Prime Ideal is Prime Filter in Dual Lattice

Theorem

Let $L = \struct {S, \vee, \wedge, \preceq}$ be a lattice.

Let $X$ be a subset of $S$.

Then

$X$ is a prime ideal in $L$

if and only if:

$X$ is a prime filter in $L^{-1}$

where $L^{-1} = \struct {S, \succeq}$ denotes the dual of $L$.

Proof

Sufficient Condition

Let $X$ be a prime ideal in $L$.

Then

$X$ is an ideal in $L$.

By Ideal is Filter in Dual Ordered Set:

$X$ is a filter in $L^{-1}$.

Let $x, y \in S$ such that

$x \vee' y \in X$

where $\vee'$ denotes the join in $L^{-1}$.

By Join is Dual to Meet:

$x \wedge y \in X$

Thus by Characterization of Prime Ideal:

$x \in X$ or $y \in X$

Hence $X$ is a prime filter

$\Box$

Necessary Condition

This follows by mutatis mutandis.

$\blacksquare$

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_7:16