Is the cardinality of an ultrapower generally the same as that of the set of functions? For any structure $M$ and ultrafilter $U$ on a set $\Omega$, let $M^{\langle U\rangle}=M^\Omega/U$ be the conventional ultrapower on $M$.
Is it the case that $|M^{\langle U\rangle}|=|M^\Omega|=|M|^{|\Omega|}$?
In all cases, it is obvious that $|M^{\langle U\rangle}|\le|M^\Omega|=|M|^{|\Omega|}\le|M|^{|U|}$.
I'm told that in the case that $U$ is non-principal, we have $|M^\Omega|\le|M^{\langle U\rangle}|$ as well, so $|M^{\langle U\rangle}|=|M|^{|\Omega|}$. I have yet to see a proof, but it seems plausible enough.
Less plausible is the idea that $|M^{\langle U\rangle}|=|M|^{|\Omega|}$ when $U$ is a principal ultrafilter. I've tried to prove that this isn't necessarily the case, but I keep getting stuck. Whatever the answer, it seems to relate to the properties of ultrafilters in a nontrivial way and I don't think that I can spend the time to really figure it out, especially since all I actually need is an "is true/is false, see ([Author], 19xx) for proof."
That said, I would very much like to understand the proof one way or the other.
 A: You've been told wrong in the nonprincipal case.
For principal ultrafilters, the equivalence relation we're quotienting by is just that $[f]=[g]$ iff $f(i)=g(i),$ where $i$ is the generator of the ultrafilter, so the cardinality of the ultrapower is the same as the cardinality of $M$. (And in fact, similar reasoning shows that the ultrapower is isomorphic to $M$ in this case.)
A: I just want to add a few things to spaceisdarkgreen's clear answer. The canonical reference for this question is Chang and Keisler's book Model Theory, section 4.3, pp. 250-253. This book contains an enormous amount of information about ultraproducts and ultrapowers.
I write $M^U$ for the ultrapower of $M$ by the ultrafilter $U$ on $\Omega$.
The obvious bounds are $|M|\leq |M^U|\leq |M|^{|\Omega|}$. The first inequality is because the map $a\mapsto [(a,a,a,\dots)]_{\sim_U}$ is injective (in fact, it is an elementary embedding, by Łoś's theorem). The second inequality is because $M^U$ is a quotient of $M^\Omega$ by the equivalence relation $\sim_U$.
If $M$ is finite, the canonical map $M\to M^U$ is an isomorphism (because every elementary embedding with finite domain is an isomorphism), so $|M^U| = |M|$.
If $U$ contains a set $X\subseteq \Omega$ with $|X|<|\Omega|$, then $U' = \{X\cap Y\mid Y\in U\}$ is an ultrafilter on $X$ and $M^{U'}\cong M^U$, so we can reduce to the case where $|X|  = |\Omega|$ for all $X\in U$.
As a special case of this last observation, note that if $U$ is principal, we have $M^U\cong M$, so $|M^U| = |M|$.
The last basic result about cardinalities of ultrapowers takes a bit more work to state. For a cardinal $\kappa$, we say that $U$ is $\kappa$-regular if there is a set $E\subseteq U$ with $|E| = \kappa$ and for all $x\in \Omega$, $\{X\in E\mid x\in X\}$ is finite.
Theorem: (Proposition 4.3.7 in Chang & Keisler) Suppose $U$ is a $|\Omega|$-regular ultrafilter on $\Omega$ and $M$ is infinite. Then $|M^U| = |M|^{|\Omega|}$.
It's worth noting that every non-principal ultrafilter on a countably infinite set is $\aleph_0$-regular. So we have:
Corollary: Suppose $|\Omega| = \aleph_0$. If $M$ is finite or $U$ is principal, then $|M^U| = |M|$. Otherwise, $|M^U| = |M|^{\aleph_0}$.
