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Let ${\cal U}$ be a non-principal ultrafilter on $\omega$, and for each $n\in\omega$, let $p_n$ denote the $n$th prime, that is $p_0 = 2, p_1=3, \ldots$.

Next we introduce the following standard equivalence relation on $\big(\prod_{n\in\omega}\mathbb{Z}/p_n\mathbb{Z}\big)$: we say $a \simeq_{\cal U} b$ for $a,b \in \big(\prod_{n\in\omega}\mathbb{Z}/p_n\mathbb{Z}\big)$ if and only if $$\{n\in\omega:a(n) = b(n)\}\in {\cal U}.$$

I think I have proved that $\big(\prod_{n\in\omega}\mathbb{Z}/p_n\mathbb{Z}\big)/\simeq_{\cal U}$ is a field. Is that field isomorphic to $\mathbb{Q} $ ?

Dominic van der Zypen
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  • If you use the algebraic closure of $\mathbf Z/p_n\mathbf Z$ in the $n$th component then the resulting ultraproduct is an algebraically closed field of characteristic $0$ but it almost certainly is not the algebraic closure of $\mathbf Q$. – KCd Jul 22 '20 at 08:26
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    @KCd: In that case it is in fact the algebraic closure of $\Bbb R$! :-) – Asaf Karagila Jul 22 '20 at 08:54
  • @KCd If $\mathcal{V}$ contains the set of $n$ such that $p_n \equiv 3 \mod 4$ then the field does not contain $i$ so is not algebraically closed. Are you saying you can choose $\mathcal{V}$ so that the resulting field is algebraically closed? – tkf Jul 22 '20 at 09:24
  • @AsafKaragila do you mean just for cardinality reasons? That ultraproduct does not contain the real numbers in a natural way as far as I can tell. – KCd Jul 22 '20 at 14:14
  • @tkf perhaps you misread what I wrote: use the algebraic closure of the field of size $p_n$ in the $n$th slot. So every component contains a solution to $x^2 + 1$, and thus the ultraproduct does as well. – KCd Jul 22 '20 at 14:17
  • @KCd: Yes. Any two algebraically closed fields of the same characteristics and cardinality are isomorphic. And yes, there is no natural way to construct this isomorphism, since it is consistent with ZF that there are no automorphisms of $\Bbb C$ other than conjugation and the identity, and such isomorphism would allow us to define such automorphism (if my memory serves me right), although in those models there are no free ultrafilters on $\omega$ anyways, so it becomes moot. – Asaf Karagila Jul 22 '20 at 14:21
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    @AsafKaragila of the same characteristic and uncountable cardinality, of course. – KCd Jul 22 '20 at 14:32
  • @KCd: Yes, and since the algebraic closure of each finite field is countable, the ultraproduct will have size $2^{\aleph_0}$ and characteristics $0$. Since algebraic closure is a first-order property, it will also be algebraically closed. So all the conditions are ripe for it to be isomorphic to $\Bbb C$. – Asaf Karagila Jul 22 '20 at 14:33
  • @AsafKaragila what I find amusing about this construction of an algebraically closed field of characteristic 0 is that it has many "concrete" automorphisms, by raising to some iterate of the $p_n$th power in the $n$th component for each $n$, although it's not clear to me how to prove any specific such automorphism is not the identity map on the subfield $\overline{\mathbf Q}$ or how to construct such an automorphism on the ultraproduct that has order $2$. – KCd Jul 22 '20 at 14:41
  • @KCd: (Caveat: I have no idea if what I'm saying is remotely true, it's too number theoretic for me.) Suppose that the set ${p\mid -1\text{ is not a quadratic residue in }\Bbb Z/p}$ is in the ultrafilter. Then the alg. closure of $\Bbb Z/p$ has an automorphism of order 2, i.e. conjugation of the $\sqrt{-1}$. And I would suspect that these automorphisms will turn into an automorphism of order 2 on the product. – Asaf Karagila Jul 22 '20 at 14:47
  • @AsafKaragila the algebraic closure of a finite field has no automorphism of order 2: if an algebraically closed field has an automorphism of finite order greater than $1$ then the field has characteristic $0$ by Artin--Schreier. I suspect you are thinking of automorphisms on a subfield without accounting for extending them to the whole field. – KCd Jul 22 '20 at 15:03
  • @KCd: Ah, you must be right about this. Anyway, that's all an interesting question, but in this case I suppose that the answer is that there is no "reasonable automorphism of order 2" because that would mean there is a "reasonable embedding of the reals" or rather "a reasonable isomorphism with $\Bbb C$", and we know that's not the case. :-) – Asaf Karagila Jul 22 '20 at 15:32
  • @KCd I see by 'algebraic closure' you meant noun, not adjective. – tkf Jul 22 '20 at 17:34
  • @AsafKaragila A simple way to see KCd's point is if $-1$ is an NQR mod $p$ then $p \equiv 3 \mod 4$ so $i^p=-i$ so the automorphism you refer to is just the Frobenius map, which of course has infinite order on $\bar{\mathbb{F}_p}$. – tkf Jul 22 '20 at 21:38
  • @tkf the situation is not that simple. Finding an example of an automorphism of infinite order sending $i$ to $-i$ in characteristic $p$ does not prove there can't be an automorphism of finite order sending $i$ to $-i$. Further reasoning is needed. Moreover, it is a nontrivial theorem that if an algebraically closed field has an automorphism with finite order greater than $1$ then the field has characteristic $0$. – KCd Jul 22 '20 at 21:54
  • @KCd Yes - I was just referring to why the construction given in the comment, of taking an arbitrary automorphism swapping $i$ and $-i$ in each co-ordinate would not work. – tkf Jul 22 '20 at 21:57

3 Answers3

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This is a simple cardinality argument. Ultraproducts of finite sets are either finite or uncountable. Since the ultrafilter is free, and the sets are all increasing in size, it is not finite.

To see why the ultraproduct is indeed uncountable, take an almost disjoint family of subsets of $\Bbb N$, and consider their characteristics functions. These functions agree on finitely many points with each other, so their equivalence classes are different in the ultraproduct (if two functions have the same equivalence class in the ultraproduct, they must have agreed on infinitely many values). Now recall that there are almost disjoint families of size $2^{\aleph_0}$ and finish the proof.

Asaf Karagila
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  • Thanks Asaf! - Is the resulting field isomorphic to $\mathbb{R}$ or is it something different? – Dominic van der Zypen Jul 22 '20 at 08:53
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    Of course not. It's very very easy to show that it's not ordered. – Asaf Karagila Jul 22 '20 at 08:53
  • @AsafKaragila I think I am missing the 'very very easy' argument for general $\mathcal{V}$ - will it fit into a comment? – tkf Jul 22 '20 at 21:42
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    @tkf: The very very easy is a remnant of "non-Arcihmedean", which would also be true, even if the field is ordered. Less easily, I suppose, either $-1$ or $-3$ are the sum of squares. – Asaf Karagila Jul 22 '20 at 21:54
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    @AsafKaragila Thanks (+1), I see: Lagrange's four-square theorem implies that $p-1$ can always be written as the sum of 4 integers squares, so $-1$ is the sum of 4 squares. – tkf Jul 22 '20 at 23:28
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Let's call your ultraproduct $K$. What's clear is that $K$ is a field of characteristic $0$ (by Łoś's theorem) of cardinality $2^{\aleph_0}$ (by the argument in Asaf's answer).

What's less clear is that we can push Łoś's theorem a bit further: $K$ is an ultraproduct of finite fields, and there are sentences that do not follow from the field axioms, but which are true in every finite field, and hence are true in $K$. Ax axiomatized the theory of finite fields in 1968. The axioms are:

  1. The field axioms.
  2. The field is perfect.
  3. The field has a unique algebraic extension up to isomorphism of each positive degree.
  4. The field is pseudo algebraically closed (PAC): every absolutely irreducible variety over the field has a point in the field.

The infinite models of this theory are called pseudo-finite fields. Note that it's not obvious that these axioms are expressible by first order schemas - it takes some work to show that they are.

Since your field $K$ is pseudo-finite, it is certainly not isomorphic to $\mathbb{R}$ or $\mathbb{C}$. For example, there exists an irreducible polynomial in $K[x]$ of degree $3$.

Alex Kruckman
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  • Thanks for your comments - beautiful! – Dominic van der Zypen Jul 22 '20 at 12:29
  • @AlexKruckman Your last sentence is equivalent to saying that there is a finite set of cubics $q_1,\cdots,q_k$ over the integers, such that for any integers $x_1,\cdots,x_k$ , the integers $q_1(x_1),\cdots,q_k(x_k)$ are coprime (otherwise you can construct an ultrafilter $\mathcal{V}$ to falsify the statement) . That seems like a very concrete statement for such abstract reasoning. I am curious, can you list the cubics? – tkf Jul 23 '20 at 00:42
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    @tkf I don't see the equivalence - maybe you could explain more? Note that the coefficients of the cubic come from $K$, not $\mathbb{Z}$. The more obvious equivalent to "there exists an irreducible cubic in $K[x]$" is "for all but finitely many primes $p$, there exists an irreducible cubic in $\mathbb{Z}/p\mathbb{Z}[x]$" (of course it's a standard theorem about finite fields that this is actually true for all primes $p$). – Alex Kruckman Jul 23 '20 at 03:35
  • @AlexKruckman That is exactly what I did - I took coefficients in $\mathbb Z$ instead of $K$. Thanks for clearing that up (+1's for comment and answer). Any idea off the top of your head if their must be an irreducible cubic (or quadratic) with integer coefficients? – tkf Jul 23 '20 at 03:42
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    @tkf It depends on the choice of ultrafilter. Any subfield of the algebraic closure of Q that has at most one algebraic extension of each degree can appear as the relative algebraic closure of Q inside an ultraproduct of the Z/pZ. In particular, that field may or may not have a degree 3 algebraic extension. See Proposition 7 (p. 260) of Ax's paper "The Elementary Theory of Finite Fields" - the proof uses the Čebotarev Density Theorem. – Alex Kruckman Jul 23 '20 at 13:31
  • To answer my question above it suffices to show that $K$ may contain $\bar{\mathbb Q}$, which Ax does in just one paragraph of that proof (top of page 261), so the Čebotarev density theorem is not needed for this and your reference is self-contained (you can get $\alpha(f)$ infinite by restricting Euler's product formula and using $\sum_{r=m}^\infty {f(r)^{\frac {-1}{n+1}}}$ diverges for a degree $n$ polynomial $f$ with positive leading coefficient). Actually, even that paragraph can be written in more elementary terms, though the primitive element theorem seems necessary. Thanks again. – tkf Jul 30 '20 at 06:10
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It is certainly a field which contains $\mathbb{Q}$. If an element $x$ is $0$ on a large set then $x=0$ and otherwise we may let $y$ be the inverse of $x$ on a large set so that $xy$ agrees with $1$ on a large set and $xy=1$. (Here a subset of $\omega$ is said to be large if it lies in $\mathcal V$).

If $\mathcal V$ contains the set of $n$ for which $p_n\equiv 1 \mod 4$, then your field contains $i$, so is not $\mathbb{Q}$.

Thus your field need not be $\mathbb{Q}$.

tkf
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