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I was hoping for an elementary method of approaching this.

My attempt: D is a division ring so Noetherian/ Artinian. Then $M_n(D)$ is Noetherian/ Artinian as a matrix ring over Noetherian/ Artinian ring.

Then a Module over a Noetherian ring is Noetherian iff finitely generated. Now this would imply that it is Artinian. I am also sure that any Artinian module over an Artinian ring is finitely generated. I saw this answer: Is every Artinian module over an Artinian ring finitely generated? but it explicitly uses commutativity of the ring which we don’t have. I am sure it is true in general but the digression takes me far from the context in which the question was asked, where $R \text{ has the really nice form } M_n(D)$

Is there any way of doing this using the nice form of $R$?

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$M_n(D)$ is semisimple, so every module $M$ over $M_n(D)$ is a direct sum of simple modules. Both Noetherian and Artinian are equivalent to $M$ being a finite direct sum of simple modules.

To see this, suppose $M= \bigoplus_{i=1}^nM_i$ where each $M_i$ is simple. Then we can write down a composition series for $M$, where we repeatedly mod out a single simple summand.

If $M = \bigoplus_{i \in I} M_i$ where each $M_i$ is non-zero and $I$ is infinite, then we can take an infinite sequence of elements $i_1, i_2, \dots$ in $I$ and we have an ascending chain $M_{i_1} \subset M_{i_1}\oplus M_{i_2} \subset \dots$ and an infinite descending chain $M \supset \bigoplus_{i \in I\setminus\{i\}}M_i \supset \bigoplus_{i \in I \setminus \{i_1, i_2\}}M_i \supset$.

The above observation is a step in the proof of a more general result by Hopkins and Levitzki.

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  • $\begingroup$ Sorry, why is it the case that a module over a semi simple ring is semi simple? $\endgroup$ – SEWillB Feb 3 '18 at 22:54

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