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

if and only if:

$\forall x, y \in S: \paren {x \wedge y \in I \implies x \in I \lor y \in I}$

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:

$x \wedge y \notin \relcomp S I$

By definition of prime ideal:

$\relcomp S I$ is filter in $L$.

By Filtered in Meet Semilattice:

$x \notin \relcomp S I$ or $y \notin \relcomp S I$

Thus by definition of relative complement:

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

$\Box$

Necessary Condition

Assume that

$\forall x, y \in S: \paren {x \wedge y \in I \implies x \in I \lor y \in I}$

Define $F := \relcomp S I$.

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

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

$\Box$

We will prove that:

$F$ is an upper section.

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

$x \preceq y$

By definition of lower section:

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

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 $L$.

$\blacksquare$

Sources

• 1980: G. Gierz, K.H. Hofmann, K. Keimel, J.D. Lawson, M.W. Mislove and D.S. Scott: A Compendium of Continuous Lattices

• Mizar article WAYBEL_7:18