Question

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?

Answer 1

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 povides 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 $\{i\,\vert\,n_i>m\}\in U$. 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.

Answer 2

Every finite field is an ultrafield (an ultrapower of itself). But a countable infinite field is not an ultrafield (an ultraproduct cannot be countable). Now take the field of fractions $\mathbb{F}_2(x_1,x_2,...)$ . It is a transcendental extension of a (finite) ultrafield which is not an ultrafield.