How many nonprincipal ultrafilters exists on $X$ such that $|X|^2=|X|$? Let $X$ be an infinite set, idemmultiple ($|X|^2 = |X|$) if that helps.  I am looking for a proof that $X$ has more than $|X|$-many ultrafilters.  It has $|X|$-many principal ultrafilters. Without at least some AC I know of no proof that $X$ must have any non-principal ultrafilters.  So let us assume BPI, the Prime ideal theorem.  If we assume BPI can we prove $|\beta X| > |X|$?   ($\beta X$ is of course the set of ultrafilters on $X$).  If we have AC then it is standard that $|\beta X| = 2^{2^{|X|}}$.  I don't much mind which set theory we work in (the question actually arose in NF, with $X = V$) co's it's the construction (if any) that I am after.
Summary: i am interested in a proof (no theory specified) that BPI implies that an idemmultiple (infinite) set $X$ has more than $|X|$-many ultrafilters
 A: Recall the usual proof of $|\beta X|=2^{2^{|X|}}$ using AC.  To sketch the argument, you replace $X$ with the set $Y$ of pairs $(A,S)$ where $A$ is a finite subset of $X$ and $S$ is a finite set of finite subsets of $X$.  Then, you explicitly construct a family of $2^{2^{|X|}}$ pairwise incompatible filters on $Y$, and extend them each to ultrafilters.  This gives $2^{2^{|X|}}$ different ultrafilters on $Y$, and hence on $X$ since $|X|=|Y|$.
With a bit of care, you can make this argument still work using only BPI and $|X|^2=|X|$.  First, we still have $|X|=|Y|$.  To prove this, note that we can totally order $X$ by BPI, and so from $|X|^2=|X|$ we can obtain a family of injections $[X]^n\to X$ for each finite $n$.  Also, $|X|^2=|X|$ implies $\aleph_0\leq |X|$ so $|X|\times\aleph_0\leq |X|^2=|X|$.  Thus $|X|\leq [X]^{<\omega}\leq|X|\times\aleph_0=|X|$, and so $|Y|=|[X]^{<\omega}\times[[X]^{<\omega}]^{<\omega}|=|X|^2=|X|$.
Now, using BPI, we can extend each of our filters on $Y$ to an ultrafilter.  In fact, we can actually extend them simultaneously to get a family of $2^{2^{|X|}}$ different ultrafilters on $Y$, proving that $|\beta X|=|\beta Y|=2^{2^{|X|}}$.  More generally, suppose you have a family $(B_i)_{i\in I}$ of nonzero Boolean algebras and you wish to find ultrafilters on all of them simultaneously.  Let $B$ be the free product of the $B_i$.  Then $B$ is a nonzero Boolean algebra: if we had $1=0$ in $B$, then we would have $1=0$ in the free product of some finitely many $B_i$ (since $B$ is the direct limit of the finite free products), but that is impossible (for instance, by picking ultrafilters on those finitely many $B_i$ to get a homomorphism from their free product to $\{0,1\}$).  Thus by BPI, $B$ has an ultrafilter, and it can be pulled back along the canonical homomorphisms $B_i\to B$ to get a family of ultrafilters on each $B_i$. (Thanks to Andreas Blass for providing a variant of this argument in the comments.)
