Characterization of Prime Ideal

Theorem
Let $L = \left({S, \wedge, \preceq}\right)$ be a meet semilattice.

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

Then
 * $I$ is a prime ideal


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

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 $y \in I$

{{qed|lemma}}

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:
 * $x \wedge y \in F$

By Meet Precedes Operands:
 * $x \wedge y \preceq x$ and $x \wedge y \preceq y$

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

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:
 * $y \in F$

Thus by definition
 * $F$ is a filter in $L$.