Maximal Ideal WRT Filter Complement is Prime in Distributive Lattice

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Theorem

Let $L$ 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

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 $F$.

$\Box$


Without loss of generality, we can thus suppose that:

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


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


Lemma

$N$ is an ideal in $L$.


Proof

Let:

$x \in N$
$y \in L$
$y \le x$

Then by the definition of $N$ there exists an $m \in M$ such that:

$x \le m \vee a$

Since $y \le x$, it follows that:

$y \le m \vee a$

so $y \in N$.


Let:

$x \in N$
$y \in N$

Then there exist $m_x$ and $m_y$ in $M$ such that:

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

Then:

$x \vee y \le \paren {m_x \vee a} \vee \paren {m_y \vee a} = \paren {m_x \vee m_y} \vee a$

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

$x \vee y \in N$

$\Box$


Lemma

$M \subsetneq N$


Proof

Let $m \in M$.

Then:

$m \le \paren {m \vee a}$

so $m \in N$.

Thus $M \subseteq N$.

We have:

$a \le \paren {m \vee a}$

so:

$a \in N$

but:

$a \notin M$

Thus:

$M \subsetneq N$

$\Box$


Lemma

$N \cap F = \O$


Proof

Aiming for a contradiction, 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 $\forall m \in M: m \vee a \notin F$.

$\Box$


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.

$\blacksquare$