TITLE: Cardinality of ultraproducts
QUESTION [8 upvotes]: This might be a trivial question, but I couldn't see how to do this now. 
Given an infinite model $M$ and infinite cardinal $\kappa > |M|$ and a non-principal ultrafilter $F$ over $\kappa$, is it always true that $N = M^\kappa/F$ has cardinality $\geq \kappa$? If not, is it sometimes true?
Thanks a lot!

REPLY [3 votes]: I think $|N|$ could be smaller than $\kappa$ for some trivial reason: if $F$ contains a set $X$ of size $\lambda$ such that $|M|^\lambda<\kappa$, then $N=M^\kappa/F$ is the same as $M^\lambda/F'$ where $F'$ is the restriction of $F$ on $X$, so has size at most $|M|^\lambda<\kappa$.
If $F$ is uniform then the question becomes more interesting (probably more difficult too). If $|M|$ is countable and $F$ is countably incomplete, then $|N|$ is at least continuum, see section 5 of this survey.
Here is another short note which I vaguely remember but it took some time to find it again...If $F$ is regular then $N$ has the largest size possible, namely $|M|^\kappa$. Also from here it seems the general question is indeed quite difficult.