Let $f$ be an irreducible polynomial and $h(f)$ the polynomial with coefficients reduced modulo a prime $p$. Then if $\deg(f)=\deg(h(f))$ and $h(f)$ is irreducible then $f$ is irreducible as an element of $\mathbb{Q}[x]$.
What I know: Since the degrees are the same that means that the leading coeffcient of $f$ doesn't vanish if reduced modulo $p$. A polynomial being irreducible means that it has no roots.
Unfortunately I don't know to continue from this. Thanks for any help!
This is a standard argument. I assume you mean $f\in\Bbb{Z}[x]$, otherwise it makes no sense to talk about reducing modulo $p$. Let us proceed by contraposition: If $f$ is reducible in $\Bbb{Q}[x]$, it is easy to check that it is also reducible in $\Bbb{Z}[x]$. In particular, $f=gk$ for non-constant polynomials $g,k\in\Bbb{Z}[x]$. Then, as reducing modulo $p$ is a ring homomorphism $h:\Bbb{Z}[x]\to (\Bbb{Z}/p\Bbb{Z})[x]$, we get that $h(f)$ factors: $h(f)=h(g)h(k)$. If $h(f)$ has the same degree as $f$, it is clear that the same must be true for $h(g)$ and $h(k)$, in particular, neither are constant after reducing modulo $p$. Thus, we have a proper factorisation of $h(f)$, showing that it is reducible.
