Birkhoff prime ideal theorem
Birkhoff Prime Ideal Theorem. Let $L$ be a distributive lattice^{} and $I$ a proper lattice ideal of $L$. Pick any element $a\notin I$. Then there is a prime ideal^{} $P$ in $L$ such that $I\subseteq P$ and $a\notin P$.
Proof.
If $I$ is prime, then we are done. Let $S:=\{J\mid J\text{is an ideal in}L\text{, and}a\notin J\}$. Then $I\in S$. Order $S$ by inclusion. This turns $S$ into a poset. Let $C$ be a chain in $S$. Let $K=\bigcup C$. If $x,y\in K$, then $x\in {J}_{1}$ and $y\in {J}_{2}$ for some ideals ${J}_{1},{J}_{2}\in C$. Since $C$ is a chain, we may assume that ${J}_{1}\subseteq {J}_{2}$, so that $x\in {J}_{2}$ as well. This means $x\vee y\in {J}_{2}\subseteq K$. Next, assume $x\in K$ and $y\le x$. Then $x\in J$ for some ideal $J\in C$, so that $y\in J\subseteq K$ also. This shows that $K$ is an ideal. If $a\in K$, then $a\in J$ for some $J\in C\subseteq S$, contradicting the definition of $S$. So $a\notin K$ and $K\in S$ also. This shows that every chain in $S$ has an upper bound. We can now appeal to Zorn’s lemma, and conclude that $S$ has a maximal element^{}, say $P$.
We now want to show that $P$ is the candidate that we are seeking: $P$ is a prime ideal in $L$ and $a\notin P$. Since $P\in S$, $P$ is an ideal such that $a\notin P$. So the only thing left to prove is that $P$ is prime. This amounts to showing that if $x\wedge y\in P$, then $x\in P$ or $y\in P$. Suppose not: $x,y\notin P$. Let ${Q}_{1}$ be the ideal generated by elements of $P$ and $x$, and ${Q}_{2}$ the ideal generated by $P$ and $y$. Since ${Q}_{1}$ and ${Q}_{2}$ properly contain $P$, $a\in {Q}_{1}$ and $a\in {Q}_{2}$. Write $a\le {p}_{1}\vee x$ and $a\le {p}_{2}\vee y$, where ${p}_{1},{p}_{2}\in P$. Then $a\vee {p}_{2}\le ({p}_{1}\vee {p}_{2})\vee x$ and $a\vee {p}_{1}\le ({p}_{1}\vee {p}_{2})\vee y$. Take the meet of these two expressions, and we obtain $(a\vee {p}_{2})\wedge (a\vee {p}_{1})\le (({p}_{1}\vee {p}_{2})\vee x)\wedge (({p}_{1}\vee {p}_{2})\vee y)$. Since $L$ is distributive, on the left hand side, we get $a\vee ({p}_{1}\wedge {p}_{2})$. On the right hand side, we have $({p}_{1}\vee {p}_{2})\vee (x\wedge y)\in P$. As the left hand side is less than or equal to the right hand side, we get that $a\vee ({p}_{1}\wedge {p}_{2})\in P$. Since $a\le a\vee ({p}_{1}\wedge {p}_{2})\in P$, $a\in P$, a contradiction^{}. Therefore, $P$ is prime and the proof is complete^{}. ∎
In the proof, we use the fact that, an element $a\in L$ belongs to the ideal generated by ideals ${I}_{k}$ iff $a$ is less than or equal to a finite join of elements, each of which belongs to some ${I}_{k}$.
Remarks.

1.
The theorem^{} can be generalized: if we use a subset $S\cap I=\mathrm{\varnothing}$ instead of an element $a\notin I$, there is a prime ideal $P$ containing $I$ but excluding $S$.

2.
Birkhoff’s prime ideal theorem has been shown to be equivalent^{} to the axiom of choice^{}, under ZF.
Title  Birkhoff prime ideal theorem 

Canonical name  BirkhoffPrimeIdealTheorem 
Date of creation  20130322 17:02:18 
Last modified on  20130322 17:02:18 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  11 
Author  CWoo (3771) 
Entry type  Theorem 
Classification  msc 06D05 
Classification  msc 03E25 
Related topic  DistributiveLattice 