Commutativity of a prime ring

Question

Question :

Show that $R$ is a prime ring containing two commuting non-zero left ideals $I$ and $J$ $\implies$ $R$ is commutative, where "commuting ideals” means $ij=ji$ for all $i \in I$, $j \in J$.

My attempt : Let $x, y \in R, \quad xIyJ=yJxI$ and I need a hint to accomplish the proof.

Answer 1

Since $R$ is prime, and $I,J$ are nonzero left ideals, we know that $xI=0$ or $xJ=0$ implies $x=0$.

Take any $x\in R$, $i\in I$ and $j\in J$. We have $$ i(xj)=(xj)i=x(ji)=x(ij) $$ so that $(ix-xi)j=0$. Thus $(ix-xi)J=0$, and hence $ix=xi$.

Now take any $x,y\in R$ and $i\in I$. We have $$ y(xi)=(xi)y=x(iy)=x(yi) $$ so that $(xy-yx)i=0$. Thus $(xy-yx)I=0$, and hence $xy=yx$.

We conclude that $R$ is commutative.