Let $R$ be a semisimple ring, $I$ be a right ideal of $R$ and $J$ be a left ideal. We wish to show $IJ = I \cap J$.
Clearly $IJ \subseteq I \cap J$. Since $R$ is semisimple and $J$ is a left $R$-modules and a submodule of the left regular module we have
$$R=J \oplus L$$
for some complement $L \subseteq R$ of $J$. Further, $IJ$ and $I \cap J$ are both submodules (in fact, they are ideals) so we have
$$R = I \cap J \oplus K \quad \text{and} \quad R=IJ \oplus S$$
Further, $I \cap J$ is semisimple and $IJ$ is a submodule of $I \cap J$ so we have
$$I \cap J= IJ \oplus Q$$
I've tried various ways of playing with the above equalities and I believe that I need to find a complement in a clever way such that it shows the $Q$ above must be 0. But I am quite stuck.
Best Answer
It is clear the $IJ \subseteq I \cap J$. Now, we have
$$R = J \oplus L$$ since $R$ is a minimal ideal. Hence, for some $j \in J$ and $l \in L$ we have
$$1=j+l$$.
Now, let $x \in I \cap J$ and conisder $x=xj+xl$. Then, in particular, $x \in J$ so $xl=0$. Hence
$$x=xj \in IJ$$