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I know simple ring implies primitive ring. I was looking to show the converse doesn't hold by taking the ring $M_{2}(\mathbb{Z})$. But I couldn't construct a left maximal ideal which doesn't contain proper ideal for it. Was it claim wrong or is there some other technique to justify?

rschwieb
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wow
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1 Answers1

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The ring $M_2(\mathbb Z)$ won't work: being right primitive is a Morita invariant property, so $\mathbb Z$ would also have to have that property. But a commutative primitive ring is a field, so $\mathbb Z$ is not primitive.

A good example to try which will work would be $End(V_K)$ where $V$ is a countably infinite dimensional $K$ vector space. I'll leave it to you to explore the details.

rschwieb
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