Let $R$ be a ring with $1$. Let $M$ be a left $R$-module and $e$ an idempotent in $R$. Then
(1) ${\rm Hom}_R(R,M)\cong M$ (isomorphic abelian groups).
(2) ${\rm Hom}_R(Re,M)\cong eM$ (isomorphic abelian groups).
The proof of (1) is: a $\varphi\in{\rm Hom}_R(R,M)$ is uniquely determined from $\varphi(1)$; and for any $m\in M$, $\varphi(1)=m$ defines a member in ${\rm Hom}_R(R,M)$.
- Q.0 Is this (proof of (1)) correct?
For (2), I am not sure about (correctness of) my proof. But, I was thinking, it is similar to (1): any $\psi\in {\rm Hom}_R(Re,M)$ is uniquely determined by $\psi(e)$; because $\psi(re)=r\psi(e)$. But I didn't get,
- Q.1 why for arbitrary $m\in M$, the map $\psi(e)=m$ doesn't extend to a map in ${\rm Hom}_R(Re,M)$?
(I mean, we take $em$ to be arbitrary in $eM$, rather than $m$ arbitrary in $M$ so that, to get isomorphism in (2). Or more precisely the map $e\mapsto em$, for any $m\in M$, extends to a member in ${\rm Hom}_R(Re,M)$; but the map $e\mapsto m$ does not necessarily extend to member in ${\rm Hom}_R(Re,M)$; why? )
- Q.2 In (2), replace $e$ by arbitrary element. Does then isomorphism holds: $${\rm Hom}_R(Ra,M)\cong aM?$$
Your proof of (1) is correct, and generalises to give a proof of (2). Let $e$ be an idempotent, so $e^2=e$. We have a map $\mathrm{Hom}_R(Re,M)\to M$ sending $\phi\mapsto\phi(e)$. Note that the image lies in $eM$, since $\phi(e)=\phi(e^2)=e\phi(e)$. Moreover, this assignment is additive:,the zero map is sent to the zero element of $eM$, and $\phi+\psi$ is sent to $\phi(e)+\psi(e)$.
To check that it is an isomorphism, it is enough to show that it is bijective, or equivalently to construct the (set-theoretic) inverse. This must send $m\in eM$ to $\phi\colon re\mapsto rem=rm$, so it remains to prove that this $\phi$ is indeed an $R$-module homomorphism. This is then clear: we have $\phi(re+se)=rem+sem=\phi(re)+\phi(se)$ and $\phi(rse)=rsem=r\phi(se)$.
Thus we have an isomorphism of additive groups $\mathrm{Hom}_R(Re,M)\xrightarrow\sim eM$.
What about for general $a\in R$? We are actually using two ideas in the above proof. Firstly, we have a surjective $R$-module homomorphism $R\to Ra$, $r\mapsto ra$, and so composition yields a (necessarily injective) map $$\mathrm{Hom}_R(Ra,M)\rightarrowtail\mathrm{Hom}_R(R,M)\cong M, \quad \phi\mapsto\phi(a).$$ Conversely, we have the inclusion $Ra\leq R$, and composition yields the (in general neither injective nor surjective) map $$ M\cong \mathrm{Hom}_R(R,M) \to \mathrm{Hom}_R(Ra,M), \quad m \mapsto (\phi\colon ra\mapsto ram). $$ For an idempotent $a=e$, these maps actually compose to give the identity on $\mathrm{Hom}_R(Re,M)$, and we can compute the image as being $eM$. This comes from the fact that we have a direct sum decomposition $R=Re\oplus R(1-e)$. For general $a$ this all breaks down.