Maximal Ideal WRT Filter Complement is Prime in Distributive Lattice

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Theorem

Assertions

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 $F$.

Then $M$ is a prime ideal.

Proof

Aiming for a contradiction, suppose $M$ is not a prime ideal.

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

$a \wedge b \in M$

$a \notin M$

$b \notin M$

Lemma 1

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

$\Box$

Without loss of generality, we can thus suppose that:

$\forall m \in M: m \vee a \notin F$

Let $N = \set {x \in L: \exists m \in M: x \le m \vee a}$.

Lemma 2

$N$ is an ideal in $L$.

$\Box$

Lemma 3

$M \subsetneq N$

$\Box$

Lemma 4

$N \cap F = \O$

$\Box$

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

This contradicts the maximality of $M$.

Thus $M$ is a prime ideal.

$\blacksquare$