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Let $R$ be a commutative artinian ring with identity. It is true that for $n>0$ the matrix ring $M_n(R)$ is left and right artinian?

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Yes. This is because the Artinian condition is a Morita invariant condition. This means that $R$ and any ring Morita equivalent to it (including its matrix rings) is left and right Artinian.

You can prove it without the full brunt of Morita theory, though.

The way to connect (generalized by Morita) $R$ and $M_n(R)$ is to examine $R^n$ as an $M_n(R)$ module with the obvious (matrix multiplication) module action.

If you can get ahold of it at a library, or take a look in googlebooks, Chapter 7 of Lam's Lectures on Modules and Rings has a very accessible introduction to seeing this connection between matrix rings and their base rings

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Right, but shouldn't you include a proof or a reference to this claim? It doesn't follow trivially from the definition of Artinian. – Qiaochu Yuan Sep 7 '12 at 20:13
What rschwieb means is that there is an equivalence of categories between $A$-$\mathrm{Mod}$ and $M_n(A)$-$\mathrm{Mod}$, which happens to map the regular left $A$-module $A$ to the regular left $M_n(A)$-module $M_n(A)$. Since an equivañlence preserves the lattice of subobjects, the claim follows immediately from this. This should be discussed pretty much anywhere where Morita equivalences are discussed, like the books by Anderson-Fuller Modules and rings (iirc the title...) or Pierce Associative Algebras – Mariano Suárez-Alvarez Sep 7 '12 at 20:17
@QiaochuYuan Yes kind sir but I need more than three minutes to find it. – rschwieb Sep 7 '12 at 20:21
I wrote the comment while your answer was in its previous state. – Mariano Suárez-Alvarez Sep 7 '12 at 20:23
@qiaochu I sure hope there isn't (and never will be) a rule that answers have to be completely edited and fully cited before the first post! – rschwieb Sep 7 '12 at 20:28

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