Let $L, K$ be complete lattices. A lattice homomorphism $f: L\to K$ is said to be *incomplete* if there is an infinite set $S \subseteq L$ such that $f(\bigvee\_L S) \neq \bigvee\_K f(S).$


Suppose that $L$ contains a [prime ideal](https://en.wikipedia.org/wiki/Ideal_%28order_theory%29#Prime_ideals) $P$ such that there is $S\subseteq P$ with $\bigvee\_L S \notin P$. Then we can define $f:L\to K$ by sending $P$ to the smallest element of $K$ and $L\setminus P$ to the largest element of $K$ and easily see that $f$ is an incomplete lattice homomorphism.


Does the converse of the above hold? More exactly: Assume $L, K$ are complete lattices and $f:L\to K$ is an incomplete lattice homomorphism. Does this imply that $L$ contains a prime ideal $P$ such that there is $S\subseteq P$ with $\bigvee\_L S \notin P$?



Incomplete lattice homomorphisms between complete lattices
Let $h:\mathbb R\to \mathbb R$ be an incomplete lattice homomorphism, e.g. the identity below $0$, and $x\mapsto x+1$ on $\mathbb R\_{\ge 0}$. 


Let $X\_1$ and $X\_2$ be disjoint copies of the reals, and let $X:=X\_1\cup X\_2 \cup \{\infty, -\infty\}$ be the horizontal sum of those two linear orders, with new last and greatest elements. 


Then $X$ is a complete lattice. 


Let $f:X\to X$ be defined as follows: $f\upharpoonright X\_1 $ is the identity, 
$f\upharpoonright X\_2:X\_2\to X\_2$ is defined using $h$. Then $f$ is an incomplete lattice homomorphism. But the only prime ideals on $X$ are $X\_1\cup \{-\infty\}$ and $X\_2\cup \{-\infty\}$. (And, depending on your definition, also $\emptyset$ and $X$.)


