Characterization of Pseudoprime Element when Way Below Relation is Multiplicative

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Theorem

Let $L = \struct {S, \vee, \wedge, \preceq}$ be a bounded below continuous lattice such that

$\ll$ is multiplicative relation

where $\ll$ denotes the way below relation of $L$.

Let $p \in S$.


Then $p$ is pseudoprime element if and only if

$\forall a, b \in S: a \wedge b \ll p \implies a \preceq p \lor b \preceq p$


Proof

Sufficient Condition

Let $p$ be pseudoprime element.

Let $a, b \in S$ such that

$a \wedge b \ll p$

By definition of meet:

$\inf \left\{ {a, b}\right\} \ll p$

By Characterization of Pseudoprime Element by Finite Infima:

$\exists c \in \left\{ {a, b}\right\}: c \preceq p$

Thus

$a \preceq p$ or $b \preceq p$

$\Box$


Necessary Condition

Suppose

$\forall a, b \in S: a \wedge b \ll p \implies a \preceq p \lor b \preceq p$

Aiming for a contradiction, suppose:

$p$ is not a pseudoprime element.

By Prime is Pseudoprime (Order Theory):

$p$ is not a prime element.

By definition of prime element:

$\exists x, y \in S: x \wedge y \preceq p$ and $x \npreceq p$ and $y \npreceq p$

By definition of continuous:

$\forall z \in S: z^\ll$ is directed.

and

$L$ satisfies axiom of approximation.

By Axiom of Approximation in Up-Complete Semilattice:

$\exists u \in S: u \ll x \land u \npreceq p$

and

$\exists v \in S: v \ll y \land v \npreceq p$

By Way Below Relation is Auxiliary Relation:

$\ll$ is auxiliary relation.

By Multiplicative Auxiliary Relation iff Congruent:

$u \wedge v \ll x \wedge y$

By definition of transitivity:

$u \wedge v \ll p$

By assumption:

$u \preceq p$ or $v \preceq p$

This contradicts $u \npreceq p$ and $v \npreceq p$.

$\blacksquare$


Sources

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