# 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?)

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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.

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This is helpful, Andreas. Thanks! –  Jon Bannon Sep 17 '10 at 20:56

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.)

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Since the original question was asked about (two-sided) ideals rather than one-sided ideals, you might as well restrict your argument to that case. Then your observation above greatly simplifies: $I \cap J = IJ + JI$ for ANY pair of two-sided ideals $I$ and $J$. This makes it clear that the original question is asking for situations in which $IJ + JI = IJ$ - that is, situations in which $JI \subseteq IJ$. –  Manny Reyes Sep 18 '10 at 17:11
Hmm, you have a good point. I must admit to being somewhat unfamiliar with the conventions regarding non-commutative rings. My initial reading was that he just said `ideals' without any further qualification and so I chose to examine the most restrictive instance of this class. However, making the assumption that the ideals are two-sided does greatly simplify things. –  Mikola Sep 18 '10 at 22:07