solvarmedw

2022-10-07

Consider the language $L=\left\{+,\cdot ,0,1\right\}$ 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 $p\ge N$. 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 $0$.

Carson Mueller

The four square theorem yields the following theorem in positive characteristic:
$\mathrm{\exists }a,b,c,d:{a}^{2}+{b}^{2}+{c}^{2}+{d}^{2}+1=0$

Tatiana Cook

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.

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