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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 $L$?

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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.

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  • $\begingroup$ @JosephVanName: Your first sentence isn't right; for instance, if $L$ is finite but not distributive, every ideal is principal, so finite meets distribute over joins of ideals (and $f$ is an isomorphism). You need to assume the lattice is (finitely) distributive to begin with. $\endgroup$ Jul 13, 2015 at 13:36
  • $\begingroup$ I see. You are correct. That was my silly error. $\endgroup$ Jul 13, 2015 at 13:46
  • $\begingroup$ @JosephVanName: Oh, it's a shame that you deleted your comments entirely though. The connection with compactifications of locales in the case that $L$ is distributive was still interesting! $\endgroup$ Jul 13, 2015 at 13:47
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    $\begingroup$ I must add that for distributive lattices (and clearly only for distributive lattices), the condition that finite meets distribute over joins of ideals if and only if finite meets distribute over arbitrary joins. i.e. if $L$ is a complete lattice, then $f:Id(L)→L$ is a lattice homomorphism if and only if $L$ satisfies $x\wedge\bigvee_{i\in I}y_{i}=\bigvee_{i\in I}(x\wedge y_{i})$, and this infinite distributivity law states that $L$ is a frame. If $L$ is a frame, then $f$ is a frame homomorphism. Furthermore, $Id(L)$ is a compact frame and the right adjoint of $f$ is a dense localic embedding. $\endgroup$ Jul 13, 2015 at 13:52
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    $\begingroup$ In fact, the Stone-Cech compactification of a completely regular frame $L$ is a subframe of $Id(L)$: If $L$ is a completely regular frame, then let $R(L)⊆Id(L)$ be the subframe consisting of all ideals $I\subseteq L$ such that if $x∈I$ then there is some $y∈I$ with $x≪y$. Then the mapping $f_{|R(L)}:R(L)\rightarrow L$ is the point-free analogue to the Stone-Cech compactification mapping. $\endgroup$ Jul 13, 2015 at 13:54

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