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Let $R$ be a commutative ring with 1 and let $M$ be a left $R$-module. Show that $Hom_R(M,M)$, the set of $R$-homomorphisms from $M$ to itself is a left $R$-module if we set $(r·f)(m) = rf(m)$ for all $r ∈ R,m ∈ M,f ∈ Hom(M,M)$.

Here is my proof of this, I'd appreciate it if someone looked over the details, I have some questions scattered about here and there, which I marked with letters for anyone who wants to help me out.

So first things first I need to show that $Hom_R(M,M)$ is an abelian group; to this extent, let $f,g in Hom_R(M,M)$, $m,n in M$.

Then $(f+g)(m)=f(m)+g(m)=g(m)+f(m)=(g+f)(m)$.

My first question $(a)$: Did what I wrote above prove that commutativey condition require dfor $Hom_R(M,M)$ to be an abelian group? Or rather do we need to get another element of $M$ involved to show that $f,g$ are morphisms of abelian groups? Something like:

$(f+g)(m+n)=f(m+n)+g(m+n)=f(m)+f(n)+g(m)+g(n)=f(m)+g(m)+f(n)+g(n)=(f+g)(m)+(f+g)(n)$

This is question $(a)$, I wouldn't mind if one of you genius's shed some light in general on what's going on this for me. Thanks!

Next, Let $0 in Hom_R(M,M)$ s.t. $0(m)=0_m$ $forall m in M$, it is trivial to show that this is the identity element for $Hom_R(M,M)$ as an abelian group.

Next, if $f in Hom_R(M,M)$, define $-f$ by $(-f)(m)=-f(m)$ $forall m in M$. It is trivial that $-f$ is the inverse of $f$ for the binary operation of $Hom_R(M,M)$

Now, let $r in R$. Then we have:

$(f+g)(rm) = f(rm) + g(rm) = rf(m) + rg(m)=r(f(m)+g(m))=r(f+g)(m)$ and therefore $(f+g)$ is $R$-linear, and thus $(f+g) in Hom_R(M,M)$ .

Question (b): By this point, have I proven that $Hom_R(M,M)$ is an abelian group? Where there any steps that were unnecessary or any that could have been more clear?

Now is the part that is a little tricky for me, showing that $Hom_R(M,M)$ is a left $R$ module using the natural action of $r*f(m)=rf(m)$. I get a little confused on this, because a part of me thinks it should be true automatically since $f$ is $R$-linear, but the truth is you also need the commutativity condition on $R$.

So, I must show that $g=rf in Hom_R(M,M)$. To this extent, let elements be defined as above:

$g(m+n) = rf(m+n) = rf(m)+rf(n) = g(m)+g(n)$. Thus $g$ a homomorphism.

But is $g$ $R$-linear? Let $s in R$.

$g(sm)=rf(sm)=r(sf(m))=(rs)(f(m))$

but also, $sg(m)=s(rf(m))=(sr)(f(m))$

so $g=rf$ is $R$-linear $iff$ $R$ is commutative.

Again, i'd appreciate it if somebody when through this for me and told me where I could be more explicit, efficient, or if I forgot something completely. I just realized I didn't show associativity, but i'd probably put that right along with the identity function and inverse functions as trivial.

Thanks in advance!