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Let $A$ be a finite dimensional semisimple algebra over $\mathbb{C}$ and $M$ is a finitely generated A-module. Prove that $M$ has only finitely many submodules iff $M$ is a direct sum of pairwise non-isomorphic simple modules.

Any hints for how to solve it?

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@jspecter, $\mathbb{C}^2$ has infinitely many subspaces! – Bruno Joyal May 7 '12 at 4:01
@CC_Azusa, use Artin-Wedderburn to reduce to the case where $A=M_n(\mathbb{C})$, and use the fact that any simple $A$-module is isomorphic to $\mathbb{C}^n$ with the natural action of $M_n(\mathbb{C})$. Show that $\mathbb{C}^n\oplus \mathbb{C}^n$ has infinitely many submodules. – Bruno Joyal May 7 '12 at 4:07
@Bruno Please consider converting your comment into an answer, so that this question gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see here, here or here. – Julian Kuelshammer Jun 10 '13 at 20:32

This is a CW answer composed in an effort to remove this question from the unanswered queue. It is based on Bruno's suggestions in the comments.

Suppose $R$ is a semisimple Artinian ring, which can be expressed as a finite direct sum $R\cong \oplus R_i$ where the $R_i$ are simple Artinian rings. For any right $R$ module $M$, $MR_i$ is a right $R$ submodule of $M$ (as well as a right $R_i$ module), and moreover $M=\oplus MR_i$. This is just the decomposition of $M$ into homogeneous pieces, and indeed the distinct $MR_i$ are sums of distinct isotypes of simple $R$-submodules of $M$.

The decomposition also establishes that every submodule of $M$ is going to be a direct sum of submodules of the $MR_i$. So, the product of these $R$ modules will have finitely many submodules if each individual $MR_i$ has finitely many submodules.

We will show that for a simple Artinian $\Bbb C$-algebra $A$, an $A$-module has finitely many submodules iff the module is simple. Consider $S\oplus S$ for a simple $A$-module $S$. Observe that $(x,\lambda x)A$ is a submodule of $S\oplus S$ for any $\lambda\in \Bbb C$ and $x\neq 0$.

Now if $(x,\alpha a)A=(x,\beta x)A$ for some $\alpha,\beta$, it follows that $(xY,\alpha xY)=(x,\beta x)$ for some $Y\in A$. But then $x= xY$, and $\alpha x=\alpha x Y=\beta x$. Since $x\neq 0$, $\alpha=\beta$. This shows that for distinct choices of $\lambda$, $(x,\lambda x)A$ are distinct submodules. Since there are infinitely many choices for $\lambda$, $S\oplus S$ has infinitely many submodules. Any nonsimple $A$ module would have to contain at least two copies of $S$, and hence infinintely many submodules.

So in summary, if $M$ has finitely many submodules, then each of the $MR_i$ is actually a simple $R$ (and simple $R_i$) module. Since each $MR_i$ corresponds to a distinct homogeneous piece of $R$, they are pairwise nonisomorphic. In this case it is even possible to count the submodules: if there are, say, $k$ of the $MR_i$ which are nonzero in the decomposition, then $M$ has $2^k$ submodules

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