TITLE: Does an ultrapower of an Aronszajn tree have an $\omega_{1}$-branch?
QUESTION [14 upvotes]: Throughout this question, I shall let $A^{\mathcal{U}}$ denote the ultrapower of a structure $A$ by an ultrafilter $\mathcal{U}$. Suppose that $T$ is an Aronszajn tree and $\mathcal{U}$ is an ultrafilter on a countable set $I$. Let $T_{\alpha}$ denote the $\alpha$-th level of a tree $T$. Let $T^{(\mathcal{U})}$ denote the tree where the $\alpha$-th level of $T^{(\mathcal{U})}$ is the ultrapower $(T_{\alpha})^{\mathcal{U}}$ and where if $(x_{i})_{i\in I}/\mathcal{U}\in(T_{\alpha})^{\mathcal{U}},(y_{i})_{i\in I}/\mathcal{U}\in (T_{\beta})^{\mathcal{U}}$, then $(x_{i})_{i\in I}/\mathcal{U}<(y_{i})_{i\in I}/\mathcal{U}$ iff $\{i\in I|x_{i}<y_{i}\}\in\mathcal{U}$. Then $T^{(\mathcal{U})}$ is clearly a tree. In other words, we trim the ultrapower $T^{\mathcal{U}}$ so that it becomes a tree $T^{(\mathcal{U})}$.
Does the tree $T^{(\mathcal{U})}$ have a branch of length $\omega_{1}$? What about when we assume that $T$ is a Suslin tree? What happens when we consider ultraproducts of countably many Aronszajn trees instead of ultrapowers?

REPLY [6 votes]: This is a footnote to the solution by Blass, and provides a high level explanation (at least when CH holds). The reasoning provided below also takes care of the ultraproduct version of the question.
Theorem. Assume CH. Then the ultrapower $T^\cal{U}$, viewed as a ranked tree, has a proper elementary end extension.
(see top of p.333 of here for the official definition of a ranked tree).
Note that the above immediately implies that the subtree $T^\cal{(U)}$ of $T^\cal{U}$ has an $\omega_1$-branch because the set of predecessors of any node in the end extension, when intersected with $T^\cal{(U)}$, will result in an $\omega_1$-branch.
Outline of the proof of the above theorem: By classical model theory, $T^\cal{U}$ is $\omega_1$-saturated. On the other hand, using CH and elementary cardinal arithmetic, we see that the cardinality of $T^\cal{U}$ is $\omega_1$, and therefore $T^\cal{U}$ is a saturated structure. 
We then observe that since $T$ is an $\omega_1$-tree, $T$ satisfies the regularity scheme (which asserts that definable maps from the tree onto a bounded subset of the tree can be made constant by restriction to an unbounded definable subset of the tree). This shows that the ultrapower $T^\cal{U}$ inherits the regularity scheme from $T$.
Recall that saturated structures are resplendent. Our proof is now completed by noting that resplendent ranked trees satisfying the regularity scheme have proper elementary end extensions. The proof of this latter fact uses a compactness argument that succeed, thanks to a theorem of Keisler that asserts that countable tree structures satisfying the regularity scheme have proper elementary end extensions. QED
Remark: Keisler´s aforementioned theorem is established using the same sort of omitting types argument that is used to show that countable models of ZF have proper elementary end extensions (Chang-Keisler, Model Theory, theorem 2.2.18).