If $R$ is commutative then $\operatorname{Hom}_R(R,M)$ is isomorphic with $R$-module $M$. Is the converse also true?

Question

I'm reading Elements of Abstract and Linear Algebra by E.H. Connell and I'm wondering if the converse of the following theorem is true (p.71):

Theorem Suppose $M=M_R$ and $f,g: R\to M$ are homomorphism with $f(\underline{1})=g(\underline{1})$. Then $f=g$. Further, if $m\in M$, $\exists!$ homomorphism $h:R\to M$ with $h(\underline{1})=m$. In other words, $\operatorname{Hom}_R(R,M)\approx M$.

Proof Suppose $f(\underline{1}) = g(\underline{1})$. Then $f(r)=f(\underline{1}\cdot r) = f(\underline{1})r=g(\underline{1})r= g(\underline{1}\cdot r) = g(r)$. Given $m\in M$, $h: R \to M$ defined by $h(r)=mr$ is a homomorphism. Thus evaluation at $\underline{1}$ gives a bijection from $\operatorname{Hom}_R(R,M)$ to $M$, and this bijection is clearly a group isomorphism. If $R$ is commutative, it is an isomorphism of $R$-modules.

Based on the comment in the proof of this theorem I believe the converse is only true for commutative rings.

Proposition Suppose $M$ is an abelian group and $R$ is a commutative ring, and $\operatorname{Hom}_{\mathbb{Z}}(R,M)$ the set of all group homomorphisms. If the evaluation $\Phi:\operatorname{Hom}_{\mathbb{Z}}(R,M), f\mapsto f(\underline{1})$ is a group isomorphism then $M$ can be made into a $R$-module.

Proof. As $\operatorname{Hom}_{\mathbb{Z}}(R,M)$ is an abelian group under pointwise addition we define a scalar multiplication to make it an $R$-module as $ f\cdot r: R\to M, s\mapsto f(rs) $.

1. $f\cdot r$ is a group homomorphism: $(f\cdot r)(s + t) = f(r(s+t))=f(rs+rt)=f(rs)+f(rt)=(f\cdot r)(s)+(f\cdot r)(t)$
2. Distributivity with respect to group addition in $M$ $((f_1+f_2)\cdot r)(s) =(f_1+f_2)(rs) = f_1(rs)+f_2(rs)= (f_1 \cdot r)(s) + (f_2 \cdot r)(s) = (f_1 \cdot r + f_2 \cdot r)(s) $
3. Distributivity with respect to ring addition $(f\cdot(r_1+r_2)(s)=f((r_1+r_2)s)=f(r_1s+r_2s)=f(r_1s +r_2s)=f(r_1s)+f(r_2s)=(f\cdot r_1)(s)+(f\cdot r_2)(s)=(f\cdot r_1+f\cdot r_2)(s)$
4. Compatibility of scalar multiplication $(f\cdot(r_1 r_2))(s)=f(r_1 r_2 s)=f(r_1(r_2 s))=(f\cdot r_1)(r_2 s)=((f\cdot r_1) \cdot r_2)(s)$
5. Identity of scalar multiplication $(f\cdot \underline{1})(s)=f(\underline{1}s)=f(s)$

We define scalar multiplication in $M$ via bijection $\Phi$ as $m\cdot r = (\Phi^{-1}(m) \cdot r)(\underline{1})$ that turns a group homomorphism $f\in \operatorname{Hom}_{\mathbb{Z}}(R, M)$ to a $R$-module homomorphism. One needs to show $f(ar)=f(a)\cdot r$. From the bijectivity of $\Phi$ there exisits a unique $g$ such that $g(\underline{1})=f(a)$ on the other hand $(f\cdot a)(\underline{1})=f(a\underline{1})=f(a)$ and $g=f\cdot a)$. Therefore $f(a)\cdot r = ((f\cdot a)\cdot r)(\underline{1})= (f \cdot (ar))(\underline{1})=f(ar\underline{1})=f(ar)$. $\blacksquare$

Question, I don't see commutativity of $R$ being used in my proof. Is something wrong with the proposed proposition?

Edit 1 I've improved the notation to distinguish between the set of all group homomorphisms and the set of all $R$-module homomorphisms.

Edit 2 I'm exploring the definition of a $R$ module. A scalar multiplication defines a ring homomorphism from the $R$ into the endomorphism ring of the group $M$.$$\Psi: R \to \operatorname{End}(M), r \mapsto \Psi(r):M\to M, m \to m\cdot r.$$

So defining a scalar multiplication amounts to defining an ring endomorphism.

On the other hand a scalar multiplication defines a group homomorphism: $$\tilde{\Phi}: M \to \operatorname{Hom}_{\mathbb{Z}}(R,M), m\mapsto \tilde{\Phi}(m):R\to M, r \mapsto m\cdot r$$

I'm trying to characterise a scalar multiplication in terms of this type of mapping.

Edit 3 (2022-10-16) After having read replies to Why is $\operatorname{Hom}(M,N)$ not necessarily an $R$ module? the comment in the original proof that requires $R$ to be commutative is too strong. In general the following is true:

Theorem Suppose $R$ is a ring. Suppose $M$ is a $R$-bimodule and $N$ is a right $R$-module. Then the set of all right $R$-linear maps $\operatorname{Hom}_R(M,N)$ is a right $R$-module with scalar multiplication given by $f\cdot r : M \to N, m\mapsto f(rm)$.

Since every ring $R$ is a $(R,R)$-bimodule the left action of $R$ on $R$ induces a right action of $R$ on $\operatorname{Hom}_R(R,M)$ via $ f\cdot r: R\to M, s\mapsto f(rs) $. To check that this defines right $R$-module on $\operatorname{Hom}_R(R,M)$ one needs to check that $f\cdot r$ is a right $R$-linear:

$$ (f\cdot r)(s a) = f(r s a) = f(r s) a = (f \cdot r)(s) a.$$

The calculations analogical to 1-5 above check remaining axioms.

The evaluation at $\underline{1}$, $\Phi:\operatorname{Hom}_{R}(R,M), f\mapsto f(\underline{1})$, is right $R$-linear as well

$$ \Phi(f\cdot r) = (f \cdot r) (\underline{1}) = f(r\underline{1}) = f(r) = f(\underline{1} r) = f(\underline{1})r = \Phi(f)r.$$

Therefore it is an isomorphism of $R$-modules in general case.

Theorem Suppose $R$ is a ring and $M=M_R$ is a right $R$-module. Then $\operatorname{Hom}_R(R,M)$ is a right $R$-module isomorphic to $M$.

This can be easily extended to following

Theorem Suppose $R$ is a ring and $M=M_R$ is a right $R$-module. Then $\operatorname{Hom}_R(R^n,M)$ is a right $R$-module isomorphic to $M^n$.

Now there is no surprise then that the proof of my proposition does not use commutativity of $R$. I agree that the proposition itself is not particularly interesting.

Edit 4 (2022-12-23) I've changed $\operatorname{Hom}_{\mathbf{Grp}}(R,M)$ to a more standard notation $\operatorname{Hom}_\mathbb{Z}(R,M)$.

Answer 1

I think your statement is a little weird. A much easier statement is this: if $R$ is any ring and $M$ an abelian group, then $\operatorname{Hom}_{\mathbb{Z}}(R,M)$ is a left $R$-module through $(r\cdot f)(x)=f(xr)$, and also a right $R$-module with $(f\cdot r)(x)=f(rx)$. When $R$ is commutative these structures are equal.

Then of course, if $\operatorname{Hom}_{\mathbb{Z}}(R,M)$ happens to be isomorphic to $M$ (that seems rare, but why not), you can transfer these module structures to $M$.