Fields of finite characteristic are not axiomatizable.

Question

"Prove that fields of finite characteristic are not axiomatizable by the 1st order theory." How to prove this statement?

Characteristic is a smallest number of units such that it's sum is equal to zero.

I found similar theorems in "Handbook of Mathematical Logic" by Jon Barwise. (1st book, "Model Theory", page 16.)

But it's a complicated question for me, I didn't manage to catch an analogy with torsion-free groups.

There is a formula in "weak second order logic": $$\exists n\geqslant 1. \,[n1=0],$$ where $n1:=\underbrace{1+...+1}_{n \text{ times}}$.

I can axiomatize fields of zero characteristic. That is just axioms of fields plus infinite many of such axioms: $1+1\neq 0, 1+1+1\neq 0, 1+1+1+1\neq 0, \dots .$

Answer 1

Suppose, toward a contradiction, that $T$ were a theory whose models are exactly the fields of finite characteristic. Let $Z$ be the set of axioms you gave for "characteristic $0$". So $T\cup Z$ has no models. Therefore, by compactness, there are finite subsets $T_0\subseteq T$ and $Z_0\subseteq Z$ such that $T_0\cup Z_0$ has no models. Since $Z_0$ is finite, let $p$ be a prime number greater than all the $n$'s for which $Z_0$ contains the sentence $1+1+\cdots+1=0$ with $n$ $1$'s. Then all the sentences in $Z_0$ are true in the finite field $\mathbb Z/p$. All the sentences of $T_0$ (and in fact all the sentences of $T$) are also true in $\mathbb Z/p$ because this is a field of finite characteristic. So $\mathbb Z/p$ is a model of $T_0\cup Z_0$; contradiction.

Answer 2

Let $\mathcal U$ be a nonprincipal ultrafilter on $\Bbb N$, i.e. no finite set belongs to $\mathcal U$, and consider $$M:=\prod_n\Bbb F_{p_n}/\mathcal U$$ where $p_n$ is the $n$th prime and $\Bbb F_p$ is the field of $p$ elements.
Now, if $T$ is a theory of fields such that each $\Bbb F_p$ is its model, then so is $M$, while the characteristic of $M$ is zero, because for any prime $q$, except for finitely many (namely, one) $p$, all $\Bbb F_p\models \overbrace{1+\dots+1}^q\ne 0$, hence it also holds in $M$.