The lattice of ideals of a distribituve lattice is itself distributive

Question

I have found myself stuck on a problem and would appreciate a hint. The problem is to show that the lattice of ideals $I(L)$ of a distributive lattice $L$ is itself distributive. This is question 2 in section 3 of the first chapter in the universal algebra book by Burris and Sankappanavar.

Answer 1

Hint: to show that $$(I \vee J) \wedge K = (I \wedge K) \vee (J \wedge K),$$ where $I, J$ and $K$ are ideals of the lattice $L$, notice that meet is given by intersection, and $$I \vee J = \{ a \in L : \exists i\in I\;\exists j \in J\;(a \leq i \vee j) \}.$$ Additionally, you might find useful the fact that if $k \leq i \vee j$ then $k = k \wedge (i \vee j)$, and use distributivity.

I don't think that I can give you any more hints without spoiling it. Good luck!