Is $mathbb{Q};cong; (prod_{ninomega}mathbb{Z}/p_nmathbb{Z})/simeq_{cal U}$?

Question

Let ${cal U}$ be a non-principal ultrafilter on $omega$, and for each $ninomega$, 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 $big(prod_{ninomega}mathbb{Z}/p_nmathbb{Z}big)$: we say $a simeq_{cal U} b$ for $a,b in big(prod_{ninomega}mathbb{Z}/p_nmathbb{Z}big)$ if and only if $${ninomega:a(n) = b(n)}in {cal U}.$$
I think I have proved that $big(prod_{ninomega}mathbb{Z}/p_nmathbb{Z}big)/simeq_{cal U}$ is a field. Is that field isomorphic to $mathbb{Q} $ ?

Alex Kruckman · Answer

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:

The field axioms.
The field is perfect.
The field has a unique algebraic extension up to isomorphism of each positive degree.
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 $3$.

Asaf Karagila · Answer

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.

tkf · Answer

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 $p_nequiv 1 mod 4$, then your field contains $i$, so is not $mathbb{Q}$.
Thus your field need not be $mathbb{Q}$.

Is $mathbb{Q};cong; (prod_{ninomega}mathbb{Z}/p_nmathbb{Z})/simeq_{cal U}$?

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