Find Simple submodules of Algebra Module I have the following algebra: 
$$ A=\left\{\begin{pmatrix} 
             a_{11} & a_{12} & 0\\
             0      & a_{22} & 0\\
             0      & 0      & a_{33}\\
             \end{pmatrix}:a_{ij}\in F\right\}$$
Where $F$ is an arbitrary field.
What are the simple submodules of $A$, is $A$ semi-simple?  
I thought I had a composition series:
$$ \{0\} \subset \left\{\begin{pmatrix} 
             0      & a_{12} & 0\\
             0      & 0      & 0\\
             0      & 0      & 0\\
             \end{pmatrix}\right\}
\subset \left\{\begin{pmatrix} 
             0      & a_{12} & 0\\
             0      & a_{22} & 0\\
             0      & 0      & 0\\
             \end{pmatrix}\right\}
\subset \left\{\begin{pmatrix} 
             0      & a_{12} & 0     \\
             0      & a_{22} & 0     \\
             0      & 0      & a_{33}\\
             \end{pmatrix}\right\}
\subset A$$
Which I thought would give me the following composition factors: 
$$\left\{\begin{pmatrix} 
             0      & a_{12} & 0\\
             0      & 0      & 0\\
             0      & 0      & 0\\
       \end{pmatrix}\right\} \text{first term},
\left\{\begin{pmatrix} 
             0      & 0      & 0\\
             0      & a_{22} & 0\\
             0      & 0      & 0\\
       \end{pmatrix}\right\} \text{second term},
\left\{\begin{pmatrix} 
             0      & 0      & 0\\
             0      & 0      & 0\\
             0      & 0      & a_{33}\\
       \end{pmatrix}\right\} \text{third term},
\left\{\begin{pmatrix} 
             a_{11} & 0      & 0\\
             0      & 0      & 0\\
             0      & 0      & 0\\
       \end{pmatrix}\right\} \text{last term}$$.
But it seems to me that $\left\{\begin{pmatrix} 
             0      & 0      & 0\\
             0      & a_{22} & 0\\
             0      & 0      & 0\\
       \end{pmatrix}\right\}$ isn't even a submodule. What has gone wrong here??
As for whether or not it is semi-simple; I would want to say that it is the direct sum of these composition factors but I can't because one of these factors doesn't seem to be a submodule??
Thanks in advance!
 A: It is not semisimple. It has a nonzero nilpotent ideal:
$$ I=\left\{\begin{pmatrix} 
             0& x & 0\\
             0      & 0 & 0\\
             0      & 0      & 0\\
             \end{pmatrix}\middle |\,x\in F\right\}$$  
Of course, every nilpotent ideal is contained in $J(A)$: so, $I\subseteq J(A)$. But on the other hand, $A/I\cong F\times F\times F$, which is semisimple, so $J(A)\subseteq I$. (The Jacobson radical is the smallest ideal $I$ such that $A/I$ is has Jacobson radical zero.)
So in fact the Jacobson radical $J(A)=I$.
The simple right modules of $A$ as an $A$ module are precisely the same as the simple modules of $A/J(A)$ as an $A/J(A)$ module (this is a property of the Jacobson radical.)
Since $F\times F\times F$ is a semisimple ring, it already contains copies of its simple modules, namely $0\times 0\times F$,$0\times F\times 0$, and $F\times 0\times 0$. Lifting this action up to $A$, you have all three isotypes of simple $A$ module.

But it seems to me that $\left\{\begin{pmatrix} 
             0      & 0      & 0\\
             0      & a_{22} & 0\\
             0      & 0      & 0\\
       \end{pmatrix}\right\}$ isn't even a submodule. What has gone wrong here??  

It does not need to be a submodule. The quotient module of the first two modules you provided isn't a subset of $A$ at all (remember, they are cosets, not matrices). However, you can pick representatives that look like things in $A$. This module you are talking about is naturally isomorphic to $0\times F\times 0$ I mentioned above.
If you multiply something of that form on the left with a matrix from your ring, you of course get something that looks like $\begin{bmatrix}0&x&0 \\ 0&y&0 \\0&0&0\end{bmatrix}$, but mod $I$ this is the same thing as something of the form $\begin{bmatrix}0&0&0 \\ 0&y&0 \\0&0&0\end{bmatrix}$.
Incidentally, you can identify your ring as $T_2(F)\times F$ with the block diagonals. The Jacobson radical will be the sum of the radicals of both rings (and perhaps you know about the upper triangular matrices a little better.) It's also easy to find out the simple modules of the product if you know the simple modules of the rings in the product.
