Searching Explanation/References for a certain Corollary to Chebotarev Density Theorem

Question

I am reading “Notes on the model theory of finite and pseudo-finite fields” by Zoé Chatzidakis https://www.math.ens.psl.eu/~zchatzid/papiers/Helsinki.pdf and on page 16 (or in this newer version of the same notes on page 15 https://www.math.ens.psl.eu/~zchatzid/papiers/Singapore.pdf) she uses “a consequence of Chebotarev's theorem” that essentially states

$\textbf{(4.11)}$ Let $f_1(T), \dots, f_m(T), g(T) \in \mathbb{Z}[T]$, $T$ a single variable. Let $L$ be the Galois extension of $\mathbb{Q}$ obtained by adjoining all roots of the polynomials $f_i(T)$, $i = 1, \dots, m$. Assume that there is a subfield $E$ of $L$ such that $\text{Aut}(L/E)$ is cyclic and that $E$ contains at least one root for each of the $f_i$’s but no root for $g$. Then there are infinitely many prime numbers $p$ such that $\mathbb{F}_p$ has the same property as $E$, i.e. each $f_i$ has at least one root in $\mathbb{F}_p$ but it contains no root of $g$.

I unfortunately do not know enough about number theory to see how this is connected to the statement of the Chebotarev Density Theorem. Chatzidakis only references the statement of the Chebotarev Density Theorem in Fried and Jarden's Field Arithemic but I am not able to see how one could deduce this consequence from that?

I think I understand that a polynomial has a root over $\mathbb{F}_p$ if and only if $p$ splits completely in a splitting field of the polynomial. And the Chebotarev Density Theorem tells us that infinitely many primes split totally, hence, our polynomial should have roots over infinitely many $\mathbb{F}_p$'s. (As was used in https://math.stackexchange.com/a/2885423 I think.)

Moreover, maybe one could use a somewhat similar argument to https://math.stackexchange.com/a/608997 in some way to get that $g(T)$ does not have a zero for infinitely many primes?

But still, even if I had these two separate results how could I combine them to the Corollary?

References are very welcome, as I anyway would prefer to rather cite this Corollary than to include a proof. Though I'm also very interested to see how it follows from the theorem.

Answer 1

And the Chebotarev Density Theorem tells us that infinitely many primes split totally

This is only a corollary of Chebotarev. The full theorem tells us much more than this and we need more of the full theorem to get this result.

What Chebotarev actually says is the following. Let $G = \text{Gal}(K/\mathbb{Q})$ be the Galois group of a finite Galois extension $K$ of $\mathbb{Q}$. For all but finitely many primes $p$ we can write down something called a Frobenius element $\text{Frob}_p \in G$, well-defined up to conjugacy, which, somewhat loosely speaking, is defined by the property that $\text{Frob}_p(\alpha) \equiv \alpha^p \bmod p$ for suitable $\alpha \in K$. In particular, it follows that if $f(x) \in \mathbb{Z}[x]$ splits completely in $K$, then (again, for all but finitely many $p$) the factorization of $f(x) \bmod p \in \mathbb{F}_p[x]$ can be read off from the cycle decomposition of $\text{Frob}_p$ acting on the roots of $f(x)$ in $K$, and is the same as the factorization of $f(x)$ over the fixed field $E_p = K^{\text{Frob}_p}$.

Taking Frobenius elements gives us a map from primes $p$ (with finitely many exceptions) to conjugacy classes in $G$, and what the Chebotarev density theorem says is that the density of primes whose Frobenius element lies in a fixed conjugacy class $C$ is $\frac{|C|}{|G|}$; as a corollary, there are infinitely many primes whose Frobenius element lies in a fixed conjugacy class. The statement about split primes comes from taking $C = \{ e \}$ to be the conjugacy class of the identity, but here we will need to take a non-identity conjugacy class.

Now we consider the situation of 4.11. Let $G = \text{Gal}(L/\mathbb{Q})$. By hypothesis, $E$ is a subextension such that $\text{Aut}(L/E)$ is cyclic; let $g$ be a generator and let $C$ be its conjugacy class. Chebotarev tells us that there are infinitely many primes $p$ such that $\text{Frob}_p \in C$; what this means, as above, is that the factorizations of the polynomials $f_i(x), g$ are the same over $E_p = L^{\text{Frob}_p}$ (which is either $E$ or one of its Galois conjugates, and this doesn't affect the factorizations) as they are $\bmod p$, and we conclude.