails to hold if $R$ is not commutative? Or is there a problem in defining the action of $R$ on $rM$? Here $R$ is a ring with unity.

Answer 1

If $M$ is a two-sided $R$-module, then $rM$ is a right $R$-module. But of course, you probably mean that $M$ is only a left $R$-module. Then, how do you know that for $m\in M$ and $x\in R$, $x\cdot rm=rn$ for some $n\in M$? However, if $r$ is a central element of $R$, then yes, $rM$ is a left $R$-module. (In particular, if $R$ is commutative, every element of $R$ is central, and this gives the result you already know.)

Note that I assume that you want to have the action of $R$ on $rM$ to be induced by the action of $R$ on $M$. If you want an arbitrary action, then there probably is, depending on $R$ and $M$ (the trivial $R$-action $x\cdot rm=0$ for all $x\in R$ and $m\in M$ is one possible choice, but you probably don't want that).

Consider $R$ to be the ring of $2\times 2$ real matrices. Let $M$ be the left ideal (which is a left module) consisting matrices of the form $$\begin{pmatrix}*&0\\*&0\end{pmatrix}.$$ (The asterisks need not be equal.)

Take $r=\begin{pmatrix}1&0\\0&0\end{pmatrix}$. Then, $rM$ consists of all matrices of the form $$\begin{pmatrix}*&0\\0&0\end{pmatrix}.$$ It is an easy exercise to show that this is not a left ideal (or a left module) with the action of $R$ inherited from $M$.

If you want to put a different $R$-action on $rM$, then note that every unitary left $R$-module is a direct sum of copies of the standard $R$-module $V=\Bbb{R}^2$ (where $R$ acts on $V$ by matrix multiplication). This means: for a vector space $U$ of finite dim $k$, there exists a unitary $R$-module structure on $U$ if and only if $k$ is even. Our set $rM$ is a $1$-dim vector space, so it cannot have a unitary left $R$-action. This means $rM$ can only have the trivial $R$-action.

Answer 2

It fails in general stability for scalar multiplication: you should have, for any $\lambda\in R$ and any $m\in M$, $$\lambda(rm)=r m'$$ for some $m'\in M$. This is true if the ring is commutative $(m'=\lambda m)$ or if $M$ is a a divisible $R$-module.