# If Ideal and Filter are Disjoint then There Exists Prime Ideal Including Ideal and Disjoint from Filter

## Theorem

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

Let $I$ be an ideal in $L$.

Let $F$ be a filter on $L$ such that

$I \cap F = \O$

Then there exists a prime ideal $P$ in $L$: $I \subseteq P$ and $P \cap F = \O$

## Proof

Define $X := \set {P \in \map {\operatorname {Ids} } L: I \subseteq P \land P \cap F = \O}$

where $\map {\operatorname {Ids} } L$ denotes set of all ideals in $L$.

$I \in X$

We will prove that

$\forall Z: Z \ne \O \land Z \subseteq X \land \paren {\forall Y_1, Y_2 \in Z: Y_1 \subseteq Y_2 \lor Y_2 \subseteq Y_1} \implies \bigcup Z \in X$

Let $Z$ such that

$Z \ne \O$ and
$Z \subseteq X$ and
$\forall Y_1, Y_2 \in Z: Y_1 \subseteq Y_2 \lor Y_2 \subseteq Y_1$

By definition of $X$:

$Z$ is set of subsets of $S$.

By definition of $X$:

$\forall Y \in Z: Y$ is lower set.
$J := \bigcup Z$ is lower set.

By definition of non-empty set:

$\exists Y: Y \in Z$

By definition of subset:

$Y \in X$

By definition of $X$:

$I \subseteq Y$
$Y \subseteq J$
$I \subseteq J$

We will prove that

$\forall A, B \in Z: \exists C \in Z: A \cup B \subseteq C$

Let $A, B \in Z$.

By assumption:

$A \subseteq B$ or $B \subseteq A$
$A \cup B = B$ or $A \cup B = A$

Thus by Set is Subset of Itself:

$\exists C \in Z: A \cup B \subseteq C$

$\Box$

By definition of $X$:

$\forall Y \in Z: Y$ is directed.
$J$ is directed.

By definition of ideal in ordered set:

$J$ is an ideal in $L$.

We will prove that

$J \cap F = \O$

Let $x \in J$.

By definition of union:

$\exists A \in Z: x \in A$

By definition of subset:

$A \in X$

By definition of $X$:

$A \cap F = \O$

Thus by definitions of intersection and empty set:

$x \notin F$

$\Box$

Thus by definition of $X$:

$\bigcup Z \in X$

$\Box$

By Zorn's Lemma:

there exists $Y \in X$: $Y$ is maximal set of $X$.

Then by definition of $X$:

$Y \in \map {\operatorname {Ids} } L$ and $I \subseteq Y$ and $Y \cap F = \O$

We will prove that

$Y$ is a prime ideal.

Let $x, y \in S$ such that

$x \wedge y \in Y$

$x \notin Y$ and $y \notin Y$

Define $P_y = \map {\operatorname{finsups} } {Y \cap \set y}^\preceq$

$Y \cup \set y \subseteq P_y$
$Y \subseteq Y \cup \set y$
$Y \subseteq P_y$
$I \subseteq P_y$

By definition of singleton:

$y \in \set y$

By definition of union:

$y \in Y \cup \set y$

By definition of subset:

$y \in P_y$

We will prove that

$P_y \cap F \ne \O$

$P_y \cap F = \O$

By definition of $X$:

$P_y \in X$

By definition of minimal set:

$Y = P_y$

A contradiction between $y \notin Y$ and $y \in P_y$

$\Box$

By definitions of non-empty set and intersection:

$\exists v: v \in P_y \land v \in F$

Define $P_x = \map {\operatorname{finsups} } {Y \cap \set x}^\preceq$

Analogically:

$\exists u: u \in P_x \land u \in F$
$\exists u' \in Y: u \preceq u' \vee \sup \set x$
$\exists v' \in Y: v \preceq v' \vee \sup \set y$
$\paren {v' \vee u'} \vee x = v' \vee \paren {u' \vee x}$
$\paren {v' \vee u'} \vee x \succeq u' \vee x$
$\sup \set x = x$

By definition of transitivity:

$u \preceq v' \vee u' \vee x$

By definition of upper set:

$u' \vee v' \vee x \in F$

Analogically:

$u' \vee v' \vee y \in F$
$\paren {u' \vee v' \vee x} \wedge \paren {u' \vee v' \vee y} \in F$

By definition of distributive lattice:

$\paren {u' \vee v'} \vee \paren {x \wedge y} \in F$
$u' \vee v' \in Y$
$\paren {u' \vee v'} \vee \paren {x \wedge y} \in Y$

This contradicts $Y \cap F = \O$

$\blacksquare$