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We suppose that all rings are left Artinian simple rings and all modules over a ring are of finite length.

Let $M \neq 0$ be a left module over a ring $R$. By Wedderburn theorem, $R$ is a matrix ring over a division ring $D$. We denote $D$ by $D(R)$. We denote the set of $R$-submodules of $M$ by $L(M)$. We regard $L(M)$ as an ordered set by the inclusion relation.

Let $N \neq 0$ be a left module over a ring $S$. Suppose length $M$ = length $N$ and $D(R)$ is isomorphic to $D(S)$. Is $L(M)$ isomorphic to $L(N)$?

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  • $\begingroup$ What does $D(M)$ mean? $\endgroup$ – Dune Sep 24 '14 at 6:47
  • $\begingroup$ @Dune I edited the obvious typos. $\endgroup$ – user178399 Sep 24 '14 at 6:49
  • $\begingroup$ Question asker seems to have disappeared. Strange. $\endgroup$ – Jyrki Lahtonen Sep 24 '14 at 11:25
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    $\begingroup$ @JyrkiLahtonen Maybe he was transported by an equivalence of categories to a Morita equivalent universe? $\endgroup$ – Jeremy Rickard Sep 24 '14 at 13:43
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Since $D(R) \cong D(S)$, the rings $R$ and $S$ are Morita equivalent, i.e., there is an equivalence of categories between $R$-modules and $S$-modules. In particular, if $M$ and $N$ are the same length, then they are matched by tensoring with some $R-S$-bimodule followed by a module isomorphism. This induces an isomorphism $L(M) \cong L(N)$ of the posets of submodules.

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To every $D$-submodule $V$ of $D^{\oplus k}$, associate the $(D^n)^{\oplus k}$-submodule consisting of all $k$-tuples in $(D^n)^{\oplus k}$ satisfying every $D$-linear dependence relation with coefficient vector in $V$. The resulting function is an order-reversing lattice isomorphism between $L(D^{\oplus k})$ and $L((D^n)^{\oplus k})$.

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