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Every proper lattice ideal of a Boolean algebra extends to a prime lattice ideal. More generally, an ideal disjoint from a lattice filter extends to a prime ideal still disjoint from that filter. This Boolean case of the Stone prime filter theorem is equivalent to the ultrafilter lemma over Zermelo–Fraenkel set theory and is weaker than the axiom of choice.

The Stone prime filter theorem says that if a lattice filter $F$ and a lattice ideal $I$ of a distributive lattice are disjoint, then there is a prime filter of a distributive lattice $P$ such thatEquivalently, whenever $a\nleqq b$, there is a prime filter containing $a$ and omitting $b$.

For a prime filter $P\in \hat{H}$, first suppose $a\Rightarrow b\in P$. If $Q\supseteq P$ and $a\in Q$, then $a\Rightarrow b\in Q$ andso $b\in Q$. Thus no prime filter above $P$ belongs to $a^{\ast }\setminus b^{\ast }$, and

Conversely, suppose $a\Rightarrow b\notin P$. The lattice filter generated by $P\cup \{a\}$ is disjoint from the principal lattice ideal $\downarrow b$. Indeed, an intersection would give some $p\in P$ with $p\wedge a\le b$, whence $p\le a\Rightarrow b$ and then $a\Rightarrow b\in P$, a contradiction. The Stone prime filter theorem therefore extends this filter to a prime filter $Q\supseteq P$ that omits $b$. Then $Q\in a^{\ast }\setminus b^{\ast }$, so $P\in \uparrow (a^{\ast }\setminus b^{\ast })$.

We have proved, for every $P$,which is the required identity

The Stone map sends an element $a$ of a distributive lattice to the clopen up-set $a^{\ast }=\{P:a\in P\}$ of its Priestley dual space. The Stone prime filter theorem makes this lattice homomorphism injective.