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A $(R,S)$-bimodule $M$ is the same as either a ring map $S^{op}\to \mathrm{hom}_R(M,M)$ or a ring map $R\to \mathrm{hom}_{S^{op}}(M,M)$ (i.e. $M$ can be viewed as $(S^{op},R^{op})$-bimodule).  Since any ring $R$ is an $(R,R)$-bimodule, there are ring maps $R^{op}\to \mathrm{hom}_R(R,R)$ and $R\to \mathrm{hom}_{R^{op}}(R,R)$; it turns out that these maps are isomorphisms, hence $\mathrm{hom}_R(R,R)^{op}=\mathrm{hom}_{R^{op}}(R,R)$.

Now, given a $(R,R)$-bimodule $M$, is it true that the maps $R^{op}\to \mathrm{hom}_R(M,M)$ and $R\to \mathrm{hom}_{R^{op}}(M,M)$ are dually equivalent, and in particular $\mathrm{hom}_R(M,M)^{op}=\mathrm{hom}_{R^{op}}(M,M)$? Thanks for your help

John
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1 Answers1

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Let $R= \Bbb Q[x]$. Then the datum of a $(R,R)$-bimodule is equivalent to a vector space $V$ equipped with a pair of commuting endomorphisms $(f,g)$ of $V$. (Assume all vector spaces are over $\Bbb Q$). Then $\mathrm{End}_R(V)$ is the set of all $\Bbb Q$-linear endomorphisms of $V$ that commute with $f$, wheras $\mathrm{End}_{R^{op}}(V)$ is the set of all endomorphisms that commute with $g$. It's not hard to construct examples of commuting endomorphisms $f$ and $g$ such that these two sets of endomorphisms don't have the same dimension, e.g. take $V$ to be finite-dimensional of dimension $\geq 2$ and take $f$ to be a scalar multiple of the identity and $g$ any endomorphism that is not a scalar multiple of the identity.

Lukas Heger
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  • Thank you, so if I understand: considering the space $\mathbb Q^2$, with structure of bimodule given by any transformation (identified by a $T\in \mathrm{Mat}2(\mathbb Q)$) and the transformation $A:=a\mathrm{Id}$, where $a\in \mathbb Q^\times$ and $\mathrm{Id}$ is the identity matrix. Since for any $M\in \mathrm{Mat}_2$, holds $MA=AM$, the bimodule structure is well defined, and $\mathrm{hom}{\mathbb Q[X]}(\mathbb Q^2,\mathbb Q^2)=\mathrm{Mat}_2(\mathbb Q)$. – John Nov 13 '22 at 00:45
  • Instead not every $M$ satisfies $MT=TM$, so $\mathrm{hom}{\mathbb Q[X]^{op}}(\mathbb Q^2,\mathbb Q^2)$ is a subspace of (hence has different dimension from) $\mathrm{Mat}_2(\mathbb Q)$. But in general, if a ring $R$ is an $S$-module, $R^{op}$ is an $S$-module and $R^{op}\cong R$ as $S$-modules. Then in particular $\mathrm{hom}{\mathbb Q[X]}(\mathbb Q^2,\mathbb Q^2)$ and $\mathrm{hom}_{\mathbb Q[X]}(\mathbb Q^2,\mathbb Q^2)^{op}$ must have equal dimension. – John Nov 13 '22 at 00:45
  • @JerryScott yes, it seems you understand my answer correctly – Lukas Heger Nov 13 '22 at 00:50