2. Ultraproducts

An ``ultraproduct'' of algebras is their direct product modulo a congruence relation constructed from a nonprincipal ultrafilter. The congruence relation tends to collapse the product down to something that looks like a ``generic'' copy of the individual algebras, reflecting whatever features they have in common.

The construction is set-theoretic and actually works for sets with relations as well as for algebras. In detail:

Definition. Let $I$ be an infinite index set. Let algebras $A _ i, i \in I$ be given. Choose a nonprincipal ultrafilter ${\cal U}$ on $I$. On the direct product $\prod _ {i \in I} A _ i$, define an relation $\equiv$ by saying a$\equiv$   b when a and b agree on a large set of indices. The ultraproduct of the $A _ i$ is the direct product modulo $\equiv$:

$A ^ * = (\prod _ {i \in I} A _ i)/\equiv$, or more simply $A ^ * = \prod _ {i \in I} A _ i/{\cal U}$.

There are several things to consider here:

• Does the phrase ``agree on a large set of indices'' mean that there is some large set $J \subseteq I$ of indices such that $a _ j = b _ j$ for all $j \in J$, or that the set of all $i \in I$ with $a _ i = b _ i$ is large? By the properties of large sets, it doesn't matter; the meanings are the same.

• It must be checked that $\equiv$ is an equivalence relation. This follows from the properties of large sets.

• It must also be checked that $\equiv$ is a congruence relation, so that the ultraproduct is an algebra.

• We say ``the'' ultraproduct even though the result does depend on the choice of ${\cal U}$.

Ultraproducts have some startling properties:

1. Any $n$-ary relation common to the $A _ i$ has a reasonable definition on their ultraproduct.

2. Any first-order sentence true in the $A _ i$ is true in their ultraproduct. (This extends to first-order formulas.)

3. An ultraproduct of fields is a field. (Why?)

4. The ultraproduct is unchanged (up to isomorphism) if finitely many factors are omitted. (Why?)

5. If all the $A _ i$ are finite and isomorphic, then $A ^ *$ is a copy of the same algebra. (Why?)

Examples.

(a) The ultraproduct of countably many copies of the field R of reals is the field R$^ *$ of ``nonstandard reals''. It is possible to do calculus using ``infinitesimals'' in R$^ *$.

(b) The ultraproduct of countably many copies of the ring Z of integers is the ring Z$^ *$ of ``nonstandard integers''. Some of them are ``infinite''.

(c) The ultraproduct Z$_2 \times$   Z$_3 \times$   Z$_5 \times \cdots/{\cal U}$ is a field of characteristic 0.

(d) The ultraproduct of chains 1$\times$   2$\times$   3$\times \cdots/{\cal U}$ is an infinite chain. (What does it look like?)