Theorem: Let $R$ be a Wedderburn ring, and let $\mathcal{S}$ be a representative set of the simple right $R-$modules. For each $S \in \mathcal{S}$, let $I_S$ be the sum of all those minimal right ideals of $R$ that are isomorphic to $S$. Then:
$|\mathcal{S}|$ is finite.
Each $I_S$ is a minimal (nonzero, two-sided) ideal of $R$.
$R$ is the direct sum of the ideals $I_S$.
Each ideal $I_S$ is a simple right artinian subring of $R.$
If $S,T \in \mathcal{S}$, with $S \not \cong T$, then $T I_S = 0$.
In particular, I have a question about the minimality argument of part (2).
We suppose that $I$ is an ideal of $R$ so that $I < I_S$. Since $I$ is proper, we can choose a minimal right ideal $J \leq I_S$ with $J \cong S$ and $J \not \subseteq I$. Then $J \cap I$ is a right ideal of $R$ properly contained in $J$, and hence $J \cap I = 0$. However, $JI \subseteq J \cap I$ since $I$ is a left ideal, so $JI = 0$.
Al this is fine with me. Next Isaacs writes: It follows that $I_S I = 0$ since $I_S$ is a sum of modules isomorphic to $J$, and each is annihilated by I.
But we have only shown that one of the modules isomorphic to $J$ is annihilated by $I$. Why does it follow that all of them are annihilated by $I$?
My thoughts: If $J'$ is another minimal right module isomorphic to $J$, then $J J' \subseteq J$. If $J J' > 0$, then $J J' = J$ since $J$ is right simple. So if $I$ contains any right module isomorphic to $J$, then it contains all of them - from which the result would follow. I am not sure how to argue that $J J' > 0$, or even if it is true...
Am I looking in the right direction?
The answer was very simple: $J$ and $J'$ are by definition both isomorphic as $R$-modules, so naturally if $I$ annihilates one then it annihilates the other. From this it follows that $I_S I = 0$, so $R I = 0$, implying that $I = 0$.