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


*

*What is the intuition behind this fact?

*How to actually prove it?

*Is there a "constructive version" avoiding prime ideals?

 A: *

*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 $a$.

*The proof follows in the end.

*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.
