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Can anyone tell me why the endomorphism ring of a finite-length module is artinian? Bonus points if you can do it without using the radical, semisimplicity, Fitting's lemma or anything fancy. If you have to or it makes the proof easier, that's OK too, but I have reason to believe that there's a simple proof (namely Dennis and Farb give this as an exercise in Chapter 0 of their book Noncommutative Algebra).

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Just to clarify: I think you want to say "right Artinian." In general the endomorphism ring does not have to be left Artinian. Namely, suppose $A$ is a ring that is left Artinian but not right Artinian (it is known that such rings exist). View $A$ as a module over itself using the left multiplication action. This module has finite length. The endomorphism ring is isomorphic to the opposite ring of $A$. That ring is right Artinian but not left Artinian. –  senti_today Aug 1 '10 at 0:26
This is actually a cautionary tale that applies to all questions of the type "tell me why/when X is true": context (i.e. the origin of the question along with precise definitions or links to them) needs to be given. –  Victor Protsak Aug 1 '10 at 0:48
I think that the origin of the question was specified in the formulation, wasn't it? –  senti_today Aug 1 '10 at 1:34
Perhaps this is a cautionary tale that applies to giving exercises in textbooks! –  senti_today Aug 1 '10 at 1:36
@VP You might be able to view it with Google Books; however, there does not appear to be any essential difference in the statement. –  Dylan Moreland Aug 1 '10 at 17:25

2 Answers 2

up vote 3 down vote accepted

Never mind. I was more successful with Google this time. It turns out that the statement is simply false. In the comment above I gave an example showing that the endomorphism ring need not be left Artinian. The following paper contains a (much less trivial) example showing that the endomorphism ring also need not be right Artinian:

"Ring of Endomorphisms of a Finite Length Module" R. N. Gupta and Surjeet Singh Proceedings of the American Mathematical Society, Vol. 94, No. 2 (Jun., 1985), pp. 198-200.

Incidentally, the module in that example appears to be a nontrivial self-extension of a simple module.

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Good job clearing it up! –  Victor Protsak Aug 1 '10 at 0:45
I can't edit, so I'll mention that this article is freely available at ams.org/journals/proc/1985-094-02/S0002-9939-1985-0784161-2/… –  Dylan Moreland Aug 1 '10 at 1:58
Thanks for that. I'm not sure whether I'm more relieved or annoyed. In any case, I'll let Professor Farb know... –  Justin Campbell Aug 1 '10 at 2:03
To completely put the matter to rest, I'll point out that in the example by Gupta and Singh, the ring is not right Artinian, but it is left Artinian. However, suppose we have rings $A$ and $B$ and modules $M$ over $A$ and $N$ over $B$ such that $R:=End_A(M)$ is not left Artinian and $S:=End_B(N)$ is not right Artinian. Then the product $C:=A\times B$ naturally acts on $M\times N$, and $End_C(M\times N)\cong R\times S$ is neither left nor right Artinian. –  senti_today Aug 1 '10 at 2:04

What is true is that the endomorphism ring of any finite length module is semiprimary.

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