On the irreducibility of a polynomial and Gauss lemma

Question

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.

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.