Matrices and dimension question Let $k$ be an algebraically closed field and consider the $k$-algebra $M_{n}(k)$ (the set of all $n \times n$ matrices with entries in $k$). Let $T$ be an indecomposable $M_{n}(k)$ module, why must $T$ have length $1$ and dimension $n$ as a $k$-vector space?
 A: The "right way" to do this is to say that $A=M_n(k)$ is a semi-simple ring, which means that all the $A$-modules are semi-simple (and that is equivalent to $A$ being a semi-simple $A$-module e.g. on the left).
In your case, this holds because $A$ is isomorphic (as an $A$-module) to $n$ copies of $k^n$ (obvious action of $A$ on $k^n$), and $k^n$ is easily seen to be a simple $A$-module.
Moreover, any simple $A$-module is a quotient of the $A$-module $A$, so it is isomorphic to $k^n$.
$T$ is simple (because it is semi-simple and indecomposable), so it is isomorphic to $k^n$.
You can also achieve this "by hand", if you know a little linear algebra/reduction theory.
Giving yourself a (finite-dimensional) $A$-module is equivalent to giving yourself a morphism of $k$-algebras $f : A \rightarrow M_m(k)$ for some $m$ (equal to the dimension of the module over $k$, of course).
In $A$ you have a "full system of orthogonal idempotents", the $E_{i,i}$: they satisfy $E_{i,i}^2=E_{i,i}$, $E_{i,i}E_{j,j}=0$ if $i \neq j$, and $\sum_i E_{i,i} = 1$.
If we let $e_i = f(E_{i,i})$, the $e_i$ satisfy these relations also.
Linear alg/reduction tells you that after conjugation by an invertible $m \times m$ matrix, the $e_i$ are diagonal matrices with zeros and ones on the diagonal, the places where the ones are are "disjoint" and cover the whole diagonal.
Next $f(E_{i,j}) = f(E_{i,i}E_{i,j}E_{j,j})=e_if(E_{i,j})e_j$, so the $f(E_{i,j})$ have a special form, and working out their relations you get that after conjugation by an invertible matrix, $f$ maps a matrix $M$ to a block-diagonal matrix having $m/n$ (necessarily an integer) copies of $M$ on the diagonal.
That is a concrete way of saying that any $A$-module of finite dimension over $k$ is a direct sum of copies of $k^n$.
