# Prime Ideal in Lattice

## Theorem

Let $\left({L, \le}\right)$ 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 $\min \left\{{a, b}\right\}$, 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

- $a \wedge b \in L \setminus I$

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.

$\blacksquare$