Let $K=\lim(K\_{i})$ be an ultrafield (over a non-principal ultrafilter), and let $K\hookrightarrow K'$ be a field extension of $K$. 


When the field $K'$ is finite over $K$ it is also an ultrafield by Łoś's theorem. What can be said when the trascendence degree of $K'$ over $K$ is infinite?



Is every field extension of an ultrafield an ultrafield?
[EDIT on March 26, 2019: There are issues with this argument, found by @YCor who also gives a correct and more general answer.]


Let me show that if $k$ is algebraically closed, and $X=(x^{(\alpha)})$ any nonempty family of indeterminates, then $k(X)$ is not an ultrafield (which provides a lot of counterexamples since there are algebraically closed ultrafields). In fact I shall only assume that for some $r>1$, every element of $k$ is an $r$-th power.


Fix a prime $p$ not dividing $r$. Assume $k(X)=\lim(K\_i)$ (for a nonprincipal ultrafilter $U$ on an infinite set $I$). Let $x$ be one of the indeterminates. Then $x$ is the class of a family $(x\_i)\_{i\in I}$. Take an infinitely large integer, i.e. a family $(n\_i)\_{i\in I}\in\mathbb{N}^I$ such that for each $m\in\mathbb{N}$ we have that $\{i\,\vert\,n\_i>m\}\in U$. 
[EDIT: the problem is here. We need to know that $I$ is "reasonable" or at least that $U$ is not $\sigma$-complete, see YCor's answer.]


Let $z$ be the class of $(x\_i^{p^{n\_i}})\_{i\in I}$. Then $z$ is a $p^n$-`th power in $k(X)$ for all $n$, hence $z\in k$. In particular, $z$ is an $r$-th power, which means that for all $i$ in some $J\in U$, $x\_i^{p^{n\_i}}$ is an $r$-th power, and therefore so is $x\_i$ because $r$ is prime to $p$. We conclude that $x$ is an $r$-th power in $k(X)$, contradiction.


