Is the $k$-algebra of upper triangular matrices an indecomposable algebra? 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?
 A: 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.
