Krdytkyu

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 $varphi:Rlongrightarrow mathrm{End}_R(M)$.

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 $varphi:Rlongrightarrow mathrm{End}_R(M)$, then we can construct both left and right $R-$modules.

Is this right? Please, explain your answers please.

Thank you.

An abelian group $M$ can be given the structure of $R$-module if and only if there exists a ring homomorphism
$$
varphicolon Rtooperatorname{End}_{mathbb{Z}}(M)
$$
If $M$ is given a structure of $R$-module, then the homomorphism $varphi$ is defined by $phi(r)colon xmapsto rx$. 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 $Rtooperatorname{End}_{R}(M)$ (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 $Rtooperatorname{End}_{mathbb{Z}}(M)$, 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 $psicolon R^{mathrm{op}}tooperatorname{End}_{mathbb{Z}}(M)$, where $R^{mathrm{op}}$ is the opposite ring (with the same addition and multiplication $(r,s)mapsto sr$).

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 $xin R$, there is a map
   begin{align}
   μ_x:M&longrightarrow M, \
   m& longmapsto μ_x(m):=xcdot 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):=xcdot m\
end{align}
is a ring homomorphism.

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

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