Currently I am working on understanding Dedekind's Theorem about $\operatorname{Gal}(F/\mathbb{Q})$ and the permutations it contains if all the hypothesis are met.
Somewhere in my notes appears the following statement:
If $\operatorname{red}_2(f(x))$ is irreducible over $\mathbb{F}_2$ then $f(x)$ is irreducible over $\mathbb{Q}$.
Where $$\operatorname{red}_p: \mathbb{Z} \to \mathbb{F}_p : f(x) \mapsto \operatorname{red}_p(f(x))$$ is its reduction modulo $p$ with $p$ being a prime integer.
I am trying to understand where this statement comes from but I can no figure it out. Could anyone point it out, please? Any reference will be highly appreciate it.
Let $f\in \mathbf{Z}[X]$. We need the extra assumption that $p$ doesn't divide the leading coefficient of $f$. This is easy to see: consider $f=2X^2+X\in \mathbf{Z}[X]$, the reduction modulo $2$ is $\overline{f}=X\in\mathbf{F}_2[X]$: $f$ is irreducible in $\mathbf{F}_2[X]$ but certainly not in $\mathbf{Z}[X]$.
Now assume this and let $\overline{f}\in\mathbf{F}_p[X]$ be the reduction modulo $p$. The argument goes as follows by the contrapositive: a factorisation $f=gh$ in $\mathbf{Z}[X]$ would give rise to a factorisation $\overline{f}=\overline{g}\overline{h}$ in $\mathbf{F}_p[X]$.
Can you generalise this to a polynomial $f\in \mathbf{Q}[X]$?