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Let $K$ be a commutative ring. Then $$K \text{ is a field } \iff \text{ the only ideals of K are } \{0\} \text{ and } K$$

As a remark in my course, it is said that it does not hold if $K$ is not commutative. Can anyone give a counterexample ? It will surely be a ring $A$ with ideals $\{0\} \text{ and } A$ but that is not a field, I have been looking for something with matrices but the struggle is real.

Kilkik
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    I've duplicated this against an existing question that explains why square matrix rings over fields are examples of simple rings, which answers the question asked here with details. Some of them might also detail why they are noncommutative, but that's already pretty easy to see. – rschwieb Jun 09 '22 at 13:47
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    Of course, Hamilton's quaternions are also an example of a noncommutative ring with exactly two ideals. – rschwieb Jun 09 '22 at 13:51
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    Looking for something with matrices indeed. A full ring of square matrices over a field is what most people would give as an example. The question is: what stopped you from drawing a conclusion? – rschwieb Jun 09 '22 at 13:53
  • @rschwieb honestly : laziness. Computing ideals with matrices and not being sure if it will work made me wanna ask for a quick response lol. Lazy but honest. – Kilkik Jun 13 '22 at 00:20

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The only two sided ideals of the ring $Mat_{n \times n}(k)$ of dimension $n$ square matrices over a field $k$ are zero and everything.

A noncommutative ring whose only left ideals are zero and everything is necessarily a division ring.

Dustan Levenstein
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