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?
Matrix Rings over Artinian commutative Rings
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ring-theory
1 Answers
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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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0Right, but shouldn't you include a proof or a reference to this claim? It doesn't follow trivially from the definition of Artinian. – 2012-09-07
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0What 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* – 2012-09-07
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0@QiaochuYuan Yes kind sir but I need more than three minutes to find it. – 2012-09-07
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0I wrote the comment while your answer was in its previous state. – 2012-09-07
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0@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! – 2012-09-07
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0@Mariano Ok thank you :) Gonna remove my comment then. – 2012-09-07
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0Thanks rschwieb, your answer is very complete. – 2012-09-07