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I am trying to understand and prove theorems about polynomial factorization mod $q$.

Theorem Let $P(X) \in \mathbb Z[X]$ be an irreducible polynomial of degree $p$ a prime. There exists a prime $q$ such that $P(X)$ is irreducible mod $q$.

They came from the following source: https://www.isibang.ac.in/~sury/aparnaramesh.pdf the proof of the 2nd theorem is on page 24. I don't understand the proof.

Can anybody explain roughly how this is proved?

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    As stated, your source shows that $P(X)=X^4-10X^2+1$ is a counterexample to the second claim, so why are you trying to prove it? – N. S. Jun 13 '20 at 18:47
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    See the proof here: "if $f$ is an irreducible polynomial of degree at least $2$ then there are infinitely many primes $p$ such that $f$ does not have a root $\bmod p$. ". – Dietrich Burde Jun 13 '20 at 18:51
  • @N.S. IMy apologizes I had a mistake in my transcription, I have corrected it now. –  Jun 13 '20 at 18:59
  • I split the first part of this into its own question https://math.stackexchange.com/questions/3719076/proving-that-an-irreducible-integer-polynomial-takes-on-values-with-infinitely-m –  Jun 14 '20 at 08:35
  • When we view the Galois group $G$ of $P(X)$ as a permutation of the $p$ roots, irreducibility of $P(X)$ implies that $p\mid |G|$. Therefore $G$ contains at least one $p$-cycle $\sigma$. By Chebotarev's theorem there exists infinitely many primes $q$ such that $\sigma$ (or its conjugcay class) is the Frobenius element corresponding to $q$. This means that the Frobenius element has order $p$, and consequently $P(X)$ must be irreducible modulo $q$. – Jyrki Lahtonen Jun 16 '20 at 19:06
  • But have you covered the theory appearing in my comment? – Jyrki Lahtonen Jun 16 '20 at 19:07

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