Assume that $R$ is a ring with $1$ such that every $R$ module is injective. I've proved that $R$ is the finite direct sum of left ideals. I want to write $R$ as a ring direct product of $2$-sided ideals which have no sub-$2$-sided-ideals (so the factors are all simple/irreducible modules). The first step is to choose a minimal $2$-sided ideal $R_1$. Since $R$ is injective, we can decompose it as the direct sum of left ideals $R_1\oplus R'$. Now, I need to show that $R'$ is also a right ideal, which is where I'm stuck.
So say $r\in R'$, $s\in R$. In the direct sum, we would decompose as follows $r'=0+r'$. If we right multiply both sides by $s$, then we get $r's=0+r's$. Now, since $r's\in R$, then we can decompose it as $a+b$, where $a\in R_1$, $b\in R'$. Then we have $r's=a+b$, and now I'm not sure how to proceed. I need to prove that $r's=b$, or equivalently, that $a=0$, but I don't see any obvious reason why this must be true.