What can be said about $I\cap J$ when $I$ is a Left Ideal of $R$ and $J$ is a Right Ideal?

Question

Is there anything one can say about $I\cap J$ when $I$ is a left ideal of $R$ and $J$ is a right ideal?

All I can think of is that $IJ\subseteq I\cap J$.

Edit:

Here is a counter example to show that it needn't be a left, right or two sided ideal. Let $R$ be a ring, and condsider the two by two $R$-matrices. We have left/right ideals $$I=\left\{ \left(\begin{array}{cc} a & 0\\ b & 0 \end{array}\right)\vert a,b\in R\right\} ,J=\left\{ \left(\begin{array}{cc} a & b\\ 0 & 0 \end{array}\right)\vert a,b\in R\right\} $$ but the intersection is not an ideal.

However, it is easy to show that $I\cap J$ is always a subring.

Answer 1

The property of the intersection does not even hold.

Take the left ideal $L$ from $M_2(F)$ whose elements are all zero in the right column, and the right ideal $T$ of matrices with zeros on the bottom row. Then $LT\nsubseteq T$.

But $TL\subseteq T\cap L$, naturally.

From this example you can also see the intersection is not an ideal on either side.

You can say the intersection is an additive subgroup, but not anything else really.

The product $LT$ meaning the set of finite sums of pair wise products, is a two sided ideal too.

Answer 2

I just wanted to add to this that there is one interesting property: $I\cap J$ is a $J$-$I$ bimodule, i.e. if $K=I\cap J$ then

$$KI\subset K$$ $$JK\subset K$$

Answer 3

It is an additive subgroup, from $I$ and $J$ being additive subgroups themselves. Apart from this I noticed, for any element $x\in I\cap J$, $$xl\in I \quad \forall \quad l\in I $$ which follows from multiplicative closure of $I$. Also $$xl\in J$$ as $J$ is a right ideal. This tells us $$xl\in I\cap J$$ Similarly , $$rx\in I\cap J\quad \forall \quad r\in J$$ So $I\cap J$ sort of behaves like a right ideal for the elements of $I$ and a left ideal for the elements of $J$. I'm not sure if this is what the question expects, but I hope it provides some insight.