Maximal Ideal WRT Filter Complement is Prime in Distributive Lattice/Lemma 1

Lemma for Maximal Ideal WRT Filter Complement is Prime in Distributive Lattice

Let $\struct {S, \vee, \wedge, \preceq}$ be a distributive lattice.

Let $F$ be a filter in $L$.

Let $M$ be an ideal in $L$ which is disjoint from $F$ such that:

no ideal in $L$ larger than $M$ is disjoint from $F$.

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

Proof

Aiming for a contradiction, suppose such exist.

Since:

$m \vee \paren {n \vee b} \ge n \vee b$
$n \vee b \in F$
$F$ is a filter

it follows that:

$m \vee \paren {n \vee b} \in F$

Applying associativity yields:

$\paren {m \vee n} \vee b \in F$

By the same argument:

$\paren {m \vee n} \vee a \in F$

By the definition of a filter:

$\paren {\paren {m \vee n} \vee b} \wedge \paren {\paren {n \vee m} \vee a} \in F$

Distributing:

$\paren {m \vee n} \wedge \paren {b \vee a} \in F$

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:

$\paren {m \vee n} \wedge \paren {b \vee a} \in M$

contradicting the supposition that $M$ is disjoint from $F$.

$\blacksquare$