Let p be a fixed prime, and for a polynomial $f$ let $\overline{f}$ denote the reduction of $f$ mod p, so that $\overline{f}$ is a polynomial with coefficients in $\mathbb{F}_p$.
Now, suppose $n\in\mathbb{N}$, and let $f,g\in \mathbb{Z}[x]$ such that $f$ is monic and $\overline{f}$ is irreducible in $\mathbb{F}_p[x]$, $\overline{f}$ and $\overline{g}$ have no common non-constant factors in $\mathbb{F}_p[x]$, and the polynomial $h(x)=f(x)^n+p\cdot g(x)$ has $\deg(h)>\deg(g)$. Show that h is irreducible in $\mathbb{Q}[x]$
This is part of an assignment, so I will show the work I've done.
First, I make some observations about $f$ and $g$. First off, $f$ is irreducible in $\mathbb{Z}[x]$ since it is monic, so any proper factors would also be monic and hence be a non-trivial factorization when reduced to polynomials in $\mathbb{F}_p[x]$, contradicting that $\overline{f}$ is irreducible in $\mathbb{F}_p[x]$.
$f$ and $g$ must also have no common non-constant factors, since any such common factor would be divisible by p (since it would have to disappear when reducing to $\mathbb{F}_p[x]$, but since f is monic any common factor of $f$ and $g$ must also be monic.
Now, I proceed by contradiction and assume h is reducible. Then by Gauss' lemma it is reducible in $\mathbb{Z}[x]$. Let us write $h=kl$, where $k,l\in\mathbb{Z}[x]$, $k,l$ both monic. It is clear from the definition of $h$ that $\overline{h} = \overline{f}^n$, and since reduction mod p is a ring homomorphism both $\overline{k}$ and $\overline{l}$ must divide $\overline{f}^n$. Since a polynomial ring over a field has unique factorization, this means that $\overline{k} = \overline{f}^s$ and $\overline{k} = \overline{f}^r$ for some $s,r<n$.
From here on, I'm stuck. If I could prove that all this implies that $k$ divides $f$, I'd be done. However, I'm not sure that's true at all, and if it's not true I would probably need a completely different proof strategy. This is for a Galois theory course, and it makes me a little suspicious that I haven't actually used anything about field extensions.
Best Answer
From $\overline k = \overline f^r, \overline l = \overline f^s$ we get $k=f^r+p\cdot k_1$ and $l=f^r+p\cdot l_1$. Using this and the definition of h on $h=kl$ we get the equation $f^n + p\cdot g = f^{r+s} + p(f^s k_1 + f^rl_1 + pk_1l_1)\Rightarrow \overline g = \overline f^r \overline k_1+ \overline f^s \overline l_1$ which contradicts $gcd(\overline f, \overline g)=1$.