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# 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 $S:=\{J\mid J\mbox{ is an ideal in }L\mbox{, and }a\notin J\}$. 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 $y\leq x$. 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 $a\leq p_{1}\vee x$ and $a\leq p_{2}\vee y$, where $p_{1},p_{2}\in P$. Then $a\vee p_{2}\leq(p_{1}\vee p_{2})\vee x$ and $a\vee p_{1}\leq(p_{1}\vee p_{2})\vee y$. Take the meet of these two expressions, and we obtain $(a\vee p_{2})\wedge(a\vee p_{1})\leq((p_{1}\vee p_{2})\vee x)\wedge((p_{1}\vee p% _{2})\vee y)$. 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 $a\leq a\vee(p_{1}\wedge p_{2})\in P$, $a\in P$, 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 $S\cap I=\varnothing$ 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.

## Mathematics Subject Classification

06D05*no label found*03E25

*no label found*

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