I am looking at this polynomial:$f(x)=x^{5}-x-1$, the textbook says it is irreducible over $\mathbb{Q}[x]$ because it is irreducible over $\mathbb{F}_{3}[x]$.
Why can we reduce to $\mathbb{F}_{3}[x]$? Do we use Hensel's Lemma here? And in $\mathbb{F}_{3}[x]$ how to determine it is irreducible?
Thanks!
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Because it's a monic polynomial, if it is reducible in $\Bbb Q[x]$, then it's reducible in $\Bbb Z[x]$. And if it's reducible in $\Bbb Z[x]$, then by mapping each of those factors to the corresponding elements of $\Bbb F_3[x]$ you get a factorisation of $x^5-x-1$ over $\Bbb F_3[x]$.
Now, what happens to this chain of logic if there is no factorisation over $\Bbb F_3$?
Note that the converse doesn't apply. There are irreducible polynomials in $\Bbb Z[x]$ whose reduction to any $\Bbb F_p[x]$ is reducible. $x^4 + 1$ is one example.
Why can we reduce to $\mathbb{F}_3$? Because it works! And once it is irreducible over $\mathbb{F}_p$ for some prime $p$, is is irreducible over $\mathbb{Z}$ and $\mathbb{Q}$ as well.
A reference from this site (out of many):
A question from the mod p irreducibility test's proof
Over $\mathbb{F}_3$ it is easy to see that the polynomial cannot decomposed as $(x^2+ax+b)(x^3+cx^2+dx+e)$ by comparing coefficients and solving easy equations over $\mathbb{F}_3$. A linear factor is impossible by the rational root test from the beginning.