Elements of bounded distributive lattice belonging to same prime ideals are equal?

Question

I have read in a paper that by an easy application of Zorn's lemma one may show that two elements of a bounded distributive lattice are equal iff they are contained in exactly the same prime ideals of the lattice.

1. What is the intuition behind this fact?
2. How to actually prove it?
3. Is there a "constructive version" avoiding prime ideals?

Answer 1

1. The intuition is that if you have $a \neq b$, then $a \nleq b$ or $b \nleq a$ (possibly both); in the first case, there is an ideal $I$ containing $b$ but not $a$ (for example, $(b]$, the ideal generated by $b$), and it can be extended to a prime ideal $J$ which doesn't contain $a$.
2. The proof follows in the end.
3. I don't think there is what we could call a constructive version. Perhaps it doesn't take the full power of AC/ZL, but it certainly needs something similar (not sure what...)

Now the proof.
Suppose $L$ is a distributive lattice (if I got it right, it doesn't need to be bound), and $a,b\in L$ are such that $a \nleq b$. Define $\mathcal{I}(L)$ to be the set of ideals of $L$. Let $$\mathcal{X} = \{ K \in \mathcal{I}(L): b \in K, a \notin K \}.$$ $(b] \in \mathcal{X}$, whence $\mathcal{X} \neq \varnothing$.
Let $\mathcal{C} = \{K_i:i \in I\}$ be a chain of elements from $\mathcal{X}$, and let $K = \bigcup_{i \in I} K_i$.
It's clear that $a \notin K$ and $b \in K$ and it is easy to see that $K$ is an ideal. So $K \in \mathcal{X}$.
By LZ, there exists a maximal element $I$ in $\mathcal{X}$.

Let us see that $I$ is a prime ideal.
Suppose (for a contradiction) that $u, v \in L \setminus I$ are such that $u \wedge v \in I$.
Let $I_u = (I \cup \{u\}]$. By the maximality of $I$ in $\mathcal{X}$, we have that $I_u \notin \mathcal{X}$.
Since $b \in I_u$, we conclude that $a \in I_u$.
Hence, by the definition of $I_u$, there exists $k_u \in I$ such that $a \leq u \vee k_u$;
similarly, there exists $k_v \in I$ such that $a \leq v \vee k_v$. Now $$a \leq (u \vee k_u \vee k_v) \wedge (v \vee k_v \vee k_u) = (u \wedge v) \vee (k_u \vee k_v) \in I,$$ so that $a \in I$, a contradiction.