A module $M_{R}$ is called semi-artinian if every nonzero image of $M$ contains a simple submodule. Given $m\in M$ and $a_1,a_2,...$ in J(R). Why $ma_1a_2...a_{n-1}a_n=0$ for some $n\geq 1$.

(Edit by KennyTM: The above is OP's original question. The latest, completely changed question follows:)

if $R$ is regular what is the relation between:

1) J(R)

2) a left $R$-module has a projective cover?

  • $\begingroup$ Are you assuming R is noetherian? Not every semiartinian ring is T-nilpotent, that is, not every semiartinian ring is perfect. $\endgroup$ – Jack Schmidt Dec 14 '10 at 0:39
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    $\begingroup$ Dear Arash, Where did this question come from? The argument that Mariano links to below is quite subtle, and so I'm curious at what level this question was assigned as an exercise (if indeed it was so assigned). Regards, $\endgroup$ – Matt E Dec 14 '10 at 2:41
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    $\begingroup$ This question completely changed with the edit 3 mins ago. All the above comments and the answer below apply to the previous version, which can be read in the edit history. $\endgroup$ – Matt E Dec 14 '10 at 20:34
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    $\begingroup$ @Arash: Why are you changing your questions in this manner? Something strange is happening to Arash's questions. I've flagged it for moderator attention (this is not the only question to which this has happened, and some of the previous edits were to silly questions) $\endgroup$ – Arturo Magidin Dec 14 '10 at 20:35
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    $\begingroup$ @Arash: Please visit the meta thread quoted above and explain why you made the radical edits to the questions. Current consensus is to roll-back your edits if you log in and don't address the problem, or don't log in by the end of today. $\endgroup$ – Arturo Magidin Dec 15 '10 at 17:37

This is proved in Askar A. Tuganbaev' s Semidistributive modules and rings, page 153, which you can read on googlebooks.

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    $\begingroup$ Dear Mariano, I tried to prove this and failed, and then looked at your reference. As I thought, the argument uses the socle filtration. But when I tried to argue along these lines, I got confused, because the socle filtration can be infinite in length. In your reference, this is gotten around by a tricky descending induction on ordinals. This is one of (perhaps the only) time I have seen an ordinal induction argument like this that I don't immediately see how to rephrase another way. Do you see how to write the argument without this explicit descent on ordinals? $\endgroup$ – Matt E Dec 14 '10 at 2:38
  • $\begingroup$ @Matt: I don't seem to be able to get rid of that induction. The idea of the socle filtration is very natural... $\endgroup$ – Mariano Suárez-Álvarez Dec 14 '10 at 14:33

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