Prove that any $R$-module $M$ is isomorphic to $\mathrm{hom}_R(R,M)$

This is an exercise of Advanced linear algebra third edition of Steven Roman:

Prove that any $$R$$-module $$M$$ is isomorphic to $$\mathrm{hom}_R(R,M)$$

My work so far:

We want to show that $$M\approx\mathrm{hom}_R(R,M)$$, where $$M$$ is a $$R$$-module. If the ring have an unity then it is easy to check that any homomorphism between $$R$$ and $$M$$ have the form $$\varphi_v(r):=r\cdot v$$ for any arbitrary $$v\in M$$. Hence there is a bijection $$v\mapsto \varphi_v$$ between $$M$$ and $$\mathrm{hom}_R(R,M)$$, that respect module operations, so the statement holds for rings with unity.

Now we can see that the maps $$\varphi_v$$ are homomorphisms also for non-commutative rings without unity, however I can't show that they are the unique kind of homomorphisms to conclude the exercise.

I need some help.

• Can you establish $Hom(M \times R, M) \approx Hom(M, M^R)$? – user359302 Apr 19 at 4:43
• That should be either $\text{Hom}_R (M \otimes R, M)$ or $\text{Bilinear}_R (M \times R, M)$. – Dean Young Apr 23 at 22:41

The statement that $$M\cong\operatorname{Hom}_R(R,M)$$ is generally false for rings without unit. Take $$R=\mathbb{Z}\oplus\mathbb{Z}$$ with the trivial multiplication. Any abelian group with action $$rx=0$$ is a module over $$R$$. Take $$M=\mathbb{Z}$$. Then $$\operatorname{Hom}_R(R,M)\cong M\oplus M$$ which is not isomorphic to $$M$$.

Notice also that, in this case, the map $$\varphi_v$$ is the zero map, so the map $$M\to\operatorname{Hom}_R(R,M)$$ is definitely not injective.

• I dont follow exactly what you mean. The ring $\Bbb Z^2$ with pointwise addition and multiplication have the identity $(1,1)$. Also the ring $\Bbb Z$ have the identity $1$ – Masacroso Apr 23 at 22:52
• @Masacroso I consider the trivial multiplication $(a,b)(c,d)=(0,0)$, as specified. – egreg Apr 23 at 22:59
• ok, I think I understand. So $R$ is some arbitrary ring with trivial multiplication, right? However I dont understand how you knows that $\operatorname{Hom}_R(R,M)\cong M\oplus M$. – Masacroso Apr 23 at 23:10
• @Masacroso No, $R$ is not “arbitrary”, I chose $R=\mathbb{Z}\oplus\mathbb{Z}$. A module homomorphism $\mathbb{Z}\oplus\mathbb{Z}\to\mathbb{Z}$ (in the given module action) is simply an abelian group homomorphism. – egreg Apr 23 at 23:19
• sorry, I didnt read carefully yesterday. Thank you for your time and explanations – Masacroso Apr 24 at 8:51

Your approach seems to be showing injectivity and surjectivity of a certain map. An alternative is to construct homomorphisms both ways.

Note: we make $$\text{Hom}_R(R, M)$$ into a right $$R$$-module, just like $$M$$ is, by declaring $$f \cdot a : R \rightarrow M$$ to sent $$b$$ to $$f(b)a$$.

Consider the map $$\phi : \text{Hom}_R(R, M) \rightarrow M$$ sending $$f$$ to $$f(1)$$. $$\phi$$ is a homomorphism. Indeed, $$\phi(f \cdot a) = f(1)a = f(a)$$, and $$\phi(f + g) = (f+g)(1) = f(1) + g(1)$$.

Consider also the map $$\psi : M \rightarrow \text{Hom}_R(R, M)$$ sending $$m$$ to the map $$\psi(m) : R \rightarrow M$$ sending $$r$$ to $$rm$$. $$\psi$$ is a homomorphism.

These maps are inverse. Indeed, for $$f : R \rightarrow M$$, $$\psi ( \phi (f))(r) = \psi (f(1))(r) = rf(1) = f(r)$$. And, for $$m \in M$$, $$\phi ( \psi (m)) = \psi(m)(1) = 1 \cdot m = m$$.