A sentence false in a field of characteristic $0$ but true in all fields of positive characteristic?

Question

Consider the language $L=\{+,\cdot, 0, 1\}$ of rings. It is easy to show using compactness that if $\sigma$ is a sentence that holds in all fields of characteristic $0$, there is some $N\in \mathbb N$ such that $\sigma$ holds for all fields of characteristic $p\geq N$. A sort of converse would be, if $\sigma$ is a sentence that holds in all fields of positive characteristic, $\sigma$ is true in all fields of characteristic $0$. I have no idea how to come up with a counterexample or a proof of this.

Thanks for any help.

Answer 1

The four square theorem yields the following theorem in positive characteristic:

$$\exists a,b,c,d : a^2 + b^2 + c^2 + d^2 + 1 = 0$$

Answer 2

The canonical solution is to use the fact that $x^2 + y^2 = -1$ has a solution in every finite field, which can be proved by a simple counting argument. That equation is solvable in any field of positive characteristic, because it is solvable in the prime subfield. But not in $\mathbb R$.

This is the same idea as the other answer, but the proof of solvability is easier than the 4-squares theorem.

For the general question of what can be done in first-order theory of rings to separate characteristic 0 and characteristic $p$, there are brilliant results on this (and several other matters) in Bjorn Poonen's paper at http://arxiv.org/abs/math/0507486 .