Does ${\cal Id}(L) \cong {\cal Id}(K)$ imply $L\cong K$?

Question

For any lattice $L$ we denote the complete lattice of the ideals of $L$ by ${\cal Id}(L)$. Are there non-isomorphic lattices $L\not \cong K$ such that ${\cal Id}(L) \cong {\cal Id}(K)$?

Answer 1

No; in fact, we can canonically recover $L$ from $\mathcal{Id}(L)$ as the sublattice of compact elements (that is, elements $x$ such that whenever $x=\bigvee S$, there is a finite subset $F\subseteq S$ such that $x=\bigvee F$). It is clear that any principal ideal is compact; conversely, any compact ideal is a finite join of principal ideals and hence principal itself.