Ultraproduct of polynomial rings There are several instances of families $ (K_i)_{i \in I}$ and $(L_i)_{i \in I}$ of fields $K_i$ and $L_i$ such that the ultraproducts $\prod_{i \in I} K_i/\mathcal U$ and $\prod_{i \in I} L_i/\mathcal U$ are isomorphic. For instance, this is the case in the Ax-Kochen-Ershov principle or for ultraproducts of countable algebraically closed fields of unbounded characteristic (or characteristic $0$) on $I$.
Now, I am wondering what happens if one adds one transcendental element in each component of the ultraproduct, that is,

is there an isomorphism $\prod_{i \in I} K_i[X]/\mathcal U \cong \prod_{i \in I} L_i[X]/\mathcal U$

Is it maybe even true that these ultraproducts are polynomial rings (in "many" variables) over the ultraproducts of fields?
Thank you for your help!
 A: The answer is negative in general even for ultrapowers, i.e. when there are fields $K$ and $L$ with $K_i = K$ and $L_i = L$ for all $i$. This comes down to the surprising fact that $K\equiv L$ does not imply $K[X] \equiv L[X]$ in general.
Let $K = \mathbb{Q}^{\text{alg}}$ and $L = (\mathbb{Q}(x))^{\text{alg}}$. Let $\mathcal{U}$ be a non-principal ultrafilter on $\mathbb{N}$. Then the ultrapowers $K^\mathcal{U} = \prod_{i\in I} K/\mathcal{U}$ and $L^\mathcal{U} = \prod_{i\in I} L/\mathcal{U}$ are isomorphic, since they are algebraically closed fields of the same characteristic $0$ and the same cardinality $2^{\aleph_0}$ (in fact, they are both isomorphic to $\mathbb{C}$!).
On the other hand, $K[X]\not\equiv L[X]$. One explanation is that in a polynomial ring over a field of characteristic zero, the transcendence degree of the field over $\mathbb{Q}$ is definable (in a precise sense). See here for a proof and a reference. The book "Model Theoretic Algebra" by Jensen and Lenzing (pp. 36-38) has more details.
So for any ultrafilter $\mathcal{U}$, we have $$K[X]^\mathcal{U}\equiv K[X]\not\equiv L[X] \equiv L[X]^\mathcal{U}.$$
Since the ultrapowers are not even elementarily equivalent, they are certainly not isomorphic.
