# Model theory of fields of characteristic $0$

I am stuck on the following problem:

A sentence is valid in the class of fields of characteristic $$0$$ iff it is true for all fields of characteristic $$p>n$$ for some $$n\in\mathbb{N}$$.

Now, the one direction follows from a simple application of compactness, as the class of fields of characteristic $$0$$ is $$\Delta$$-elementary:

Let $$\phi_F$$ be the sentence expressing "is a field" and $$\chi_p$$ be the sentence expressing "has characteristic $$p$$". Take

$$\Gamma=\{\phi_F\}\cup\{\neg\chi_p\mid p\text{ prime}\},$$

then $$\mathfrak{A}\models\Gamma$$ iff $$\mathfrak{A}$$ is a field of characteristic $$0$$.

Now, suppose $$\phi$$ is valid in the class of fields of characteristic $$0$$, that is $$\Gamma\models\phi$$, then by compactness there is a finite subset $$\Phi\subset\Gamma$$ such that $$\Phi\models\phi$$ but

$$\Phi\subseteq\{\phi_F,\neg\chi_{p_1},\dots,\neg\chi_{p_n}\}\models\phi$$

where $$p_k$$ is the $$k$$-th prime. Then especially we have that $$\phi$$ is true in every field of characteristic $$>p_n$$.

Now, the converse direction is giving me problems. A hint I have received told me to assume that $$\phi$$ would be true in all characteristics $$>n$$ and then to apply compactness. Another, to apply compactness to the negation.

However, with both I have achieved almost no results. I am not sure of how to approach that.

EDIT: With the converse direction, I mean the following:

If a sentence is not valid in the class of fields of characteristic $$0$$, then for all $$n\in\mathbb N$$ there is a prime $$p>n$$ such that $$\phi$$ is not true in a field of characteristic $$p$$.

• The problem is that the converse is false. See here for examples: math.stackexchange.com/questions/332433/… Jan 23, 2020 at 4:33

Here is a solution (that actually answers this problem).

Lemma 1. The statement "$$F$$ is a field of characteristics $$0$$" is not expressible as a single sentence in the language of fields.

Proof. If it were, take an ultraproduct of all finite prime fields using a free ultrafilter. Easily, this is a field of characteristics $$0$$. If there was a single sentence asserting this, it would have to hold in infinitely many of the finite fields, which is impossible.

Remark. Note that this actually shows that no single sentence implies the characteristics of the field is $$0$$ over the theory of fields with characteristics "at least $$n$$" for any finite $$n$$.

Now suppose that $$\psi$$ is true in some field of characteristics $$0$$. Since being of characteristics $$0$$ is not equivalent to any single sentence in the language of fields, it cannot be that $$\psi$$ implies (over the theory of fields) that the characteristics is $$0$$. In particular, it means that it is consistent with a finite characteristics.

But since $$\psi$$ is also true in some field of characteristics $$0$$, it is consistent with "the characteristics of the field is at least $$n$$" for any finite $$n$$, and therefore there are arbitrarily large finite characteristics where $$\psi$$ is consistent.

Finally, take $$\psi=\lnot\phi$$.

• I think there is a problem with your last assertion that the statement is consistent with "characteristic is at least n ". Although this is true, it doesn't ensure that there is a model of finite characteristic. The problem is the "at least". The consistency is also witnessed by a model of characteristic 0. I don't see a reason why a model of finite characteristic should exist (and in fact, it doesn't always exists; see my link in my comment under the question) Jan 23, 2020 at 4:51