Non-Isomorphic Ultrapowers It is clear that given a family $(\mathfrak{A}_i)_{i\in I}$ of $L$-structures their ultraproduct may depend on the choice of the ultrafilter (for this question I am only considering non-principal ultrafilters).
Is there an easy example as to why given and $L$-structure $\mathfrak{A}$ the choice of the (nonprincipal) ultrafilter $\mathcal{U}$ on $I$ may affect the isomorphism class of $\mathfrak{A}^I/\mathcal{U}$?
 A: Let $L$ have unary predicate symbols $\dot X$ for all subsets $X$ of $\mathbb N$, and let $\mathfrak A$ be the structure with universe $\mathbb N$ in which each $\dot X$ has the obvious interpretation $X$.  In any elementary extension $\mathfrak B$ of $\mathfrak A$, define the type of any element $b$ to be the set of those $X\subset\mathbb N$ for which $\dot X(b)$ is true in $\mathfrak B$.  
Now consider ultrapowers of $\mathfrak A$ by ultrafilters $\mathcal U$ on $\mathbb N$.  Notice that $[i]_{\scr U}$, the equivalence class in $\mathfrak A^{\mathbb N}/\mathcal U$ of the identity function $i$, has type $\mathcal U$.  So every ultrafilter on $\mathbb N$ occurs as the type of an element in such an ultrapower.  But any one ultrapower realizes only $2^{\aleph_0}$ types, as it has only $2^{\aleph_0}$ elements.  And isomorphic ultrapowers realize the same types.  So to realize all $2^{2^{\aleph_0}}$ non-principal ultrafilters on $\mathbb N$, there must be $2^{2^{\aleph_0}}$ non-isomorphic ultrapowers.
If you are unhappy with my use of an uncountable language $L$, then it becomes necessary to work harder, and in particular to use uncountable index sets $I$, but it is still possible to get non-isomorphic ultrapowers of any structure, even for a countable language --- indeed even for the language consisting of just equality.  The key phrase to look up here is "regular ultrafilter".
