Irreducibility Over the Rationals (Still Lacking Extra Condition)

Question

Let $p$ be a prime and $g(x)\in\Bbb Z[x]$ be irreducible modulo $p$. Let $f(x) = g^n(x) + ph(x)$ where $n$ is a positive integer and $h(x)\in\Bbb Z[x]$. Given that $g(x)$ and $h(x)$ are relatively prime modulo $p$ and the extra condition $deg( h(x))≤ ndeg(g(x))$, I want to show that $f(x)$ is irreducible in $\Bbb Q[x]$. I tried to start by supposing $f$ is reducible so $f =ab$ but I am not sure how to reach the desired contradiction.

Answer 1

if $f=ab$ with $a,b\in \mathbb{Z}[x]$, then $ab=g^n$ modulo $p$. Since $g$ is irreducible modulo $p$, we must have $a=g^k$ and $b=g^l$ modulo $p$ with $k+l=n$ and $k,l>0$. So $f=(g^k+pu)(g^l+pv)$, where $u,v \in \mathbb{Z}[x]$. It follows that $$f=g^n+p(g^lu+g^kv)+p^2uv=g^n+ph \Rightarrow g^lu+g^kv+puv=h,$$ but then modulo $p$ we would have $g|h$ which is a contradiction.

EDIT: This proof fails in the step where it is claimed that $k,l>0$. Some additional conditions are needed, since one can have a counterexample:

$$f(x)=(1+px)(g+px)=g+p(gx+x+px^2),$$ where $g \neq x$ is any irreducible polynomial, $n=1$, and $h=gx+x+px^2$ which is coprime with $g$.