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Let $k$ be an algebraically closed field. Let $U_{n}(k)$ denote the $k$-algebra of upper triangular matrices of size $n \times n$ over $k$. Is it true that this $k$-algebra is indecomposable?

Is this because its socle is isomorphic to $k^{n}$ and $k^{n}$ is simple, so that $U_{n}(k)$ has simple socle, hence indecomposable. Is this true?

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up vote 4 down vote accepted

I think what you want to show is that $U_n$ is not a non-trivial direct product of algebras. Is that what "indecomposable" means for you?

Let the simple modules be $S_1, \ldots, S_n$ where $S_i$ is the one-dimensional module on which a matrix with diagonal elements $d_1, \ldots, d_n$ acts as $d_i$. Let $P_i$ be the projective cover of $S_i$, so $P_i = U_n E_{ii}$ where $E_{ii}$ is the idempotent matrix having zeros everywhere except for a one at position $(i,i)$. Clearly $P_i$ has a basis consisting of all elementary matrices $E_{ai}$ in $U_n$. From this you can see that $P_i$ is uniserial, with composition factors $S_i$, $S_{i-1}$,... $S_1$. There is a non-split extension of $S_i$ by $S_{i-1}$ for $i>1$, coming from the top two layers of $P_i$.

If there were a non-trivial decomposition $U_n = A \oplus B$, the simple modules would fall into two non-empty classes (those for $A$ and those for $B$) with no non-split extensions between them. Since we showed this doesn't happen, there's no such decomposition.

I'm not quite sure what you mean in your socle comment, but $k^n$ is probably not going to be simple in any interpretation.

A more elementary way would be to look at the centre. The centre of a non-trivial algebra has dimension at least one since it contains the identity, so the centre of a non-trivial direct product has dimension at least two.

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isn't it that $k^{n}$ is simple as a $M_{n}(k)$ module? anyway I was trying to use the fact that if $Soc(M)$ is simple $A$-module then $M$ is indecomposable (here $M$ is a $A$-module where $A$ is a $k$-algebra). So here we can find the computation of the socle: books.google.com.mx/… –  user10 Sep 29 '11 at 14:53
    
they show the right socle is a copy of $k^{n}$ no? please have a look at example $7.14$b). –  user10 Sep 29 '11 at 14:54
    
to me indecomposable means that if $M = N_{1} \oplus N_{2}$ where $N_{i}$ are $A$-modules with $A$ a $k$-algebra, then $N_{1}=0$ or $N_{2}=0$. –  user10 Sep 29 '11 at 14:56
2  
It's not relevant that $k^n$ is a simple $M_n(k)$-module, because you are not working over $M_n(k)$. The socle of your algebra is definitely not simple. I'm still not sure what you mean by indecomposable though, $U_n$ is definitely not indecomposable as a module over itself (it's a direct sum of projectives). –  mt_ Sep 29 '11 at 15:10

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