 Solution

Let $P_1$ and $P_3$ be the sets of primes congruent to one and three modulo four. Both are infinite. For $P_1$, if a finite list contained them all, put $N=(2p_1\cdots p_r{)}^2+1$. Any prime divisor $q$ of this odd number is new and has minus one as a square modulo $q$, hence belongs to $P_1$ by the given criterion. For $P_3$, use $N=4p_1\cdots p_r-1$. It is three modulo four, so at least one of its odd prime divisors is three modulo four; none of the listed primes divides $N$. These are elementary Euclidean proofs of infinitely many primes in each odd residue class modulo four.

On the set of all primes, the cofinite sets together with $P_1$ have the finite intersection property; so do the cofinite sets together with $P_3$. Extend the generated filters to ultrafilters ${\mathcal{U}}_1$ and ${\mathcal{U}}_3$ using the ultrafilter lemma. They are nonprincipal because they contain every cofinite set, including the complement of any singleton. FormEvery sentence of $\Sigma$ has a cofinite truth set, hence belongs to both ultrafilters. The Łoś theorem shows $F_1,F_2\models \Sigma$. In particular they are fields, since each field axiom holds in every factor.

For each fixed positive integer $n$, the sentence $n\cdot 1\ne 0$ holds in ${\mathbb{F}}_p$ unless $p$ divides $n$. Its truth set is cofinite, so both fields satisfy it. Therefore $\boxed{\mathrm{char}{F}_1=\mathrm{char}{F}_2=0}$.

Finally the sentence $\exists x(x^2+1=0)$ holds at exactly the primes in $P_1\cup \{2\}$. This set belongs to ${\mathcal{U}}_1$, whereas its complement contains $P_3$ and belongs to ${\mathcal{U}}_3$. Łoś therefore givesThus minus one is a square in the first characteristic-zero first-order model of $\Sigma$ and is not a square in the second. The construction also shows why eventual truth over all primes need not decide a sentence whose truth splits along two infinite prime subsequences.