---
tag: 984525NJS0
kind: definition
name: Prime ideal of a lattice
msc: [06B10, 06D05]
defines: QKWVXJWHP6
uses: [WH4SFAACAS, 5GW2T2AM1A, J2T75ATRV5, Z15FXM9WM1, 0HM0HQCK6N]
sources:
  - url: https://en.wikipedia.org/wiki/Ideal_(order_theory)
---

## Definition

Let $(L, \vee, \wedge)$ be a [[lattice|WH4SFAACAS]].
A *prime ideal* of $L$ is an [[ideal|5GW2T2AM1A]] $I$ of $L$ such that:

1. $I \ne L$;
2. for all $x, y \in L$, if $x \wedge y \in I$, then $x \in I$ or $y \in I$.

## Remarks

A proper ideal $I$ is prime exactly when its complement $L \setminus I$ is a [[filter|Z15FXM9WM1]]; such a filter is called a prime filter.
Example: let $X$ be a set, $p \in X$, and $I = \set{A \subseteq X : p \notin A}$ in the power set of $X$.
$I$ is an ideal, it is proper since $X \notin I$, and if $p \notin A \cap B$ then $p \notin A$ or $p \notin B$.
Non-example: in the power set of $\set{1, 2, 3}$, the principal ideal $\set{\emptyset, \set{1}}$ is proper but not prime.
Indeed $\set{2} \cap \set{3} = \emptyset$ lies in it, while neither $\set{2}$ nor $\set{3}$ does.
In a [[distributive lattice|0HM0HQCK6N]], if an ideal $I$ and a filter $F$ are disjoint, then some prime ideal contains $I$ and is disjoint from $F$.
This is the prime ideal theorem for distributive lattices, and its proof uses a form of the axiom of choice.