# Prime Ideal is Prime Element

## Theorem

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

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

Then

$I$ is a prime ideal
$I$ is a prime element in $\left({ \mathit{Ids}\left({L}\right), \precsim}\right)$

where

$\mathit{Ids}\left({L}\right)$ denotes the set of all ideals in $L$,
$\mathord\precsim = \mathord\subseteq\restriction_{\mathit{Ids}\left({L}\right) \times \mathit{Ids}\left({L}\right) }$

## Proof

### Sufficient Condition

Let $I$ be a prime ideal.

Let $x, y \in \mathit{Ids}\left({L}\right)$ such that

$x \wedge y \precsim I$

By definition of $\precsim$:

$x \wedge y \subseteq I$
$x \cap y \subseteq I$

Aiming for a contraindication suppose

$x \not\precsim I$ and $y \not\precsim I$

By definition of $\precsim$: $x \nsubseteq I$ and $y \nsubseteq I$

By definition of subset:

$\exists a \in x: a \notin I$

and

$\exists b \in y: b \notin I$
$a \wedge b \preceq a$ and $a \wedge b \preceq b$

By definition lower set:

$a \wedge b \in x$ and $a \wedge b \in y$

By definitions subset and intersection:

$a \wedge b \in I$
$a \in I$ or $b \in I$

This contradicts $a \notin I$ and $b \notin I$

$\Box$

### Necessary Condition

Let $I$ be prime element in $\left({\mathit{Ids}\left({L}\right), \precsim}\right)$

Let $x, y \in S$ such that

$x \wedge y \in I$
$X := x^\preceq$ and $Y := y^\preceq$ are ideals in $L$
$X \cap Y = X \wedge Y$

We will prove that

$X \wedge Y \subseteq I$

Let $a \in X \wedge Y$.

By definition of intersection:

$a \in X$ and $a \in Y$

By definition of lower closure of element:

$a \preceq x$ and $a \preceq y$

By definition of infimum:

$a \preceq x \wedge y$

Thus by definition of lower set:

$a \in I$

$\Box$

By definition of $\precsim$:

$X \wedge Y \precsim I$

By definition of prime element:

$X \precsim I$ or $Y \precsim I$

By definition of $\precsim$:

$X \subseteq I$ or $Y \subseteq I$

By definition of reflexivity:

$x \preceq x$ and $y \preceq y$

By definition of lower closure of element:

$x \in X$ and $y \in Y$

Thus by definition of subset:

$x \in I$ or $y \in I$

Hence by Characterization of Prime Ideal:

$I$ is prime ideal.

$\blacksquare$