# Characterization of Prime Ideal

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## Theorem

Let $L = \struct {S, \wedge, \preceq}$ be a meet semilattice.

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

Then

$I$ is a prime ideal
$\forall x, y \in S: \left({ x \wedge y \in I \implies x \in I \lor y \in I }\right)$

## Proof

### Sufficient Condition

Assume that

$I$ is a prime ideal.

Let $x, y \in S$ such that

$x \wedge y \in I$

By definition of relative complement:

$x \wedge y \notin \complement_S\left({I}\right)$

By definition of prime ideal:

$\complement_S\left({I}\right)$ is filter in $L$.

By {{Filtered in Meet Semilattice]]:

$x \notin \complement_S\left({I}\right)$ or $y \notin \complement_S\left({I}\right)$

Thus by definition of relative complement:

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

$\Box$

### Necessary Condition

Assume that

$\forall x, y \in S: \left({ x \wedge y \in I \implies x \in I \lor y \in I }\right)$

Define $F := \complement_S\left({I}\right)$.

By definition of proper subset:

$\exists x_0 \in S: x_0 \notin I$

By definition of relative complement:

$x_0 \in F$

Thus by definition

$F$ is a non-empty set.

We will prove that

$F$ is filtered.

Let $x, y \in F$.

By definition of relative complement:

$x \notin I$ and $y \notin I$

By assumption:

$x \wedge y \notin I$

By definition of relative complement:

$x \wedge y \in F$
$x \wedge y \preceq x$ and $x \wedge y \preceq y$

Thus

$\exists z \in F: z \preceq x \land z \preceq y$

$\Box$

We will prove that

$F$ is an upper set.

Let $x \in F, y \in S$ such that

$x \preceq y$

By definition of lower set:

$y \in I \implies x \in I$

By definition of relative complement:

$x \notin I$

Then

$y \notin I$

Thus by definition of relative complement:

$y \in F$

$\Box$

Thus by definition

$F$ is a filter in $L$.

$\blacksquare$