Prime Ideal is Prime Filter in Dual Lattice

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

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

Then
 * $X$ is a prime ideal in $L$


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

where $L^{-1} = \left({S, \succeq}\right)$ denotes the dual of $L$.

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 filter in $L^{-1}$.

Let $x, y \in S$ such that
 * $x \vee' y \in X$

where $\vee'$ denotes 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 prime filter

Necessary Condition
This follows by mutatis mutandis.