Maximal Ideal WRT Filter Complement is Prime in Distributive Lattice

Theorem
Let $L$ be a distributive lattice,

let $F$ be a filter in $L$, and

let $M$ be an ideal in $L$ which is disjoint from $F$ such that no ideal in $L$ larger than $M$ is disjoint from $F$.

Then $M$ is a prime ideal.

Proof
Suppose, for the sake of contradiction, that $M$ is not a prime ideal.

Then by Prime Ideals in Lattices, there are elements $a$ and $b$ of $L$ such that

$a \wedge b \in M$, $a \notin M$, and $b \notin M$.

Lemma
There do not exist $m$ and $n$ in $M$ such that $m \vee a \in F$ and $n \vee b \in F$.

Proof
Suppose such exist.

Since $m \vee \left({n \vee b}\right) \ge n \vee b$, $n \vee b \in F$, and $F$ is a filter,

$m \vee \left({n \vee b}\right) \in F$.

Applying associativity yields

$\left({m \vee n}\right) \vee b \in F$.

By the same argument,

$\left({m \vee n}\right) \vee a \in F$.

But then, again by the definition of a filter,

$\left({\left({m \vee n}\right) \vee b}\right) \wedge \left({\left({n \vee m}\right) \vee a}\right) \in F$.

Distributing,

$\left({m \vee n}\right) \wedge \left({b \vee a}\right) \in F$.

But by assumption, $b \vee a \in M$, and, by the definition of an ideal, $m \vee n \in M$,

so again by the definition of an ideal,

$\left({m \vee n}\right) \wedge \left({b \vee a}\right) \in M$, contradicting the fact that $M$ is disjoint from $F$.

Thus we can suppose without loss of generality that


 * $m \vee a \notin F$ for all $m \in M$.

Let $N = \left\{ {x \in L \mid x \le m \vee a \text{ for some }m \in M}\right\}$.

Lemma
$N$ is an ideal in $L$.

Proof
If $x \in N$, $y \in L$, and $y \le x$, then

by the definition of $N$ there exists an $m \in M$ such that $x \le m \vee a$.

Since $y \le x$, $y \le m \vee a$, so $y \in N$.

If $x \in N$ and $y \in N$, there exist $m_x$ and $m_y$ in $M$ such that

$x \le m_x \vee a$ and $y \le m_y \vee a$.

Then $x \vee y \le \left({m_x \vee a}\right) \vee \left({m_y \vee a}\right) = \left({m_x \vee m_y}\right) \vee a$.

But $m_x \vee m_y \in M$, so

$x \vee y \in N$.

Lemma
$M \subsetneq N$

Proof
If $m \in M$, then $m \le \left({m \vee a}\right)$, so $m \in N$.

Thus $M \subseteq N$.

$a \le \left({m \vee a}\right)$, so $a \in N$, but $a \notin M$.

Thus $M \subsetneq N$.

Lemma
$N \cap F = \varnothing$

Proof
Suppose $x \in N \cap F$.

Then $x \in N$, so for some $m \in M$, $x \le m \vee a$.

Furthermore, $x \in F$, so by the definition of a filter, $m \vee a \in F$.

But this contradicts our assumption that $m \vee a \notin F$ for all $m \in M$.

By assuming that $M$ is not a prime ideal, we have constructed an ideal $N$ properly containing $M$ that is disjoint from $F$, contradicting the maximality of $M$.

Thus, $M$ is a prime ideal.