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I am trying to solve the following problem:

Let $M$ and $N$ be two left $A$-modules. Prove that $Hom_A(M,N)$ has a left $Z(A)$-module structure with: $(a.f)(m)=a.f(m)$. Show $Hom_A(A,N) \cong N$ as $Z(A)$-modules.

I am a bit confused by this exercise since I could show a more general result:$Hom_A(M,N)$ is a left $A-$module rather than $Z(A)$-module (where I suppose that $Z(A)$ is the center of the ring $A$), here is the proof:

i.$(a.(f+g))(m)=a.((f+g)(m))=a.(f(m)+g(m))=a.f(m)+a.g(m)=(a.f)(m)+(b.f)(m)$ (the last equality holds since $N$ is a left $A$-module, I will use this fact to show the three remaining properties).

ii.$((a+b).f)(m)=(a+b).f(m)=a.f(m)+b.f(m)=(a.f)(m)+(b.f)(m)$

iii. $((ab).f)(m)=(ab).f(m)=a.(b.f(m))=a.((b.f)(m))=(a.(b.f))(m)$

iv. $(1_A .f)(m)=1_A. f(m)=f(m)$

So, to prove the properties I've only used the fact that $N$ is a left $A-$module, I don't understand the role of $Z(A)$ in here.

As for the last part, I don't understand what it means for two abelian groups to be isomorphic as $A-$modules; suppose I find an isomorphism between $Hom_A(A,N)$ and $Z(A)$, then it follows these two groups are isomorphic, but is this enough to affirm the two groups are isomorphic as $Z(A)-$modules.

I would appreciate if someone could clear these things up for me and suggest me how could I prove the statements of the problem.

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2 Answers 2

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If a ring $A$ is not commutative, there's no natural way to endow $\operatorname{Hom}_A(M,N)$ with the structure of $R$-module, where $M$ and $N$ are left (or right, if you prefer) $A$-modules.

You're trying to define, for $f\in\operatorname{Hom}_A(M,N)$ and $a\in A$, $$ af\colon M\to N,\quad af\colon x\mapsto af(x)=f(ax) $$ However, this is not a homomorphism of $A$-modules. Set, for ease of calculation, $g=af$. The map $g$ is easily seen to be a homomorphism of abelian groups (under $+$); so, let $b\in A$; we wish to prove that $$ g(bx)=bg(x) $$ for all $x\in M$. But this amounts to saying that $$ af(bx)=b(af(x)) $$ or, since $f$ is a homomorphism of $A$-modules, $$ abf(x)=baf(x) $$ that is $$ (ab-ba)f(x)=0. $$ You see that there's no reason for this to hold, because in general $ab-ba\ne0$ in a noncommutative ring.

However, this shows that if you suppose $a\in Z(A)$ (the center of $A$), the equality $ab=ba$ holds for all $b\in A$, by definition. Thus $\operatorname{Hom}_A(M,N)$ can be given a structure of module over $Z(A)$.

For the last part, consider the obvious map $$ v\colon \operatorname{Hom}_A(A,N)\to N $$ defined by $v(f)=f(1)$ and prove it is an isomorphism of $Z(A)$-modules.

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I think the issue occurs earlier -- have we even given a map $A \times \operatorname{Hom}_A(M, N) \to \operatorname{Hom}_A(M, N)$ that might define an $A$-module structure on $\operatorname{Hom}_A(M, N)$? You're taking $(a, f)$ to the map of sets $g\colon M \to N$ defined by $g(x) = a(f(x))$. Is this $g$ an $A$-linear map?

For the second question it's again a matter of linearity: you will produce some group isomorphism $\phi\colon \operatorname{Hom}_A(A, N) \to N$ and the final check will be verifying $\phi(cf) = c\phi(f)$ for all $c \in Z(A)$ and $f \in \operatorname{Hom}_A(A, N)$.

By the way, it's instructive to think of an example of a ring $A$ and two $A$-modules which are isomorphic as groups but not modules. $A=k[t]$ might be a good place to start.

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