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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 $P$ 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 $\bigvee_L S \notin P$?

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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 $X$.)

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