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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?

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Do you know what semi-simple means? What is your background/the context? – Plop Sep 13 '11 at 17:18
@Plop: yes, I know what semi-simple means, feel free to give any suggestions, this is self-study. Even if I don't understand the concepts/definitions you use I will search for them. – user10 Sep 13 '11 at 17:39
up vote 3 down vote accepted

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 $k^n$.

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I don't understand why $T$ is isomorphic to $k^n$ I thought that $A$ is isomorphic to $k^{n^{2}}$. Can you please explain this part? – user10 Sep 22 '11 at 3:19
Let $V_i \subset A$ be the sub vector space of matrices that are $0$ except maybe on the $i$th column. Then $V_i$ is a sub-$A$-module of the left $A$-module $A$, and $A = \oplus_{i=1}^{n} V_i$. – Plop Sep 22 '11 at 17:32

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