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Given a ring $R$, how to prove sầu that $operatornameEnd_R(R)cong R^ oithatvietphat.vnrmop$, where $R^ oithatvietphat.vnrmop$ is the opposite ring of $R$.

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I read this proposition somewhere, but I think it is wrong. Because for any given $finoperatornameEnd_R(R) $ và any $r$ in $R$, we can get $f(r)$ by $r f(1)$, and $f(1)$can only be $1$, so $operatornameEnd_R(R)$ 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 $operatornameEnd_R(R)$. Thanks!


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It is not the case that $f(1)$ must be $1$. A bản đồ in $ oithatvietphat.vnrmEnd_R(R)$ is (a) linear & (b) commutes with multiplication (on the left) from $R$. These are not ring endomorphisms (which must satisfy the multiplicative sầu property $f(ab)=f(a)f(b)$, but vì 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 $ oithatvietphat.vnrmEnd_RR$ has $f(r)=f(rcdot1_R)=rf(1_R)$ & hence is right multiplication by some $ain R$. Furthermore, if $varphi_a:xmapsto lớn xa$ for every $ain R$, then

$$varphi_avarphi_b(x)=varphi_a(xb)=xba=varphi_ba(x)$$

& hence $aullet_small oithatvietphat.vnrmEnd_RRb=ba$ (after identifying the underlying sets of $ oithatvietphat.vnrmEnd_RR$ và $R$).

Thus $ oithatvietphat.vnrmEnd_RRcong R^, m op$.

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$End_R(R)$ denotes the bản đồ of $R$-module endomorphisms of $R.$ So the condition that maps 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 $R ightarrow R$. That is, those $f: R ightarrow R$ that are additive and satisfy $f(ra) = rf(a)$. (Such a homomorphisms need not take $1$ lớn $1$).

Indeed, ring endomorphisms $R ightarrow R$ 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 $defEndoperatornameEndEnd(R)$ were to lớn denote the set of unitary ring endomorphisms, then each of its elements has to maps $1$ khổng lồ $1$, but that does not force them lớn 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 $R=K$ it is in bijection with $R$ itself (but the monoid structure is not that of the multiplicative monoid), it can be larger still.

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Second, if $End_R(R)$ denotes the endomorphisms as $R$ as a module over itself, we need lớn specify in the non-commutative sầu 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 khổng lồ have scalars act on the same side as module morphisms. But it is more natural to have sầu then act from opposite sides, as morphisms must commute with scalars, and in the non-commutative sầu (but associative) setting this is the case for multiplications from different sides. So, taking morphisms lớn 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 $fmapskhổng lồ f(1)$ in one direction & $amapsto(xmapslớn ax)$ in the other direction.


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