= 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. Form
$$
\boxed{F_1=\prod_p\mathbb F_p/\mathcal U_1,\qquad F_2=\prod_p\mathbb F_p/\mathcal U_3.}
$$
Every 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 $n\cdot1\ne0$ holds in $\mathbb F_p$ unless $p$ divides $n$. Its truth set is cofinite, so both fields satisfy it. Therefore $\boxed{\operatorname{char}F_1=\operatorname{char}F_2=0}$.
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 gives
$$
\boxed{F_1\models\exists x(x^2+1=0),\qquad F_2\models\neg\exists x(x^2+1=0).}
$$
Thus 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.
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