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There is a famous theorem in commutative ring theory which states: "Let $R$ be a commutative ring with unity and let $M$ be a (two-sided) ideal of $R$. Then, $M$ is maximal if and only if $R/M$ is a field". This is not valid if $R$ is noncommutative (the matrices form an example). My question is thus:

What condition satisfies "Let $R$ be a ring with unity and let $M$ be a maximal two-sided ideal. Then, $R/M$ is a division ring if and only if condition"?

In general, $R/M$ will be a simple ring in that case, but what else is needed for me to conclude it is a division ring?

Thanks in advance!

Gauss
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1 Answers1

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…if and only if $M$ is maximal as a right ideal of $R$.

rschwieb
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  • While I like this answer, is there a more “internal” answer? As in, can you impose a condition on $R$? If it is commutative, for instance, it does work. If it is local, I think it also works. – Gauss Aug 27 '22 at 18:01
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    The ways you mentioned, and any further way I could think of to mention, would all be simply sufficient conditions to make $M$ a maximal right ideal. Like "right (or left) quasi-duo." I mentioned the obvious necessary and sufficient condition only because I could not see any more interesting things to say. – rschwieb Aug 27 '22 at 18:06
  • In that case, I shall accept your answer. Thank you! – Gauss Aug 27 '22 at 18:27