Composition factors appearing as simple modules I just saw the proof of the following corollary of Jordan Holder's Theorem.

Let $A$ be a finite dimensional $k$-algebra. Then for every simple $A$-module $S$, $S$ appears as a composition factor in every composition series of  $A$-module $A$.

Now I am wondering the converse: 

For each composition factor $M$ in the composition series of the $A$-module $A$, must there exist a simple submodule $N$ of $A$ such that $N \cong M$?

I tried a few examples; in particular, the case for real upper triangular seems to imply the above is not true?  
 A: Good question.
A ring $R$ is called right Kasch if every simple right $R$ module is isomorphic to a minimal right ideal of $R$.
In your question, a ring with a right composition series is necessarily right Artinian, so you seek for a right Artinian ring which is not right Kasch.
Here is the query at the Database of Ring Theory that yields some examples, most of which are upper triangular matrix rings as you suspected. (And there is one that is a subring of the upper triangular matrix ring.)
Let's analyze the smallest one appearing: $R=T_2(F_2)$.
The ring has two maximal right ideals: 
$M_1=\begin{bmatrix}0&F\\0&F\end{bmatrix}$ and 
$M_2=\begin{bmatrix}F&F\\0&0\end{bmatrix}$
We'll call $S_1=R/M_1$ and $S_2=R/M_2$, which are all of the the simple right $R$ modules.
It only has two minimal right ideals:
$N_1=\begin{bmatrix}0&F\\0&0\end{bmatrix}$ and 
$N_2=\begin{bmatrix}0&0\\0&F\end{bmatrix}$.
Now, $N_1\cong N_2$ as right $R$ modules via the obvious map, and they are also isomorphic to $R/M_2$ as a right module, and so their right annihilator is $M_2$.
But the annihilator of $S_1=R/M_1$ is $\begin{bmatrix}0&F\\0&F\end{bmatrix}$, so clearly $S_1$ is a simple right $R$ module that does not appear as a minimal right ideal of $R$.
But of course, you have the composition series $\{0\}\subseteq N_2\subseteq M_1\subseteq R$ whose factors exhibit both simple modules $S_1$ and $S_2$.
