It is pretty routine to show that every simple left $R$-module is cyclically generated by any nonzero element and that these simple modules are precisely of the form $R/I$, where $I$ is some maximal ideal.
My question is what happens when you start to 'stitch' these together: let $S_1,S_2$ are distinct (in that they are not isomorphic) simple left $R$-modules. Then these are of the form $R/I, R/J$, where $I,J$ are maximal left ideals, respectively. My thought it is that $S_1 \oplus S_2 \cong R/(I \cap J)$. This makes since in that if $R$ were semisimple, then it is a sum of simple modules and its jacobson radical (the intersection of all maximal left ideals) is zero. Then extending my idea above, $I \cap J$ does seem to be the correct ideal. But I am unable to prove this.
My idea was to define a surjective map $\phi: R \to R/I \times R/J$ and use the first isomorphism theorem but the only obvious map is $r \mapsto (r,r)$ and its not clear that is surjective. [For a bit I thought it wouldn't be but perhaps it, non-obviously is?]
I could go the other way, $\psi: R/I \times R/J \to R/(I \cap J)$ via $(r,s) \mapsto rs+I\cap J$ (easy enough to show this is well-defined since this vanishes on $I \cap J$). This map is pretty clearly surjective. But injective? If $(r,s)$ were to map to zero, then $rs \in I \cap J$. But beyond that I'm not sure where to go with this. If the ring were commutative, I would have more since maximal implies prime and this would be something. (since this doesn't work, I'm really out of ideas)
Any ideas on how to prove this or can $S_1 \oplus S_2$ not be represented this way?