# On the irreducibility of a polynomial and Gauss lemma

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.

• 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. Dec 23, 2018 at 3:50
• 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 $\mathbb{Z}$? Dec 23, 2018 at 4:17
• @PerelMan A factorisation over $\Bbb Z$ induces one over $\Bbb F_p$. Dec 23, 2018 at 5:19
• 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). Dec 23, 2018 at 7:57

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 $$R[x]$$.