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.

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

  • $\begingroup$ 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 $1$ $\endgroup$ – Masacroso Apr 23 at 22:52
  • $\begingroup$ @Masacroso I consider the trivial multiplication $(a,b)(c,d)=(0,0)$, as specified. $\endgroup$ – egreg Apr 23 at 22:59
  • $\begingroup$ ok, I think I understand. So $R$ is some arbitrary ring with trivial multiplication, right? However I dont understand how you knows that $\operatorname{Hom}_R(R,M)\cong M\oplus M$. $\endgroup$ – Masacroso Apr 23 at 23:10
  • $\begingroup$ @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. $\endgroup$ – egreg Apr 23 at 23:19
  • $\begingroup$ sorry, I didnt read carefully yesterday. Thank you for your time and explanations $\endgroup$ – 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 $m \in M$, $\phi ( \psi (m)) = \psi(m)(1) = 1 \cdot m = m$.


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