Question on the equivalent definition of $R$-modules.

Question

Let $R$ be a commutative ring with $1_R$.

I found the following theorem as an equivalent definition of $R-$modules.

Theorem. An abelian group $(M,+)$ is an $R$-module iff there is a ring homomorphism $\varphi:R\longrightarrow \mathrm{End}_R(M)$.

The proof is not difficult but let me ask something.

Could we claim that the theorem is still valid, if $R$ is not commutative? So, left $R-modules are in general different from the right.

And if the answer is yes, if we have an abelian group $(M,+)$, and there exists a homomorphism $\varphi:R\longrightarrow \mathrm{End}_R(M)$, then we can construct both left and right $R-$modules.

Is this right? Please, explain your answers please.

Thank you.

Answer 1

An abelian group $M$ can be given the structure of $R$-module if and only if there exists a ring homomorphism $$ \varphi\colon R\to\operatorname{End}_{\mathbb{Z}}(M) $$ If $M$ is given a structure of $R$-module, then the homomorphism $\varphi$ is defined by $\phi(r)\colon x\mapsto rx$. Conversely, given the ring homomorphism $\varphi$, we can define $rx=\varphi(r)(x)$.

It is true that if $M$ has already been given the structure of $R$-module, then there exists a ring homomorphism $R\to\operatorname{End}_{R}(M)$ (for commutative $R$), but without a previous $R$-module structure, $\operatorname{End}_{R}(M)$ doesn't make sense, so the statement cannot be “if and only if”.

By the way, there could be different ring homomorphisms $R\to\operatorname{End}_{\mathbb{Z}}(M)$, giving rise to different structures of $R$-modules on the abelian group $M$.

If you write maps on the left, then the condition about ring homomorphisms works also without the assumption that $R$ is commutative and defines left $R$-modules.

A structure of right $R$-module is given by a ring homomorphism $\psi\colon R^{\mathrm{op}}\to\operatorname{End}_{\mathbb{Z}}(M)$, where $R^{\mathrm{op}}$ is the opposite ring (with the same addition and multiplication $(r,s)\mapsto sr$).

Answer 2

Let me give a try to answer and improve the theorem. Please feel free to edit my answer.

Theorem. Let $R$ be a ring with $1_R$ and $(M,+)$ an abelian group ($\iff \mathbb{Z}$-module). Then,

1. (i) For every $x\in R$, there is a map \begin{align} μ_x:M&\longrightarrow M, \\ m& \longmapsto μ_x(m):=x\cdot m \end{align} which is an endomorphism of abelian groups. That is, $μ_x \in \mathrm{End}_{\mathbb{Z}}(M) $.

(ii) The map \begin{align} μ:R&\longrightarrow \mathrm{End}_{\mathbb{Z}}(M),\\ r&\longmapsto μ(x):= μ_x : M \longrightarrow M, \ m \longmapsto μ_x(m):=x\cdot m\\ \end{align} is a ring homomorphism.

2. Let $ν:R\longrightarrow \mathrm{End}_{\mathbb{Z}}(M)$ a ring homomorphism. Then $(M,+)$ together with the scalar multiplication

\begin{align} \cdot :R \times M &\longrightarrow M, \\ (r,m)& \longmapsto r\cdot m:= ν(r)(m) \end{align} is an $R$-module.