Classification of modules over complex matrices Is it possible to classify all finite dimensional modules over the space of matrices over complex numbers, i.e. $M_n\left(\mathbb{C}\right)$?
I have a feeling that any simple module should be isomorphic to $\Bbb{C}^n$, but I have no clue on how to proceed.  I can say that for any nonzero element $v$ of the module $A$, we have $A = M_n\left(\Bbb{C}\right)v$ by simplicity, so we would like to send an element of the form $Xv$ to $Xe_1 \in \Bbb{C}^n$, but I don't know how to find the element $v$ to make the map well-defined.
Also, can we say that any module over $M_n\left(\Bbb{C}\right)$ is the direct sum of simple modules?  If so, how?
 A: If $R$ is any ring, the $M_n(R)$-modules $M$ are all isomorphic to some $L^n$, where $L$ is an $R$-module, and matrices from $M_n(R)$ act on column vectors in $L^n$ as follows:
$$ \begin{bmatrix} r_{11} & \cdots & r_{1n} \\ \vdots & \ddots & \vdots \\ r_{n1} & \cdots & r_{nn} \end{bmatrix} \begin{bmatrix} \ell_1 \\ \vdots \\ \ell_n \end{bmatrix} = \begin{bmatrix} r_{11}\ell_1+\cdots+r_{1n}\ell_n \\ \vdots \\ r_{n1}\ell_1+\cdots+r_{nn}\ell_n \end{bmatrix}. $$
So, matrix multiplication as usual, but the matrix entries are scalars in $R$ and the column vectors' entries are elements in the $R$-module $L$.
To prove this fact, we need to see "internally" what the summands of $M$ should be, and why they are all isomorphic as $R$-modules. In fact, we have a direct sum decomposition
$$ M=e_{11}M\oplus\cdots\oplus e_{nn}M $$
where $e_{ii}=\mathrm{diag}(0,\cdots,1,\cdots,0)$ is the diagonal matrix with $1$ in the $i$th entry and all other entries $0$. This follows using general theory from the fact the $e_{ii}$s are central orthogonal idempotents, but let's be a little more concrete.
(i) First, $M=\sum_i e_{ii}M$ because the sum includes all elements of $(\sum_i e_{ii})M$, and $\sum_i e_{ii}$ is the identity matrix; (ii) Second, the sum is direct, since any element of $\sum_{j\ne i}e_{jj}M$ is annihilated by $e_{ii}$ while the only element of $e_{ii}M$ annihilated by $e_{ii}$ is $0$ itself (in fact $e_{ii}$ is a projection $M\to e_{ii}M$); and (iii) Third, since $R\hookrightarrow M_n(R)$ (scalar multiples of the identity matrix), each $e_{ii}M$ is an $R$-module, and they are isomorphic $R$-modules since any permutation $\pi$ that swaps $i\leftrightarrow j$ furnishes a permutation matrix $P_\pi$ by whose multiplication is an $R$-module isomorphism $e_{ii}M\leftrightarrow e_{jj}M$.
Consider $R=\mathbb{C}$. Any $R$-module is of the form $L\cong\mathbb{C}^m$ for some dimension $m$, since modules over fields are vector spaces which are free modules. Thus, since $M_n(R)$-modules are isomorphic to $L^n$ for $R$-modules $L$, our $M_n(\mathbb{C})$-modules are isomorphic to $(\mathbb{C}^m)^n$ for some $m$. Concretely, if we interpret elements of $\mathbb{C}^m$ as $m$-tuples, the action is given by
$$ \begin{array}{ll} \begin{bmatrix} a_{11} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots \\ a_{n1} & \cdots & a_{nn} \end{bmatrix} \begin{bmatrix} (x_{11},\cdots,x_{1m}) \\ \vdots \\ (x_{n1},\cdots,x_{nm}) \end{bmatrix} &  = \begin{bmatrix} a_{11}(x_{11},\cdots,x_{1m})+\cdots+a_{1n}(x_{n1},\cdots,x_{nm}) \\ \vdots \\ a_{n1}(x_{11},\cdots,x_{1m})+\cdots+a_{nn}(x_{n1},\cdots,x_{nm}) \end{bmatrix} \\ & =\begin{bmatrix} (a_{11}x_{11}+\cdots+a_{1n}x_{n1},\cdots,a_{11}x_{1m}+\cdots+a_{1n}x_{nm}) \\ \vdots \\ (a_{n1}x_{11}+\cdots+a_{nn}x_{n1},\cdots,a_{n1}x_{1m}+\cdots+a_{nn}x_{nm}) \end{bmatrix} \end{array} $$
This is equivalent to $M_n(\mathbb{C})$s diagonal action on $\underbrace{\mathbb{C}^n\oplus\cdots\oplus\mathbb{C}^n}_m$ given by
$$ \begin{bmatrix} a_{11} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots \\ a_{n1} & \cdots & a_{nn} \end{bmatrix} \left(\begin{bmatrix} x_{11} \\ \vdots \\ x_{n1}\end{bmatrix} ,\cdots, \begin{bmatrix} x_{1m} \\ \vdots \\ x_{nm} \end{bmatrix}\right) $$
$$ = \left( \begin{bmatrix} a_{11} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots \\ a_{n1} & \cdots & a_{nn} \end{bmatrix} \begin{bmatrix} x_{11} \\ \vdots \\ x_{n1}\end{bmatrix}, \cdots, \begin{bmatrix} a_{11} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots \\ a_{n1} & \cdots & a_{nn} \end{bmatrix} \begin{bmatrix} x_{1m} \\ \vdots \\ x_{nm} \end{bmatrix} \right) $$
or more simply $A(v_1,\cdots,v_m)=(Av_1,\cdots,Av_m)$.
Of course, this does indeed imply $\mathbb{C}^n$ is the only simple $M_n(\mathbb{C})$-module.
