## Ultraproducts of fields, I June 16, 2008

Posted by Alexandre Borovik in Uncategorized.

My immediate research interests more and more focus on an interplay between finite and infinite in algebra (at least this is where my chats with my PhD student drift to). In particular, I have to use frequently a specific construction, an ultrafilter product of fields. It is pretty sublime in the sense of David Corfield and leads to appearance of very interesting canonical objects.

We start with a family of fields $F_i$, $i \in I$ . For simplicity assume that all fields are finite of unbounded order and that the index set $I$ is just the set of natural numbers $\mathbb{N}$ (actually this is the case most interesting to me). We form the Cartesian product $R$ of $F_i$: this is just a set of infinite strings $\{ (f_1, f_2, \dots ) \mid f_i \in F_i \}$

with componentwise operations of addition and multiplication. In what follows, we shall make frequent use of zero set of a string $f = (f_1, f_2, \dots )$, that is, the set of indices where the string components are zeroes: $zero(f) = \{ i \in {\mathbb{N}} \mid f_i =0\}$.

Obviously, $R$ is a commutative ring with unity. Let us look at its ideals. One can easily see that an ideal $I$ in $R$ is uniquely determined by the set $zero(I) = \{zero(f) \mid f \in I\}$;

indeed, non-zero components of a string $f \in I$ can be arbitrarily changed without moving $f$ outside of $I$ by multiplying by appropriate string of invertible elements. One can also instantly see that $zero(I)$ is a filter on $\mathbb{N}$, that is, it is a collection of non-empty subsets of $\mathbb{N}$ closed under taking finite intersections and supersets, and, moreover, that the correspondence $I \mapsto zero(I)$

is a one-to-one correspondence between proper ideals in $R$ and filters of $\mathbb{N}$ which preserves embedding of ideals and embedding of filters. Therefore, maximal ideals in $R$ correspond to maximal filters on $\mathbb{N}$; the former and the latter exist by the Zorn Lemma, one of the equivalent formulations of the Choice Axiom. If now $I$ is a maximal ideal in $R$, the fact that the factor ring $R/I$ is a field and, in particular, has no zero divisors, translates in the fact that a maximal filter $\mathcal{F}$ is an ultrafilter: it has the property that, for any subset $X \subseteq \mathbb{N}$ either $X$ or its complement ${\mathbb{N}} \smallsetminus X$ belongs to $\mathcal{F}$.

Given an ultrafilter $\mathcal{F}$ on $\mathbb{N}$, the ultraproduct $F = \prod F_i/\mathcal{F}$ is nothing more than the corresponding residue field $R/I$. There are obvious principal (ultra)filters on $\mathbb{N}$, they consist of all subsets containing a given element $i \in \mathbb{N}$; obviously, the corresponding ultraproduct is just the original field $F_i$.

Non-principal filters do exists. One very interesting non-principal filter on $\mathbb{N}$ is the Frechet filter consisting of all subsets with finite complements.

But if we take an ultrafilter containing a non-principal filter (it exists by the Zorn Lemma), the corresponding ultraproduct $F$ has many marvelous properties. In particular, if all $F_i$ have different characteristics, $F$ turns out to be a field of characteristic zero (I leave the proof of this fact as an exercise to the reader).

In the next instalment of this post, I will discuss the meaning of a frequently used assertion that an ultraproduct $F$ is a limit at infinity of finite fields $F_i$.

1. C Smith - June 16, 2008

Is there a link to a good “decoder ring” page for the symbology used in these posts?
As a non-academic, material like this is a bagel for the mind.
Thanks,
Chris

2. Alexandre Borovik - June 16, 2008

Wikipedia is quite a decent source for explanation of mathematical terminology.

3. Ultraproducts of fields, II « A Dialogue on Infinity - June 28, 2008

[…] June 28, 2008 Posted by Alexandre Borovik in Uncategorized. trackback I continue my post on ultraproducts. So, we want to understand in what sense an ultraproduct of finite fields of […]

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