Ideals in a noncommutative ring such that their product is their intersection? If $R$ is a commutative ring and $I$ and $J$ are ideals in $R$ such that $I+J=R$ then $I \cap J=IJ$. This is not generally true in noncommutative rings, e.g. let $R$ be the lower triangular 2 x 2 matrices over $\mathbb{Z}$ and let $I$ be the ideal generated by $E_{1,1}$ and $J$ be the ideal generated by $E_{2,2}$. 
Is this ever possible in a noncommutative ring? If so, what are rings satisfying this property called? (Where can we find them in the literature?)  
 A: This is more a remark, since I do not directly answer the question. The statement in the question is true for all ideals $I,J$ (without the condition $I+J=R$) if and only if all ideals a idempotent, i.e. $I^2=I$. Indeed,
$$I \cap J = (I \cap J)^2 \subset IJ \subset I \cap J.$$ The converse is obvious by taking $I=J$.
This holds for instance if $R$ is a $C^\star$-algebra.
A: I don't know non-commutative ring theory that well yet, but I ended up working this problem out anyway partially for my own edification.  Let us suppose that I, J are both left ideals in R.   Observe,
$(I \cap J) (I + J) = (I \cap J) I  + (I \cap J) J \subseteq IJ + JI$
If I + J = (1), then we get
$I \cap J \subseteq IJ + JI$
Now it can be shown that 
$I J \subseteq comm(I,J) \cap J$
where comm(I,J) denotes the set of all elements in I which commute with elements in J;
$comm(I, J) = ${$ x \in I : \forall y \in J; x y \in I $}
Therefore,
$I \cap J \subseteq IJ + JI \subseteq comm(I,J) \cap J + comm(J,I) \cap I$
If it is true that all of I commutes with J and vice-versa, then $comm(I,J) = I$, $comm(J,I) = J$ and we get:
$I \cap J = I J$
Now let us suppose that these bounds are not tight, or in other words there is some element $x \in I \setminus comm(I,J)$.  This implies that there exists some $y \in J$ such that $x y \not \in I$, and so $x J \not \subseteq I \cap J$.
Therefore if $I + J = (1)$, then $I \cap J = I J + J I$ if and only if $comm(I,J) = I$ and $comm(J,I) = J$.
(Of course there is probably an easier way to say this, so maybe it is also good to get an expert to weigh in on the issue.)
