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!
Best Answer
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.