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We know in a semisimple ring R, for every R-module, Noetherian is equivalent to Artinian, my question is:

If for every R-module M Noetherian is equivalent to Artinian, can we prove R is a semisimple ring?

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1 Answer 1

up vote 7 down vote accepted

Let $R= k[\epsilon]/\epsilon^2$. Then module is Artinian if and only if it is Noetherian if and only if it is finite-dimensional. This is clear, since if a module is finite-dimensional there can't be an infinite ascending or descending sequence, and if it's infinite dimensional than $M/\epsilon$ is an infinite-dimensional vector space, so it has infinite ascending and descending chains.

But $R$ is obviously not semisimple.

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The module $R$ is not semisimple, because the submodule generated by $\epsilon$ does not have a complement. The same is true for any ring with nilpotents. –  Will Sawin Dec 21 '12 at 6:15
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In fact, Noetherian is equivalent to Artinian for $R$-modules if and only if $R$ is an Artinian ring. –  Jeremy Rickard Jul 27 '13 at 12:32
    
$\mathbb{Z}_{p^2}$ with $p$ prime also work, and by the Wedderburn theorem and counting is easy to see that is not semisimple. –  Murphy Jul 27 '13 at 20:05

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