Let $p$ be a prime and $k$ is a positive integer. There is a natural projection of rings $Z_{p^k}[X] \longrightarrow Z_{p}[X]$. Suppose a monic polynomial $f(X) \in Z_{p^k}[X]$. $f(X)$ is called basic irreducible if it is irreducible in $Z_p[X]$.
If $f(X)$ is monic basic irreducible, then prove $f(X)$ is also irreducible in $Z_{p^k}[X]$.
I have tried that if $f=gh$ in $Z_{p^k}[X]$, then $f=gh$ in $Z_p[X]$. Without loss of genrality, assume $g=1$ in $Z_p[X]$. Then observing the coefficients, we can deduce that $g=1$ in $Z_{p^k}[X]$.
Given a polynomial $f\in\mathbb{Z}_{p^k}[x]$ let $\overline f$ its reduction in $\mathbb{Z}_{p}[x]$.
Suppose $\overline f$ irreducible and that $f=gh$ where both $g$ and $h$ are non-costant polynomials in $\mathbb{Z}_{p^k}[x]$. Since the reduction map is a ring homomorphism, this means that we must have, say, $\overline h=1$.
But now $$ \deg f=\deg\overline f=\deg\overline{gh}=\deg\overline g + \deg\overline h=\deg\overline g=\deg g $$ which is a contradiction.
The above chain of equalities is justified by the facts that reducing a monic polynomial doesn't change its degree, that we may assume $g$ and $h$ monic as well and that the degree of a product of (monic) polynomials is the sum of their degrees.