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By a ring I mean an associative ring with multiplicative unity, and by a division ring I mean a skew field.

My question is as follows

Let $D, S$ be two division rings and $_DM_S$ be a $(D, S)$-bimodule such that $_DM$ is simple left D-module and $M_S$ is a simple right $S$-module. Can this happen in the case when $D$ and $S$ are not isomorphic?

In such a case, M is isomorphic to $D$ as a left $D$-module and isomorphic to $S$ as a right $S$-module. But it seems that, in general, $M$ need not be isomorphic to any of these division rings as a ring or indeed have any ring structure at all.

In such a case, $D$ and $S$, respectively, are both isomorphic to the endomorphism ring of $M$ (viewed as a module over the other division ring), which seems peculiar, but I fail to use it to prove anything stronger.

rschwieb
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dmk
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  • A detail: $End(M_S)\cong S$ and $End(_DM)\cong D^{op}$ all as rings, but not necessarily $D$. – rschwieb Aug 14 '25 at 14:28
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    Sorry for a false start in the solutions: I forgot that you indeed wanted a bimodule and my instinct was this example with a module that was simply a left module over one and a left module over the other. – rschwieb Aug 14 '25 at 14:40

1 Answers1

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This condition implies that $D$ and $S$ are isomorphic. To make the argument as simple as possible, by simplicity if $m \in M$ is any nonzero element then the maps $s \mapsto ms$ and $d \mapsto dm$ are bijections $S \cong M$ and $D \cong M$ respectively (in fact isomorphisms of right $S$-modules resp. left $D$-modules). Composing these bijections produces a bijection $\varphi : S \to D$ defined by

$$\varphi(s) m = ms.$$

This gives $\varphi(1) = 1, \varphi(0) = 0$, and

$$\varphi(r+s) m = m(r+s) = mr+ms = \varphi(r) m + \varphi(s) m = (\varphi(r) + \varphi(s)) m$$ $$\varphi(rs) m = mrs = \varphi(r) m s = \varphi(r) \varphi(s) m$$

so $\varphi$ is a bijective ring homomorphism, hence an isomorphism. Loosely speaking this works out because $\varphi$ is a "conjugation" $\varphi(s) = msm^{-1}$.

Qiaochu Yuan
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