Boolean Prime Ideal Theorem


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Theorem

Let $\struct {S, \le}$ be a Boolean lattice.

Let $I$ be an ideal in $S$.

Let $F$ be a filter on $S$.

Let $I \cap F = \O$.


Then there exists a prime ideal $P$ in $S$ such that:

$I \subseteq P$

and:

$P \cap F = \O$


Element Extension Lemma

Let $\struct {B, \vee, \wedge, \le}$ be a Boolean lattice.

Let $F \subseteq B$ be a filter on $B$.

Let $a, x \in B$ such that:

$a \notin F$


Then either:

$a \vee x \notin F$

or:

$a \vee \neg x \notin F$


Proof from the Axiom of Choice

Let $T$ be the set of ideals in $S$ that contain $I$ and are disjoint from $F$, ordered by inclusion.

Let $N$ be a chain in $T$.

Then $\ds U = \bigcup N$ is clearly disjoint from $F$ and contains $I$.


Let $x \in U$ and $y \le x$.

Then:

$\exists A \in N: x \in A$

By the definition of union:

$x \in U$

Let $x \in U$ and $y \in U$.

Then:

$\exists A, B \in N: x \in A, y \in B$

By the definition of a chain:

$A \subseteq B$

or:

$B \subseteq A$

Hence $U$ is also an ideal.


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Without loss of generality, suppose $A \subseteq B$.

Then:

$x \in B$

Since $y$ is also in $B$, and $B$ is an ideal:

$x \vee y \in U$.

By Zorn's Lemma, $T$ has a maximal element, $M$.

It remains to show that $M$ is a prime ideal:

Every Boolean algebra is a distributive lattice.

So by Maximal Ideal WRT Filter Complement is Prime in Distributive Lattice, $M$ is a prime ideal.


Axiom of Choice

This proof depends on the Axiom of Choice, by way of Zorn's Lemma.

Because of some of its bewilderingly paradoxical implications, the Axiom of Choice is considered in some mathematical circles to be controversial.

Most mathematicians are convinced of its truth and insist that it should nowadays be generally accepted.

However, others consider its implications so counter-intuitive and nonsensical that they adopt the philosophical position that it cannot be true.


Proof from the Ultrafilter Lemma

Let $Q$ be the set of all ideals of $S$ that are disjoint from $F$.

For each $x \in S$, define:

$C_x = \set {q \in Q : x \in q}$


We want to construct a filter $\FF$ on $Q$ such that:

  • $C_x \in \FF$ if and only if $x \in I$
  • When the filter is extended to an ultrafilter $\UU$ by the Ultrafilter Lemma, the resulting structure contains either $C_x$ or $C_{\neg x}$ for every $x \in S$


We can satisfy the first property by letting $\FF = \set {C_x : x \in I}$, but the second may not be satisfied since we could have $C_x \notin \UU$ and also $C_{\neg x} \notin \UU$.

To remedy this, we additionally include $C_x \cup C_{\neg x} \in \FF$ for every $x \in S$.

For then, given any $x \in S$ such that $C_x \notin \UU$, we can intersect its complement with $C_x \cup C_{\neg x}$ to deduce $C_{\neg x} \in \UU$.


The full proof in detail is below.

Propositions

We will first demonstrate several properties of $C_x$.

First, let $x, y \in S$ be arbitrary, with $x \le y$.

For any $q \in C_y$, we have by definition of $C_y$ that:

$y \in q$

But then, by ideal axiom $\paren 1$:

$x \in q$

and so:

$q \in C_x$

Thus:

$\paren 1 \quad x \le y \implies C_y \subseteq C_x$.


Now, let $x, y \in S$ once again be arbitrary.

By definition of the join, we have:

$x \le x \vee y$
$y \le x \vee y$

and so by $\paren 1$:

$C_{x \vee y} \subseteq C_x \cap C_y$

Conversely, let $q \in C_x \cap C_y$ be arbitrary.

Then, $x \in q$ and $y \in q$, so by Ideal is Closed under Join:

$x \vee y \in q$

and so:

$C_x \cap C_y \subseteq C_{x \vee y}$

From the above:

$\paren 2 \quad C_{x \vee y} = C_x \cap C_y$.

Finite Intersection Property

Define:

$\BB = \set {C_x : x \in I} \cup \set {C_x \cup C_{\neg x} : x \in S}$

which will be a sub-basis of $\FF$.

For that to be the case, $\BB$ must have the finite intersection property.


Let:

$G = \set {C_{x_i} : 1 \le i \le n} \cup \set {C_{y_j} \cup C_{\neg y_j} : 1 \le j \le m}$

be an arbitrary finite subset of $\BB$.

Define $\ds X = \bigvee_{i = 1}^n x_i$

By repeatedly applying $\paren 2$ above, we deduce that:

$\ds \bigcap_{i = 1}^n C_{x_i} = C_X$

By Ideal is Closed under Join:

$X \in I$

and hence:

$X \notin F$


Now, we consider the $y_j$.

For an arbitrary $k \in \set {1, \dots, m}$ and $a \notin F$, we can apply the Element Extension Lemma to find some $z_k$ such that:

either $z_k = y_k$ or $z_k = \neg y_k$

and:

$a \vee z_k \notin F$

In particular, we will choose:

$\ds a = X \vee \bigvee_{j = 1}^{k - 1} z_j$

where the $z_j$ have already been selected inductively.


After this process is completed, we will have selected $\sequence {z_j}_{1 \le j \le m}$ such that:

$\forall j: z_j \in \set {y_j, \neg y_j}$
$\ds X \bigvee_{j = 1}^m z_j \notin F$

Define:

$\ds Z = X \bigvee_{j = 1}^m z_j$

so then by Join Succeeds Operands:

$x_i \le X \le Z$
$z_j \le Z$


Let $Z^\le$ denote the lower closure of $Z$.

By Lower Closure of Element is Ideal, we then have:

$Z^\le \in Q$

which by the above remarks satisfies:

$x_i \in Z^\le$
$z_j \in Z^\le$

Therefore:

$Z^\le \in C_{x_i}$
$Z^\le \in C_{z_j} \subseteq C_{y_j} \cup C_{\neg y_j}$

for all $i, j$.

Hence:

$\ds Z^\le \in \bigcap G$

and so the intersection is indeed non-empty.


As $G$ was arbitrary, $\BB$ has the finite intersection property.

$\Box$

Ultrafilter

We can now apply the corollary to the Ultrafilter Lemma to obtain an ultrafilter $\UU$ on $Q$ such that:

$\BB \subseteq \UU$

Define:

$P = \set {x \in S : C_x \in \UU}$

Before proving that $P$ is a prime ideal, we will first demonstrate the other conditions of the theorem.


By definition of $\BB$, for every $x \in I$:

$C_x \in \BB \subseteq \UU$

and so:

$x \in P$

Thus, $I \subseteq P$ as required.


Additionally, since for each $x \in F$:

$C_x = \O$

we cannot have $C_x \in \UU$ be definition of an ultrafilter.

Hence, $P \cap F = \O$.

$\Box$

Prime Ideal

It only remains to show that $P$ is a prime ideal.

We will first show that $P$ is an ideal.

Since $P \supseteq I \ne \O$, it follows that:

$P \ne \O$

which is required by the definition of an ideal.


Let $x, y \in S$ such that $y \le x$.

If:

$x \in P$

we have by $\paren 1$ above that:

$C_y \supseteq C_x \in \UU$

and hence by filter on set axiom $\paren {\text F 4}$:

$y \in P$

which is ideal axiom $\paren 1$.


Let $x, y \in P$.

Then, $C_x, C_y \in \UU$.

By filter on set axiom $\paren {\text F 3}$, it follows that:

$C_x \cap C_y \in \UU$

But, by $\paren 2$ above:

$C_{x \vee y} = C_x \cap C_y \in \UU$

and hence:

$x \vee y \in P$

which satisfies ideal axiom $\paren 2$.

Hence, $P$ is an ideal.


Moreover, by definition of filter:

$F \ne \O$

so since:

$P \cap F = \O$

it follows that:

$P \ne S$

Therefore, $P$ is a proper ideal.


Lastly, we will show that, for each $x \in S$, either:

$x \in P$

or:

$\neg x \in P$

If $x \in S \setminus P$ is given, it follows that:

$C_x \notin \UU$

and so by definition of an ultrafilter:

$Q \setminus C_x \in \UU$


We also have:

$C_x \cup C_{\neg x} \in \BB \subseteq \UU$

so we can apply filter on set axiom $\paren {\text F 3}$ to find:

$\paren {Q \setminus C_x} \cap \paren {C_x \cup C_{\neg x}} \in \UU$

or equivalently:

$C_{\neg x} \setminus C_x \in \UU$


But for no ideal $q \in Q$ can we have:

$\set {\neg x, x} \subseteq q$

for by Ideal is Closed under Join and the definition of complement:

$\top \in q$

which contradicts $q \cap F = \O$ by Top in Filter.


Therefore:

$C_{\neg x} = C_{\neg x} \setminus C_x \in \UU$

and thus:

$\neg x \in P$

As $x \in S$ was arbitrary, we can apply Proper Ideal is Prime iff Contains Element or Complement, from which the theorem follows.

$\blacksquare$


Proof from Stone's Representation Theorem for Boolean Algebras

We prove that the is equivalent to Stone's Representation Theorem for Boolean Algebras in ZF.


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Motivation

The Boolean Prime Ideal Theorem is weaker than the Axiom of Choice, but is similarly independent of ZF theory.

It is sufficient to prove a number of important theorems, although such proofs are often more involved than ones relying on the Axiom of Choice.

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