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Given a ring $R$. How to prove that $\operatorname{End}_R(R)\cong R^{op}$, where $R^{op}$ 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!

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The subscript $R$ to End is there to remind you that one is not talking about (unitary) ring endomorphisms; if one was, what would that subscript mean? Instead these are endomorphisms as $R$-modules, the context should determine whether left or right modules are meant. –  Marc van Leeuwen Jun 27 '13 at 12:42
    
Please do not use titles consisting only of math expressions; these are discouraged for technical reasons -- see meta. –  Lord_Farin Jun 27 '13 at 19:25

4 Answers 4

up vote 5 down vote accepted

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

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

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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.

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I'll just make two observations that I missed in other answers.

First, if $\def\End{\operatorname{End}}\End(R)$ were to denote the set of unitary ring endomorphisms, then each of its elements has to map $1$ to $1$, but that does not force them to be the identity of$~R$. Much depends on the nature of$~R$. It can be or more or less large (Galois) group, but in general is is only a monoid. For $R=K[X]$ it is in bijection with $R$ itself (but the monoid structure is not that of the multiplicative monoid), it can be larger still.

Second, if $\End_R(R)$ denotes the endomorphisms as $R$ as a module over itself, we need to specify in the non-commutative case whether we consider the left or right module structure; the opposite in the isomorphism $\End_R(R)\cong R^{op}$ is not a fatality, but an artefact of the wish to have scalars act on the same side as module morphisms. But it is more natural to have then act from opposite sides, as morphisms must commute with scalars, and in the non-commutative (but associative) setting this is the case for multiplications from different sides. So, taking morphisms to apply from the left as usual, we can let $\End_R(R)$ denote the ring of right $R$-module endomorphisms of $R$, and then one has $\End_R(R)\cong R$ with $f\mapsto f(1)$ in one direction and $a\mapsto(x\mapsto ax)$ in the other direction.

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