Prime Ideal in Lattice
Theorem
Let $\struct {L, \le}$ be a lattice.
Let $I$ be an ideal in $L$.
Then $I$ is a prime ideal if and only if:
- $\forall a, b \in L: a \wedge b \in I \implies a \in I \text{ or } b \in I$
where $a \wedge b$ denotes $\min \set {a, b}$, the meet of $a$ and $b$.
Proof
Necessary Condition
Let $I$ be a prime ideal.
Let $a, b \in L$ such that $a, b \notin I$.
Then $a, b \in L \setminus I$.
By the definition of prime ideal $L \setminus I$ is a filter.
By the definition of a filter:
- $\exists c \in L \setminus I: c \le a, c \le b$
By the definition of meet:
- $c \le a \wedge b$
Since $c \in L \setminus I$, the definition of a filter implies that
- $a \wedge b \in L \setminus I$
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Thus $a \wedge b \notin I$.
By contraposition it follows that $a \wedge b \in I$ implies that $a \in I$ or $b \in I$.
$\Box$
Sufficient Condition
For all $a$ and $b$ in $L$, let $a \wedge b \in I$ imply that $a \in I$ or $b \in I$.
Let $x, y \in L \setminus I$.
Then $x, y \notin I$.
By supposition, $x \wedge y \notin I$, so:
- $x \wedge y \in L \setminus I$
Let $p \in L \setminus I, q \in L$ such that $p \le q$.
Let $q \in I$.
Then by definition of ideal, $p \in I$.
This contradicts $p \in L \setminus I$.
Thus $q \notin I$.
Thus $q \in L \setminus I$.
Therefore, $L \setminus I$ is a filter.
$\blacksquare$