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As we know, if $A$ is Artin $R-$algebra and $M$ is finite generated,then $\operatorname{End}_A M$ is also an Artin $R-$algebra. My questions are:

  1. If $A$ is an Artin ring and $M$ is finite generated, is $\operatorname{End}_A M$ an Artin ring?

  2. If $A$ is an arbitrary ring and $M$ is a module with composition series, is $\operatorname{End}_A M$ an Artin ring?

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Jian
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  • by "Artin ring" do you mean that it is left and right Artinian? – rschwieb Sep 28 '17 at 11:17
  • @rschwieb as the answer below,if R is not left and right Artin,it has a counerexample. – Jian Sep 28 '17 at 11:25
  • I'm just asking because I do not know for sure what you mean by "Artin ring." An Artin $R$-algebra has a very specific meaning as a finitely generated algebra over a commutative artinian ring $R$. When you say "Artin ring" for $A$, I'm not positive what was intended. – rschwieb Sep 28 '17 at 12:57
  • If $A$ is right artinian and $M$ is a right $A$ module, then the first question could be simplified to say "$M$ has a composition series" since a finitely generated right $A$ module over a right Artinian ring $A$ necessarily has a composition series. – rschwieb Sep 28 '17 at 13:09
  • Maybe you already know, but it's known that the endomorphism ring of any module with a composition series is semiprimary (that is, $A/J(A)$ is Artinian, and $J(A)$ is a nilpotent ideal.) – rschwieb Sep 28 '17 at 13:10
  • @rschwieb I don't know this,can you give a short proof?please,why Jacobson radical of endmorphism ring is nilpotent?thank you – Jian Sep 28 '17 at 13:32
  • https://math.stackexchange.com/q/2407901/29335 – rschwieb Sep 28 '17 at 13:50

1 Answers1

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For (2) the answer is no. For example there exist rings $A$ such that $A$ is left Artinian but not right Artinian. Left Artinian rings are left Noetherian so $A$, considered as a left $A$-module, has a composition series. But the endomorphism ring of this module is the opposite ring $A^\text{op}$ which is not left Artinian.

Jim
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  • if we consider R is commutative Artin ring,is there a counterexample? – Jian Sep 28 '17 at 11:24
  • +1 because I like the observation: I don't intend this comment to detract from it @Sky If you insist that $A$ is an Artin $R$-algebra, then the above strategy can't be employed since an Artin $R$ algebra is finitely generated over $R$, it must be left and right Artinian. But I'm not sure if you have realized that would happen or not... – rschwieb Sep 28 '17 at 13:07
  • @rschwieb I know this property,I don't understand your meaning. – Jian Sep 28 '17 at 13:36
  • @Sky I'm saying there is no Artin $R$-algebra that is one-sided Artinian. – rschwieb Sep 28 '17 at 13:49
  • @rschwieb yeah,I know this since the underlying ring is commutative Artin ring. – Jian Sep 28 '17 at 13:51