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The following structure theorem is well known:

A commutative Artinian ring is a finite direct product of local Artinian rings.

Do we have such/similar structure theorem for noncommutative artinian rings?

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I wonder why this is called a structure theorem. It doesn't give any information about the building blocks (local artinian rings), just says that it suffices to consider them. How do local artinian rings look like? This would be the content of a structure theorem. Sorry for offtopic. – Martin Brandenburg Jul 5 '12 at 18:06
@MartinBrandenburg I think from the point of view of a noncommutative algebraist, local Artinian rings are already pretty "nice". $R/rad(R)$ is already as simple as possible, and $rad(R)$ is nilpotent. (The subclass of uniserial Artinian rings is completely worked out in detail, though.) I think there is no hope of getting fine results about structure like the commutative algebraists get :) – rschwieb Jul 5 '12 at 18:30
@MartinBrandenburg Looking at local rings in terms of Morita theory, they are already "very small" their equivalence class. You get between equivalent rings by doing a mixture of forming matrix rings and corner rings ($eRe$ where $e$ is an idempotent.) Making a corner ring with it would be pointless because there are no nontrivial idempotents, and forming a matrix ring destroys locality. – rschwieb Jul 5 '12 at 18:34
@Martin Well i called it so because Atiyah-Macdonald calls it that way, but I can understand what you mean. – messi Jul 5 '12 at 18:35
@rschwieb Sorry i dont know how to vote for an answer, it keeps telling me to register, but i am already registered, otherwise, your answer is acceptable for me. Thank you once again. – messi Jul 5 '12 at 18:52
up vote 1 down vote accepted

I think the best analogy you can get is:

A right artinian ring is a finite direct product of indecomposable right Artinian rings.

Basically, one arrives at the result you mentioned in the OP by noting that you cannot have an infinite chain of pairwise orthogonal idempotents in a commutative Artinian ring.

Every finite set of central orthogonal idempotents of $R$ which add up to 1 corresponds to a decomposition of the ring into factor rings $e_iRe_i$.

If a central idempotent $e$ cannot be written as the sum of two central orthogonal idempotents, then $eRe$ is indecomposable (in the sense of ring decompositions). Let's call this idempotent $e$ irreducible, for the duration of our conversation.

In any ring, you can try to write $1=\sum e_i$ as a sum of central irreducible idempotents, but sometimes irreducible ones do not exist, and sometimes these sets can be infinite. Under pretty mild finiteness conditions you can guarantee the existence of a finite set of central irreducible idempotents. Even "Noetherian" works, of course.

So why does "local" show up in the commutative case? Lemma: a commutative Artinian ring whose only idempotents are 0 and 1 is local. (Hence a commutative Artinian indecomposable ring is local.)

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Thank you rschwieb, I have one more question regarding non commutative localization. Let $R$ be a non-commutative ring and $S=R\setminus m$, where $m$ is a maximal ideal. I know that for a general multiplicative set $T$, $T^{-1}R$ may not be well defined. But is it defined for my specific $S$. And will the identity like $S^{-1}Hom(M,R)\cong Hom(S^{-1}M,S^{-1}R)$ hold when $M$ is a finite left $R$ module? You may add left or right adjectives wherever necessary. – messi Jul 5 '12 at 18:30
This might be a stupid question, but i dont know much about non-commutative localization. I have just seen words like Ore set and right denominator sets, but dont know much of anything in noncommutative case. – messi Jul 5 '12 at 18:33
@messi This subject is wayyy to complicated for me to fit into a comment: borrow a copy of Lam's Lectures on modules and rings ASAP. Summary: Noncommutative localization is complicated. Right denominator sets are "nice" enough to form a ring of quotients in the old fashioned way. You can't localize Artinian rings: they only consist of units and zero divisors. – rschwieb Jul 5 '12 at 18:38
I will get that book. Yes i understand that we cant localize artin rings, so i take it that there is no hope to get an isomorphism on Homs as above. Thanks – messi Jul 5 '12 at 18:46

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