I have been trying to read ring theory on my own, from Dummit and Foote, and this question has been on my mind for a while:
Let $I,J,K$ be ideals of a ring $R$ (may be non commutative, but has a $1$). Can we say the following:
$$I \cap (J + K) = I \cap J + I \cap K\ ?$$
I get that the RHS is a subset of the LHS, but I am not able to say anything about the other inclusion. Maybe someone can provide a counterexample?
Also, what about the case when $R$ does not have a $1$?
Best Answer
Let $R=F_2[x,y]/(x,y)^2$ where $F_2$ is the field of two elements. I find this ring interesting because its lattice of proper ideals is precisely the diamond lattice $M_3$.
Let $I=(x)$, $J=(y)$, $K=(x+y)$. Then $I\cap J$ and $I\cap K$ are both zero, but $I\cap (J+K)=I$.
If you really want an example like that, just pick your favorite ring without identity $S$, and use $R$ above and find the counterexample embedded in $R\times S$.
Rings which satisfy that property on right ideals are called right distributive, but I do not know if anyone has studied the definition that is restricted to two-sided ideals.