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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?

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What is an ultrafield? – Qiaochu Yuan Mar 18 '11 at 23:29
I'm following the terminology of Schouten's "The Use of Ultraproducts in Commutative Algebra": an ultrafield is simply an ultraproduct of fields. – user12940 Mar 18 '11 at 23:43
For those unfamiliar with the terminology in the book, an ultrafield is a ultraproduct of an infinite collection of fields over a nonprincipal ultrafilter, which in this case of the ultraproduct is a field. – Jason Polak Mar 19 '11 at 0:19
Your question seems to skip over the case of finite transcendence degree, which it seems to me may already be a source of counterexamples. For instance, $\mathbb{C}$ is an ultrafield (right?), but is $\mathbb{C}(t)$ an ultrafield? – Pete L. Clark Mar 19 '11 at 5:45
@Pete: Yes, $\mathbb{C}$ is an ultrafield: there's only one characteristic-0 algebraically closed field of cardinality continuum, so $\mathbb{C}$ is isomorphic to any nonprincipal ultrapower of the algebraic numbers. – Chris Eagle Mar 19 '11 at 19:35
up vote 7 down vote accepted

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.

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This is probably very dumb to be asked, but why do you say that the equation $x_{i}^{p^n_{i}}=y^r$ implies that $x_{i}=z^r$? – user12940 Mar 19 '11 at 22:01
(I mean: even when $p$ does not divide $r$ the implication is not clear for me). – user12940 Mar 19 '11 at 22:03
There are integers $u$, $v$ such that $up^{n_i}+vr=1$. So $x_i=x_i^{up^{n_i}}\,x_i^{vr}=(y^u\,x_i^v)^r$. – Laurent Moret-Bailly Mar 19 '11 at 22:16

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.

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This is correct, but having seen this counterexample it is natural to ask about extensions of infinite ultrafields. Isn't the answer still negative? – Pete L. Clark Mar 19 '11 at 5:42
I do not have time, but you can use the fact that (at least assuming Continuum Hypothesis) every ultraproduct is saturated. – Mark Sapir Mar 19 '11 at 10:17
@Mark: right, I had that thought as well, but was hoping for something a little Moret-Bailly's response below. – Pete L. Clark Mar 19 '11 at 19:10

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