# Polynomial with no zeros in ultraproduct of finite prime fields

I am interested in properties of ultraproducts of finite prime fields.

Let $$I$$ be a set, $$\mathcal U$$ a non-principal ultrafilter on $$I$$ and $$(p_i)_{i \in I}$$ a family of prime numbers. Let $$F$$ be the ultraproduct of the fields $$\mathbb F_{p_i}$$ wrt $$\mathcal U$$.

By taking quotient maps from $$\mathbb Z$$ to $$\mathbb F_{p_i}$$, we can consider polynomials over $$\mathbb Z$$ as polynomials over $$F$$.

Question: Is there always a polynomial over $$\mathbb Z$$ that has no root in $$F$$?

Of course this is true for the finite prime fields themselves, but I am not sure how one could argue it for their ultraproduct.

The answer to this is no, in general; it is in fact possible to have $$\mathbb{Q}^{\text{alg}}$$ as a subfield of a pseudo-finite field of the form you indicate. This follows from the following lemma:
If $$f(x)\in\mathbb{Z}[x]$$, then there are infinitely many primes $$p$$ such that $$f$$ splits completely mod $$p$$.
Okay, to see why this lets us construct the desired ultraproduct: for a polynomial $$f\in\mathbb{Z}[x]$$, let $$P_f$$ be the set of primes modulo which $$f$$ splits completely. For any $$f_1,\dots,f_n$$, by applying the above lemma to $$f=f_1\cdot\ldots\cdot f_n$$, we know that $$P_f=P_{f_1}\cap \dots \cap P_{f_n}$$ is infinite. In particular, if $$P$$ is the set of all primes, taking the filter on $$P$$ generated by $$(P_f)_{f\in\mathbb{Z}[x]}$$ yields a filter $$\mathcal{F}$$ containing no finite set.
We may thus complete $$\mathcal{F}$$ to a non-principal ultrafilter, and then taking the ultraproduct of the $$(\mathbb{F}_p)_{p\in P}$$ along this ultrafilter will yield a characteristic $$0$$ field in which every polynomial with integral coefficients splits completely, ie a field containing $$\mathbb{Q}^{\text{alg}}$$, as desired.