$K$-valued functions on an infinite set

Question

This post is inspired by this answer of wxu.

Let $K$ be a field, let $I$ be an infinite set, and let $A$ be the $K$-algebra of all $K$-valued functions on $I$, that is $$ A=K^I=\prod_{i\in I}K. $$ Let $\mathfrak a$ be the ideal of $A$ formed by finitely supported $K$-valued functions on $I$, that is $$ \mathfrak a=K^{(I)}=\bigoplus_{i\in I}K. $$ Let $\mathfrak m$ and $\mathfrak n$ be two maximal ideals of $A$ containing $\mathfrak a$. In particular the fields $A/\mathfrak m$ and $A/\mathfrak n$ are extensions of $K$.

Question 1. Are the fields $A/\mathfrak m$ and $A/\mathfrak n$ necessarily isomorphic?

The answer is yes if $K$ is finite. More precisely, in this case we have $A/\mathfrak m\simeq K$. Indeed, let $q$ be the cardinality of $K$ and observe that any element $x$ of $A$, and thus of $A/\mathfrak m$, satisfies $x^q=x$.

But in general we have $A/\mathfrak m\not\simeq K$. For instance if $A:=\mathbb Q^{\mathbb N}$, then

(1) $A/\mathfrak m$ has the cardinality of the continuum.

To see this, consider the set theoretical map
$$ \phi:\{0,1\}^{\mathbb N}\to A=\mathbb Q^{\mathbb N} $$ whose $n$-th component $$ \phi_n:\{0,1\}^{\mathbb N}\to\mathbb Q $$ is defined by $$ \phi_n(x)=\sum_{i=0}^n2^ix_i. $$ We claim that the composition $$ \pi\circ \phi:\{0,1\}^{\mathbb N}\to A/\mathfrak m $$ of $\phi$ with the canonical projection $\pi:A\to A/\mathfrak m$ is injective. This will imply (1).

Let $x$ and $y$ be two distinct elements of $\{0,1\}^{\mathbb N}$. It suffices to check that $a:=\phi(x)-\phi(y)$ is not in $\mathfrak m$. Writing $a_i$ for the value of $a$ at $i$, there is an $i_0\in\mathbb N$ such that $a_i\neq0$ for $i>i_0$, and thus there is a $b$ in $\mathfrak a\subset\mathfrak m$ such that $a_i+b_i\neq0$ for all $i$, that is $a+b$ is a unit of $A$. This implies $a$ cannot be in $\mathfrak m$, as was to be shown.

Going back to the general setting of Question 1, note that $K$ is algebraically closed in $A/\mathfrak m$. Indeed, let $f\in K[x]$ (where $x$ is an indeterminate) be irreducible of degree at least two, and let $a$ be in $A$. It suffices to show $f(\pi(a))\ne0$, where $\pi:A\to A/\mathfrak m$ is the canonical projection. Since $f(\pi(a))=\pi(f(a))$, we must check that $f(a)$ is not in $\mathfrak m$. But $f(a)$ is a unit of $A$ because we have $f(a)_i=f(a_i)\ne0$ for all $i\in I$.

Question 2. Is $A/\mathfrak m$ always purely transcendental over $K$?

As indicated above, this is true if $K$ is finite (the transcendence degree being $0$ in this case).

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EDIT 1. Question 2 is a duplicate of this MathOverflow Question of Lisa S. It seems to me Lisa's question is answered by a comment of Tim Campion.

More precisely, Tim says

"if $F$ is algebraically closed, then so is the ultrapower (it is elementarily equivalent). But I don't think any nontrivial purely transcendental extension of any field is algebraically closed..."

If $F$ is an algebraic closure of $\mathbb Q$, then, as explained in the question, any ultrapower has at least the cardinality of the continuum. In particular such an ultrapower is a proper extension of $F$.

I'm taking this opportunity to thank Anonymous for their answer. I'm studying this answer, and will perhaps make further edits, but I wanted to make this one right away.

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EDIT 5. I'm removing Edits 2, 3 and 4. I'm only keeping this link to a MathOverflow answer of Joel David Hamkins.

Summary of the answers: Eric Wofsey's answer shows that the answer to Question 1 is No. Tim Campion's argument mentioned in Edit 1 shows that the answer to Question 2 is also No.

Both answerers asked me generously to accept the other answerer's answer!

Answer 1

For any infinite field $K$, there exists an infinite set $I$ such that two different nonprincipal ultrapowers of $K$ by $I$ have different cardinalities (and are thus not isomorphic). Note first that by picking an ultrafilter that countains a countable subset of $I$, you get an ultrapower of cardinality at most $|K|^{\aleph_0}$ (in terms of ideals, this means there is a countable subset $J\subseteq I$ such that the maximal ideal contains all elements of $K^I$ that vanish on $J$). On the other hand, as mentioned in Anonymous's answer, if the chosen ultrafilter on $I$ is regular, then the ultrapower has cardinality $|K|^{|I|}$. So, if you choose $I$ such that $|K|^{|I|}>|K|^{\aleph_0}$, there will be two nonprincipal ultrapowers of different cardinality.

Answer 2

As discussed in the question you linked, maximal ideals of $K^I$ correspond to ultrafilters of $I$, and consequently, rings of the form $A/\mathfrak{m}$ are the same as ultrapowers of $K$. With that in mind, we can approach an answer to your first question.

A positive answer to your first question is not provable from ZFC (assuming ZFC is consistent). A well known example of the fields you discuss is a field of hyperreal numbers which is any ring of the form $\mathbb{R}^\mathbb{N}/\mathfrak{m}$ for a maximal ideal $\mathfrak{m}$ of $\mathbb{R}^\mathbb{N}$ containing $\oplus_{n=0}^\infty\mathbb{R}$. As stated by wikipedia and implied by Theorem 1 of this paper, if CH fails, then there will exist non-isomorphic ultrapowers of $\mathbb{R}$. In fact, $\mathbb{R}$ can be replaced with any real closed field, and the same incident will occur.

However, it gets worse. If $K$ is infinite and the ideal $\mathfrak{m}$ corresponds to what is called a $\textit{regular}$ ultrafilter of $I$, then the cardinality of $A/\mathfrak{m}$ will be the same as that of $A$ due to a theorem attributed to Keisler. Yet, there may exist nonregular ultrafilters. Indeed, Magidor shows that if the existence of a huge cardinal is consistent with ZFC, then it is consistent with ZFC that there exists a set $I$ and ultrafilter $U$ on $I$ such that the ultraproduct $\omega_0^I/U$ has a strictly smaller cardinality than $\omega_0^I$ does. As such, if we take $K$ to be any countably infinite field, then with the assumptions just stated, there could exist maximal ideals $\mathfrak{m}$ and $\mathfrak{n}$ for which $A/\mathfrak{m}$ and $A/\mathfrak{n}$ do not even have the same cardinality. (Edit: This is overkill. We can obtain a nonregular free ultrafilter by just choosing one which is not not uniform i.e. containing a subset of $I$ with cardinality less than that of $I$. Consequently, this shows in ZFC alone that $A/\mathfrak{m}$ and $A/\mathfrak{n}$ can have different cardinalities. Thanks for the catch Eric Wofsey.)

Edit 1: Due to a theorem of Chang and Keisler in their book "Model Theory", if $I$ is countable, $K$ is a field of cardinality at most $2^{\aleph_0}$, and CH holds, then all ultrapowers of $K$ are isomorphic. However, this does not answer the question when $K$ has cardinality greater than $2^{\aleph_0}$ or $I$ is uncountable and CH holds, so I would consider Pierre's first question still not entirely solved. As I noted earlier, the answer to Pierre's first question is negative if CH fails, but I would be interested to know if the question is provable or refutable (or neither) in ZFC+CH.

Edit 2: I will expand on Tim Campion's Math Overflow comment which yields a negative answer to the second question. It is a consequence of Los's theorem that any structure is elementarily equivalent to all of its ultrapowers, meaning that they satisfy the same first order sentences. In particular, for each positive integer $n$, the sentence $$\forall a_0\forall a_1\ldots \forall a_n\exists x(a_nx^n+\cdots+a_1x+a_0=0)$$ holds in $K$ if and only if it holds in $A/\mathfrak{m}$. Hence, if we choose $K$ to be an algebraically closed field, then $A/\mathfrak{m}$ will be as well. However, if $A/\mathfrak{m}$ were purely transcendental over $K$, we would have $A/\mathfrak{m}=K(S)$ for some algebraically independent set $S\subset A/\mathfrak{m}$. But $K(S)$ is not algebraically closed when $S$ is nonempty since the polynomial $x^2-s$ is irreducible over $K(S)$ for $s\in S$. If $S$ is empty, then $A/\mathfrak{m}=K$ which need not be the case.