For any lattice $L$ we denote the complete lattice of the ideals of $L$ by ${\cal Id}(L)$. If $L$ is complete, is there a lattice homomorphism from ${\cal Id}(L)$ onto $L$?



Lattice homomorphism from ${\cal Id}(L)$ onto $L$
Here's a counterexample. Let $L=\{0,1,x\_0,x\_1,x\_2,\dots,y\}$, where $x\_0<x\_1<x\_2<\dots$ and $y$ is incomparable with every $x\_n$. Then the only non-principal ideal in $L$ is $I=\{0,x\_0,x\_1,\dots\}$; identify $\mathcal{Id}(L)$ with $L\cup\{I\}$. It is easy to see that any surjective homomorphism $\mathcal{Id}(L)\to L$ is forced to send $I$ to $1$ and $y$ to $y$, which is a contradiction since $I\wedge y=0$.


On the other hand, for arbitrary $L$, consider $f:\mathcal{Id}(L)\to L$ defined by $f(I)=\bigvee I$. This is always surjective and preserves arbitrary joins, and preserves finite meets as long as finite meets distribute over joins of ideals in $L$. So if finite meets distribute over joins of ideals (aka directed joins) in $L$, $f$ is such a homomorphism.


