# Does this stand for $\text{Hom}_R(R/I, R/I)$ ?

Let $R$ be a commutative ring with unit and $M$ be a $R$-module.

We have that $\text{Hom}_R(R,M)\cong M$ as $R$-modules and $\text{Hom}_R(R,R)\cong R$ as rings.

Do we have then also that $\text{Hom}_R(R/I, R/I)\cong R/I$, where $I$ is an ideal of $R$?

• We can see $R/I$ as a ring and also as a $R$-module. So what do you mean $Hom_R(R/I,R/I) \cong R/I$ here? – Xiaosong Peng Dec 2 '16 at 2:13

Consider the map $\Phi: Hom_R(R/I,R/I)\rightarrow R/I$ defined by $\Phi(f)= f([1])$ where $[1]$ is the class of $1$ in $R/I$. $\Phi(f)=\Phi(g)$ implies that $f([1])=g([1])$. This implies that for every $r\in R$, $rf([1])=f([r])=rg([1])=g([r])$. Thus $\Phi$ is injective.
For every $[u]$ in $R/I$, define $f:R/I\rightarrow R/I$ by $f([r])=[ru]$, $\Phi(f)=[u]$, thus $\Phi$ is surjective.
• Do we have to show that $\Phi$ is also an homomorphism? – Mary Star May 15 '16 at 23:22
• We have that $\Phi (f+g)=(f+g)([1])=f([1])+g([1])=\Phi (f)+\Phi (g)$ and $\Phi (rf)=rf([1])=r\Phi (f)$, right? – Mary Star May 15 '16 at 23:25
• Yes, this is how you show that $\Phi$ is an homomorphism. – Tsemo Aristide May 15 '16 at 23:27
• So, we have that $\Phi$ is an homomorphisn and bijective, that means that $\Phi$ is an isomorphism, right? So, we have show that $\text{Hom}_R(R/I, R/I)\cong R/I$, right? – Mary Star May 15 '16 at 23:30