Why is $X^5-6X+3$ irreducible? In my math course, it is written that since over $\mathbb{F}_5[X]$, $P$ has no root in $\mathbb{F}_{25}$, it is irreducible over $\mathbb{Q}[X]$ by Gauss Lemma.
I don't really understand…
If we prove that $P$ is irreducible over $\mathbb{Z}[X]$, then Gauss proves it is over $\mathbb{Q}[X]$, but I don't understand what $\mathbb{F}_{25}$ is doing there…
The exact question was :" Find a prime $p$ such that the reduction of $P(X)$ over $\mathbb{F}_p[X]$ has no root in $\mathbb{F}_{p^2}$, then show that $P$ is irreducible"
 A: $x^5-6x+3$ is irreducible over $\mathbb{F}_5$ because it is an Artin–Schreier polynomial, of the form $x^p-x+\alpha$ for $p=5$: since such polynomial is irreducible over $\mathbb{F}_5$, it is irreducible over $\mathbb{Q}$, too.

$x^5-x+3$ has no root in $\mathbb{F}_5$ or $\mathbb{F}_{25}$ because the $\gcd$ between $x^{5^2}-x$ and $x^5-x+3$ is $1$: by assuming $x^5=x-3$ it follows that $x^{5^2}=(x-3)^{5}=x^5-3=x-6$ hence $x^{5^2}-x = -1$.
It follows that $x^5-x+3$ factors over $\mathbb{F}_5$ as the product of irreducible polynomials with degree $\geq 3$ and the conclusion is the same as before: $x^5-x+3$ is irreducible over $\mathbb{F}_5$, hence it is irreducible over $\mathbb{Q}$.
A: So Gauss' lemma states that irreducibility over $\mathbb{Z}$ implies irreducibility over $\mathbb{Q}$; so it suffices to show that $f = x^5 -6x + 3$ is irreducible over $\mathbb{Z}$. Now irreducibility in $\mathbb{F}_{p}$ (provided that $p$ does not divide the leading coefficient of $f$) implies irreducibility over $\mathbb{Z}$. So if $f$ was reducible, what are the possible factorizations? Well since $\mathbb{F}_{5}[x]$ admits unique factorization, there are the following options:


*

*$f$ is the product of 5 linear factors.

*$f$ is the product of 3 linear factors and an irreducible quadratic 

*$f$ is the product of 2 linear factors and an irreducible cubic. 

*$f$ is the product of a single linear factor and an irreducible quartic.

*$f$ is the product of a single linear factor and two irreducible quadratics. 

*$f$ is the product of a single irreducible quadratic and a single irreducible cubic. 


Checking that $f$ has no root in $\mathbb{F}_{5}$ rules out options 1-5. Now since $\mathbb{F}_{25}$ is the splitting field extension for all quadratics over $\mathbb{F}_{5}$, if $f$ possessed an irreducible quadratic in its factorization then $f$ would have a root in $\mathbb{F}_{25}$. 
