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Let $P (X) = X^5 − 6X + 3$.

Prove that it is irreducible over $\mathbb{Q}$.

In the solution I have for this exercice, I have litterally:

P is irreducible over $\mathbb{F}_5$, therefore, by Gauss Lemma, it is irreducible over $\mathbb{Q}$.

I am lacking the necessary knowledge about irreducibility by reducing modulo p. All I know is that there is a bijection between roots of P in $\overline{\mathbb{Q}}$ and the roots of $\overline{P}$ in $\overline{\mathbb{F}_p}$.

I also know this generalization of Gauss Lemma: in a factorial ring A, with its fraction field K, a primitive $P$ is irreducible in $A[X] \Leftrightarrow $ $P$ is irreducible in K[X].

How does this apply to P above?

Thank you for any directions or help.

Conjecture
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    Because $P$ is irreducible over $\mathbb{Z}_5 \Rightarrow P$ is irreducible over $\mathbb{Z}$. And then $P$ is irreducible over $\mathbb{Q}$ by Gauss's Lemma. – jijijojo Dec 23 '18 at 03:50
  • Thank you! is this a general fact? could please give some link to the litterature or a proof that P is irreducible over $\mathbb{F}_p\Rightarrow$ P is irreducible over $\mathbb{Z}$? – Conjecture Dec 23 '18 at 04:17
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    @PerelMan A factorisation over $\Bbb Z$ induces one over $\Bbb F_p$. – Angina Seng Dec 23 '18 at 05:19
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    PerelMan, I try and describe the use of modular reduction in irreducibility proofs here. Also observe that Eisenstein's criterion with $p=3$ works here as well (and is the go-to technique for many). – Jyrki Lahtonen Dec 23 '18 at 07:57

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There is a general result: Let $R$ be a ring and $I$ be a proper ideal. Fix a nonconstant monic polynomial $p(x) \in R[x]$. If $\overline{p(x)} \in (R/I)[x]$ is irreducible, then $p(x)$ is irreducible in $R[x]$.

You can prove this by considering the contrapositive.

Aniruddh Agarwal
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