For a ring $R$, does $\operatorname{End}_R(R)\cong R^{\mathrm{op}}$?

Given a ring $R$, how to prove that $\operatorname{End}_R(R)\cong R^{\mathrm{op}}$, where $R^{\mathrm{op}}$ is the opposite ring of $R$.

I read this proposition somewhere, but I think it is wrong. Because for any given $f\in\operatorname{End}_R(R) $ and any $r$ in $R$, we can get $f(r)$ by $r f(1)$, and $f(1)$can only be $1$, so $\operatorname{End}_R(R)$ is isomorphic to the trivil group.

Is there anything wrong with my logic? Please help me identify where I am wrong in my understanding of $\operatorname{End}_R(R)$. Thanks!

Solution 1:

It is not the case that $f(1)$ must be $1$. A map in $\mathrm{End}_R(R)$ is (a) linear and (b) commutes with multiplication (on the left) from $R$. These are not ring endomorphisms (which must satisfy the multiplicative property $f(ab)=f(a)f(b)$, but do not enjoy full $R$-linearity, only linearity over the prime subgring generated by $1_R$), but rather left $R$-module endomorphisms.

Thus, every element of $\mathrm{End}_RR$ has $f(r)=f(r\cdot1_R)=rf(1_R)$ and hence is right multiplication by some $a\in R$. Furthermore, if $\varphi_a:x\mapsto xa$ for every $a\in R$, then

$$\varphi_a\varphi_b(x)=\varphi_a(xb)=xba=\varphi_{ba}(x)$$

and hence $a\bullet_{\small\mathrm{End}_RR}b=ba$ (after identifying the underlying sets of $\mathrm{End}_RR$ and $R$).

Thus $\mathrm{End}_RR\cong R^{\,\rm op}$.

Solution 2:

$End_R(R)$ denotes the map of $R$-module endomorphisms of $R.$ So the condition that map takes $1$ to $1$ is not necessary.

Solution 3:

When one writes $End _{R} (R)$ what is meant most of the time is the ring of $R$-module homomorphisms $R \rightarrow R$. That is, those $f: R \rightarrow R$ that are additive and satisfy $f(ra) = rf(a)$. (Such a homomorphisms need not take $1$ to $1$).

Indeed, ring endomorphisms $R \rightarrow R$ do not form a ring in any natural way, since a sum of two ring endomorphisms need not be a rign endomorphism again.