ring endomorphism

Question

Can anyone offer some help?

Let $R$ be an unital ring $(1\ne 0)$. Recall than with $R^{\,\text{op}}$ we denote the opposite ring which is a ring as usual, only it haw a reverse multiplication $\left( x\square y=yx \right)$.

We denote $R$ as a left $R$-module with ${}_{R}R$. (equivalently for right R-module)

Prove that

$\operatorname{End}_{R}({}_{R}R)\cong R^{\,\text{op}}$,

$\operatorname{End}_{R}({}_{R}R)=R$.

Hint: if $f$ is an endomorphism of ${}_{R}R$, then $f(r)=f(1\cdot r)=f(1)r$

Thank you for your time

Answer 1

Hint:

The morphism $\;\operatorname{End}_R({}_RR)\longrightarrow R^{\,\text{op}}$ maps an endomorphism $f$ to $\varphi(f)=f(1)$.

You have to check that

• $\varphi(f+g)= \varphi(f)+\varphi(g)$ for all endomorphisms $f,g$;
• $\varphi(f\circ g)=\varphi(f)\varphi(g)$ in $R^{\,\text{opp}}$;
• $\varphi(\mathrm{id})=1_{R^{\mkern1.5mu\text{opp}}}$.