Let $R$ be a commutative ring with $1$. If $I+J=R$, then $IJ = I \cap J$. The post below has already given a solution. However, I am wondering what happens if $R$ is not commutative? Can anyone provide me with a counterexample? If I am not wrong, the counterexample given in the post is the case when $R$ is commutative and does not have unity. Which part of the proof uses commutativity of the ring? Thank you.
If $I+J=R$, where $R$ is a commutative rng, prove that $IJ=I\cap J$.
It suffices to find ideals $I,J$ such that $I+J=R$ and $IJ \neq JI$. In fact, then $IJ = I \cap J$ and $JI = J \cap I$ cannot hold both, so that either $(I,J)$ or $(J,I)$ is a counterexample, or both, as below.
Consider $R=\mathbb{Z}\langle x,y \rangle / (2x+2y=1)$, $I=(x)$, $J=(y)$. Then $I+J=R$, and one can prove that $I \cap J = (xy,yx)$, which strictly contains $IJ=(xy)$. (In order to simplify calculations, you may work with $R/(xy)$.)
Here is a positive result:
Proof: The inclusion $IJ +JI \subseteq I \cap J$ follows from $IJ \subseteq I$, $IJ \subseteq J$, $JI \subseteq I$, $JI \subseteq J$. Conversely, $I \cap J = (I \cap J)R = (I \cap J)(I+J) = (I \cap J)I + (I \cap J)J \subseteq JI+IJ$. $\checkmark$