Let $R$ be the ring $$R = \prod_{p\ \text{prime}} \mathbb{F}_p$$ where $\mathbb{F}_p$ is the field having $p$ elements.

Is it true that $R$ has a quotient by a maximal ideal which is a field of characteristic zero and contains $\overline{\mathbb{Q}}$?

Motivation: I like the problem and I can't solve it...

It should have something to do with the Chebotarev density theorem.

  • 1
    $\begingroup$ There is a proof that avoids the Chebotarev Density Theorem, the missing piece is in Bjorn Poonen's answer here: mathoverflow.net/questions/15220/… $\endgroup$ – François G. Dorais Feb 14 '10 at 3:16
  • $\begingroup$ The funny thing is: I have been looking at the classical paper "Diophantine problems over local fields" (Ax & Kochen) during the past few days, and in fact the answer to my question is really in this paper. $\endgroup$ – Wanderer Feb 26 '10 at 11:40
  • $\begingroup$ Related: math.stackexchange.com/questions/1090780 $\endgroup$ – Watson Feb 17 '17 at 15:05

The answer is Yes, and this is the ultraproduct construction. Let U be any nonprincipal ultrafilter on the set of primes. This is simply the dual filter to a maximal ideal on the set of primes, containing all finite sets of primes. (In other words, U contains the Frechet filter.)

The quotient R/U is a field of characteristic 0.

The ultraproduct construction is completely general, and has nothing to do with rings or fields. If Mi for i in I is any collection of first order structures, and U is an ultrafilter on the subsets of I, then we may form the ultraproduct Π Mi/U, which is the set of equivalence classes by the relation f equiv g iff { i in I | f(i) = g(i) } in U. Similarly, the structure is imposed on the ultraproduct coordinate-wise, and this is well-defined. The most important theorem is Los's theorem, which says that Π Mi/U satisfies a first order statement phi([f]) if and only if { i in I | Mi satisfies phi(f(i)) } in U.

In your case, since every Fp is a field, the ultraproduct is also a field. And since the set of p bigger than any fixed n is in U, then the ultraproduct will have 1+...+1 (n times) not equal to 0, for any fixed n > 0. That is, the ultraproduct will have characteristic 0.

Edit: I confess I missed the part initially about containing the algebraic numbers, and so there is more to be done, as Kevin points out. What Los's theorem gives you is that something will be true in R/U just in case the set of p for which F_p has the property is in the ultrafilter U.

What you need to know is that for any finite list of equations, that there is an infinite set of primes p for which the equations have a solution in Fp. This property is equivalent to asking whether every finite list of equations over Z is true in at least one Fp, since one can always add one more equation so as to exclude any particular Fp. Is this true? (I wasn't sure.)

But according to what Kevin says in the comments below, it is true, and this is precisely what you need for the construction to go through. You can form a filter containing those sets, which would form a descending sequence of subsets of primes, and then extend this to an ultrafilter. In this case, any particular equation would have a solution in Fp for a set of p in U, and so the ultrapower R/U would have a solution. In this case, the ultrapower will contain the algebraic numbers.

  • $\begingroup$ Interesting. I must confess that I do not know the techniques you are using. I "met" the problem in the context of Chebotarev's density theorem... $\endgroup$ – Wanderer Feb 12 '10 at 23:56
  • $\begingroup$ This is a method that comes from logic. Ultraproducts and ultrapowers appear all over model theory and set theory. Many large cardinal axioms are stated in terms of various ultrapower of the universe. $\endgroup$ – Joel David Hamkins Feb 13 '10 at 0:03
  • $\begingroup$ Joel: I read the question as "...and contains an algebraic closure of Q". Am I right in thinking that you read it some other way? $\endgroup$ – Kevin Buzzard Feb 13 '10 at 0:05
  • $\begingroup$ PS Joel: if I'm right and he did want Q-bar in, can you glance over my answer and instantly tell me exactly what statements about number fields one needs to finish the job? $\endgroup$ – Kevin Buzzard Feb 13 '10 at 0:07
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    $\begingroup$ @Joel: No. Every F_p has an irreducible quadratic, hence so does the ultrapower. $\endgroup$ – François G. Dorais Feb 13 '10 at 4:31

How far does the logic route get you? Here's a plan of attack. Say we have such a max ideal $m$.

1) If $S$ is a subset of the primes then there's an idempotent $e_S$ in $R$, which is 1 at $p$ if $p\in S$ and $0$ at $p$ otherwise. Now $e_S$ is an idempotent so either $e_S\in m$ or $e_S-1\in m$. Say $S$ is "big" if $e_S-1\in m$ and "small" if $e_S\in m$. I suspect that

(@) a set is big iff its complement is small (easy).

(a) $m$ is determined by the collection of $S$ such that $S$ is big

(b) the big $S$ form an ultrafilter

(c) This gives you a bijection between ultrafilters on the set of primes and maximal ideals.

The principal ultrafilters are the ones corresponding to the obvious quotients $R\to\mathbf{F}_p$. Any non-principal ultrafilter will give a quotient of $R$ of characteristic zero.

I don't know if (1) is right but I've always suspected it is. If it's true the proof will be straightforward algebra.

2) Now use the logicians point of view. A 1st order statement of field theory is true in $R/m$ iff the set of primes for which it's true in $\mathbf{F}_p$ is big. For example the statement "I am a field with a square root of 2" is true for a set of primes of density 1/2 (those which are 1 or 7 mod 8), so this set had better be big if you want the quotient to have a square root of 2.

3) Now enumerate all polynomials with integer coefficients and say "I have all my roots in $R/m$". All these sets had better be big. So now we have some sort of combinatorial question, which I think is the following. Let $K$ run through all number fields, Galois over $\mathbf{Q}$ for simplicity, and for each one let $S(K)$ denote the set of primes that split completely in $K$. Is there an ultrafilter which contains all of these $S(K)$?

4) Now a bunch of sets is contained in an ultrafilter iff it generates a filter that isn't the whole set of primes. But now this is some combinatorial statement about primes splitting completely in number fields. For example, statements of the form "if $S$ is the primes that split completely in $K$ and $T$ is the set of primes that split completely in $L$, then $S\cap T$ is the set of primes that split completely in $KL$" will be of use to you here. In fact is that all you need?? Figure this out and you're home.

It's time for bed here in the UK (well, at least if you have 3 kids it is) but I'd be interested to know what happens if you try and push this through.

  • $\begingroup$ Joel's answer basically says "you only need one direction of (1): ultrafilters give you max ideals". The comments below his answer justify that the primes that split completely do indeed extend to a non-principal ultrafilter. So the result is true! $\endgroup$ – Kevin Buzzard Feb 13 '10 at 0:39

Combining the answers of Joel and Kevin does work.

Write $\overline{\mathbb{Q}} = \bigcup_{n=0}^\infty K_n$ where the $K_n$ are an increasing sequence of finite Galois extension of $\mathbb{Q}$. Let $S_n$ be the primes that split completely in $K_n$. By Chebotarev's Theorem, this is a descending sequence of infinite sets of primes. Therefore, there is a nonprincipal ultrafilter $\mathcal{U}$ over the set of primes which contains all $S_n$. If $f(x) \in \mathbb{Z}[x]$ has a root in $K_n$, then $f(x)$ has roots in $\mathbb{F}_p$ for all $p \in S_n$ and hence $f(x)$ has a root in the ultraproduct $\prod_p \mathbb{F}_p/\mathcal{U}$.

  • $\begingroup$ Sorry for the duplication, I hadn't read the long thread of comments where this is all explained. $\endgroup$ – François G. Dorais Feb 13 '10 at 1:07
  • $\begingroup$ It's actually only a weak consequence of Cebotarev's theorem, and can be obtained in a more elementary manner. You just need that the zeta function of the number field has a pole, which is a basic fact. This is one of those algebraic-looking results whose most natural proofs are analytic. $\endgroup$ – Kevin Buzzard Feb 13 '10 at 8:44
  • $\begingroup$ Very interesting point, Kevin. Is there a proof of the infinitude of completely splitting primes that doesn't involve analysis? (I vaguely remember seeing such a thing in the Abelian case.) $\endgroup$ – François G. Dorais Feb 13 '10 at 16:54
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    $\begingroup$ There is a very simple elementary proof of the infinitude of completely split primes! mathoverflow.net/questions/15220/… $\endgroup$ – François G. Dorais Feb 14 '10 at 2:33

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