I am looking for a proof of the following statement:
Let $f \in \mathbb{Q}[X]$ be an irreducible polynomial with integer coefficients. Then, there are infinitely many primes $p$ such that the reduction of $f$ modulo $p$ splits completely in $\mathbb{Z}/p\mathbb{Z}$.
I only know a proof "by model theory" explained in an answer to this question: Non-standard proofs of standard theorems. Now, I am wondering how a "purely algebraic" proof or at least a proof without model theory might look like. Is the proof complicated or is this even a trivial fact? In the case of the former or if the proof is simply to long to be stated here, it would also be sufficient if someone can point me to the appropriate literature. Thank you!
If $a$ is a root of the polynomial. Then we can consider the extension $\mathbb{Q}(a)/\mathbb{Q}$. Then, by algebraic number theory, $f$ splitting into linear factors modulo a prime $p$ implies that the prime $p$ splits completely in the ring of integers of $\mathbb{Q}(a)$ (other than the finitely many primes that are not relatively prime to the conductor). However, by Chebotaryov's density theorem, this happens asymptotically for $\frac{1}{n}$th of the primes where $n$ is the degree of $f$ so it happens for infinitely many of them.