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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    $\begingroup$ What is an ultrafield? $\endgroup$ – Qiaochu Yuan Mar 18 '11 at 23:29
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    $\begingroup$ I'm following the terminology of Schouten's "The Use of Ultraproducts in Commutative Algebra": an ultrafield is simply an ultraproduct of fields. $\endgroup$ – user12940 Mar 18 '11 at 23:43
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    $\begingroup$ 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. $\endgroup$ – Jason Polak Mar 19 '11 at 0:19
  • $\begingroup$ 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? $\endgroup$ – Pete L. Clark Mar 19 '11 at 5:45
  • $\begingroup$ @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. $\endgroup$ – Chris Eagle Mar 19 '11 at 19:35

[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.

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  • $\begingroup$ 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 $x_{i}=z^r$? $\endgroup$ – user12940 Mar 19 '11 at 22:01
  • $\begingroup$ (I mean: even when $p$ does not divide $r$ the implication is not clear for me). $\endgroup$ – user12940 Mar 19 '11 at 22:03
  • $\begingroup$ There are integers $u$, $v$ such that $up^{n_i}+vr=1$. So $x_i=x_i^{up^{n_i}}\,x_i^{vr}=(y^u\,x_i^v)^r$. $\endgroup$ – Laurent Moret-Bailly Mar 19 '11 at 22:16
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    $\begingroup$ It's not clear that, given $U$, such $n=(n_i)_{i\in I}$ exists. If $I$ is countable it's fine. If $U$ is stable under countable intersections, then $n$ does not exist. The existence of $I$ with such $U$ (= existence of measurable cardinals) can't be proved in ZFC, but is unlikely to be disproved. Possibly the existence of $I$ and $U$ such that there's no such $n$ is a theorem of ZFC, though. $\endgroup$ – YCor Mar 25 '19 at 15:16
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    $\begingroup$ PS: it's more clear in my mind now. This proof (and also the claim) works if and only if the ultrafilter is not $\sigma$-complete (= not closed under countable intersections), which is the condition to find $(n_i)$. $\endgroup$ – YCor Mar 26 '19 at 14:34

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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  • $\begingroup$ This is correct, but having seen this counterexample it is natural to ask about extensions of infinite ultrafields. Isn't the answer still negative? $\endgroup$ – Pete L. Clark Mar 19 '11 at 5:42
  • $\begingroup$ I do not have time, but you can use the fact that (at least assuming Continuum Hypothesis) every ultraproduct is saturated. $\endgroup$ – user6976 Mar 19 '11 at 10:17
  • $\begingroup$ @Mark: right, I had that thought as well, but was hoping for something a little simpler...like Moret-Bailly's response below. $\endgroup$ – Pete L. Clark Mar 19 '11 at 19:10

I have mentioned in a comment that the accepted answer is not correct, although the argument is correct when the index set is countable.

Here's a result with no algebraic closedness assumption, which actually shows that the algebraic structure of the multiplicative group is enough to set up the argument, and also purports to clarify what we should allow on the index set, or even on the ultrafilter. I'm starting with the countable case.

Proposition. Let $K$ be a field. Then the field $K(t)$ is not isomorphic to any nonprincipal countable ultraproduct of fields. Actually, the multiplicative group $K(t)^*$ is not isomorphic to any nonprincipal countable ultraproduct of groups.

Since $K(t)^*\simeq K^*\times\mathbf{Z}^{(J)}$, where $\mathbf{Z}^{(J)}$ is the free abelian group on the nonempty set $J$ of monic irreducible polynomials, we have $\mathrm{Hom}_{\mathrm{Group}}(K(t)^*,\mathbf{Z})\neq\{0\}$. Hence the proposition follows from the following lemma.

Given a family $(G_n)$ of groups, I denote by $\prod^\star_nG_n$ the near product, i.e., the quotient of the product $\prod_n G_n$ by the finitely supported (aka restricted, aka finitely supported) product $\bigoplus_n G_n$.

Lemma. Let $(G_n)$ be a sequence of abelian groups. Then $\mathrm{Hom}(\prod^\star_n G_n,\mathbf{Z})=\{0\}$. In particular, for every nonprincipal ultrafilter $\sigma$ on $\mathbf{N}$, we have $\mathrm{Hom}(\prod^\sigma_n G_n,\mathbf{Z})=\{0\}$, where $\prod^\sigma_n G_n$ is the ultraproduct with respect to $\sigma$.

Proof: This is the classical Specker proof for $G_n=\mathbf{Z}$. The proof is an immediate adaptation. Consider a homomorphism $\prod^\star_nG_n\to\mathbf{Z}$, vanishing on $\bigoplus_nG_n$. For each $n$, choose a Bézout relation $2^na_n+3^nb_n=1$. For $x=(x_n)_n\in\prod_nG_n$, write $p^\omega y=(p^ny)_n$. Denote by $x\mapsto \bar{x}$ the quotient map $\prod\to\prod^\star$. Since for every $x$ and for $p=2,3$, the element $\overline{p^\omega x}$ is $p$-divisible (i.e., has $p^n$-roots for all $n$), so is its image by $f$, which forces $f\big(\overline{2^\omega x}\big)=f\big(\overline{3^\omega x}\big)=0$. Hence, for every $y=(y_n)_n\in\prod_nG_n$, we have $y=2^\omega ay+3^\omega by$ and hence $f(\overline{y})=0$. So $f=0$.

[Note: this also follows from the $G_n=\mathbf{Z}$ case, by a simple composition argument.]

The second assertion follows from the first since $\prod^\sigma_n G_n$ is a quotient of $\prod^\star_n G_n$.

Let me now considerably relax the countability assumption. Let me say that a set $I$ is reasonable if every ultrafilter on $I$, stable under countable intersections, is principal. (If there's a standard terminology I'd be happy to change.)

This only depends on the cardinal of $I$. It's consistent that every set is reasonable. If there's a non-reasonable cardinal, there's a minimal one, which is known as smallest measurable cardinal. All "reasonably small" cardinals are reasonable; obviously $\omega$ is, and notably $\alpha$ reasonable implies $2^\alpha$ reasonable and in particular sets of cardinal $\le \mathfrak{c}$ (continuum), $\le 2^{\mathfrak{c}}$, etc, are reasonable.

The point is that the proposition and lemma above still hold when the index set is supposed to have reasonable cardinal.

Lemma'. Let $I$ be a reasonable set, and $(G_i)$ a family of abelian groups. Then $\mathrm{Hom}(\prod^\star_{i\in I} G_i,\mathbf{Z})=\{0\}$.

Remark: using a nonprincipal ultrafilter stable under countable intersections, conversely if $I$ is not reasonable then $\mathrm{Hom}(\prod^\star_{i\in I} \mathbf{Z},\mathbf{Z})\neq\{0\}$, since a nontrivial homomorphism is obtaining by taking the limit along such an ultrafilter.

Proof: again, this follows from the case $G_i=\mathbf{Z}$: let $f:\prod_iG_i\to\mathbf{Z}$ vanish on $\bigoplus G_i$. If $f$ is nonzero, say on some element $(g_i)_i$, consider the homomorphism $\mathbf{Z}^I\to\prod_i G_i$ mapping $(n_i)_i$ to $(n_ig_i)_i$; it maps $\mathbf{Z}^{(I)}$ into $\bigoplus G_i$ and we can conclude from the fact that $\mathrm{Hom}(\prod^\star_{i\in I}\mathbf{Z},\mathbf{Z})=\{0\}$.

I don't know a reference, so let me provide a proof.

Choose, by contradiction, $f\neq 0$ as above. For every subset $J$ of $I$, denote by $R_J(f)$ the restriction of $f$ to $\prod_{j\in J}G_j$. Let $\mathbf{F}$ be the set of $J$ such that $R_J(f)\neq 0$. Clearly $J'\subset J\notin\mathcal{F}$ implies $J'\notin\mathcal{F}$. Let us say that $J\in\mathcal{F}$ is simple if it is not disjoint union of two elements of $\mathcal{F}$.

Claim: There exists a simple $J\in\mathcal{F}$.

Proof of claim: otherwise, one chooses $X'_0\in\mathcal{F}$, partition it as $X'_0=X_0\sqcup X'_1$ with $X_0,X'_1\in\mathcal{F}$, then partition $X'_1=X_1\sqcup X'_2$, etc, to obtain a partition $(X_n)$ of disjoint elements in $\mathcal{F}$. Then choosing $x_n$ with $f(x_n)\neq 0$, supported in $X_n$, for $u=(u_n)_n\in\mathbf{Z}^{\mathbf{N}}$, we define $f'(u)=f(\sum u_nx_n)$, we obtain a contradiction with the previously proved countable case.

Now choose $J$ simple, and define $\mathcal{U}=\{J'\in\mathcal{F}:J'\subseteq J\}$. Then $\mathcal{U}$ is an ultrafilter on $J$. Furthermore, it is stable under countable intersections: it is enough to show that the set of complements of elements of $\mathcal{U}$ (which is also the complement of $\mathcal{U}$) is stable under countable disjoint unions. This is, again, a straightforward consequence of $\mathrm{Hom}(\prod_{n\in\mathbf{N}}^\star\mathbf{Z},\mathbf{Z})=\{0\}$.

Corollary: let $K$ be a field. Then the field $K(t)$ is not isomorphic to any nonprincipal ultraproduct of fields indexed by any reasonable (in the above sense) set. More precisely, the group $K(t)^*$ is not isomorphic to any nonprincipal ultraproduct of groups indexed by any reasonable set.

Again, this is false if one allows non-reasonable cardinals. Indeed, if the ultrafilter is stable under countable intersections, and the field $K$ is countable, then the field $K(t)$ is isomorphic to its own ultrapower with respect to such an ultrafilter.


Fact. Let $\sigma$ be an ultrafilter on a set $I$. Equivalences:

  1. Then $\sigma$ is not stable under countable intersections.

  2. There exists a function $u:I\to\mathbf{N}$ such that $\lim_{i\to\sigma}u=\infty$

  3. There exists a strictly decreasing sequence $(I_n)$ of subsets of $I$, with empty intersection and $I_n\in\sigma$ for all $n$.

(An ultrafilter stable under countable intersections is usually called $\omega_1$-complete, or $\sigma$-complete.)

Proof: Suppose 1. So we have a sequence $(J_n)$ with $J_n\in\sigma$ and $J=\bigcap J_n\notin\sigma$. Setting $I_n=J^c\cap \bigcap_{i\le n}J_n$, we obtain 3. Supposing 3 (with $I_0=I$), we obtain 2 by defining $u(i)=\sup\{n:i\in J_n\}$. 2 implies 1 is clear, choosing $I_n=\{i:u(i)\ge n\}$. $\Box$

Therefore, the ultrafilter being non-$\sigma$-complete (rather than non-principal) is the optimal assumption for the proposition to hold. Namely:

Proposition: let $K$ be a field. Then the field $K(t)$ is not isomorphic to any non-$\sigma$-complete ultraproduct of fields; more precisely, the group $K(t)^*$ is not isomorphic to any non-$\sigma$-complete ultraproduct of groups. Conversely, if $K$ is a countable field, then the field $K(t)$ is isomorphic to any $\sigma$-complete ultraproduct of itself.

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