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.

  • 3
    $\begingroup$ 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. $\endgroup$
    – jijijojo
    Dec 23, 2018 at 3:50
  • $\begingroup$ 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 $\mathbb{Z}$? $\endgroup$
    – Conjecture
    Dec 23, 2018 at 4:17
  • 2
    $\begingroup$ @PerelMan A factorisation over $\Bbb Z$ induces one over $\Bbb F_p$. $\endgroup$ Dec 23, 2018 at 5:19
  • 2
    $\begingroup$ 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). $\endgroup$ Dec 23, 2018 at 7:57

1 Answer 1


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.


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