# Non-axiomatisability and ultraproducts

Let $T$ be a first-order theory over a language $L$, and let $\mathcal{M}$ be a subclass of the class of models of $T$. As I understand it, if there is no theory $\hat{T}$ over $L$ whose class of models is exactly $\mathcal{M}$, frequently, the "morally correct" reason is that $\mathcal{M}$ is not closed under ultraproducts. However, it is sometimes possible to obtain a simpler proof of non-axiomatisability by compactness or completeness considerations.

Example. Consider the first order theory of fields, and let $\mathcal{M}$ be the class of fields of positive characteristic. Then $\mathcal{M}$ is not axiomatisable in the language of fields since, for example, if we take the ultraproduct of all the finite fields $\mathbb{F}_p$, $p$ prime, we would obtain a field of characteristic 0.

We could also prove this directly: if $T'$ is the naïve axiomatisation of fields of characteristic 0 (i.e. the one with one axiom of the form $\underbrace{1 + \cdots + 1}_{n\text{ times}} \ne 0$ for every positive $n \in \mathbb{N}$), and $\hat{T}$ is any axiomatisation of $\mathcal{M}$, then $T' \cup \hat{T}$ is inconsistent, so there is some finite subset which is inconsistent, so there is some finite set $X$ for which $\hat{T}$ proves that there is an $n \in X$ such that $\underbrace{1 + \cdots + 1}_{n\text{ times}} = 0$; but there are fields of positive characteristic other than those $n \in X$ — a contradiction.

Question. Is it in fact always possible to translate a proof of non-axiomatisability using ultraproducts to one using compactness/completeness?

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I couldn't prove that non-abelian simple groups aren't axiomatizable without using ultraproducts on the relevant example sheet, and neither could my supervisor. (By the way, it's sort of funny that you call the second proof "direct." The most direct proof I know of the compactness theorem is using ultraproducts.) – Qiaochu Yuan May 26 '11 at 11:04
By the way, it is in fact a (difficult) theorem that $M$ is axiomatizable if and only if it is closed under ultraproducts. See, for example, Chapter 6 of Pete Clark's notes on model theory: math.uga.edu/~pete/MATH8900.html – Qiaochu Yuan May 26 '11 at 12:34
@Qiaochu: so difficult it's not true. For example, take T a theory in a countable language with infinite models, and M an uncountable model of T. Then the class of structures formable from M by repeatedly taking ultraproducts contains only uncountable structures, so can't be axiomatizable by downward Lowenheim-Skolem. – Chris Eagle May 26 '11 at 13:03
@Qiaochu: there's a correct version of this result in Chang and Keisler's Model Theory (1990 edition): a class of structures is axiomatizable iff it's closed under ultraproducts and isomorphisms, and its complement is closed under ultrapowers. – Chris Eagle May 26 '11 at 13:05
@Chris: huh. Seems I misread something. Thanks for the correction. – Qiaochu Yuan May 26 '11 at 13:06