I was reading a paper regarding the divisibility of polynomials modulo a composite number. The paper can be found here: https://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.821.1301&rep=rep1&type=pdf.
In Propositions 2.7/2.8 of the paper, the author gives a proof of the claim that "If a polynomial $f(x) \in Z_{p^k} [x]$is irreducible in $Z_p [x]$ then it is irreducible in $Z_{p^k} [x]$". However, I don't quite understand why the proof given is correct. In fact, I suspect that it is faulty but I feel like I might be missing something here.
It is proven in Proposition 2.7 that $f$ can be written as $f'+pg$, where $p$ does not divide $f'$. I believe that the proof for this proposition is valid. However, it is worthy to note that there is no guarantee here that the degree of $f'$ is greater than 0.
The proof of 2.8 reads essentially as follows. Suppose for sake of contradiction $f$ is reducible in $Z_{p^k} [x]$, then $f \equiv gh \mod p^k \Rightarrow f \equiv(g'+pg_{1})(h'+ph_2)\mod p \Rightarrow f \equiv g'h' \mod p $, at which point the proof is ended. But how do we know that $g', h'$ are not constants? I don't believe they necessarily have to be, in which case the proof breaks down.
Am I missing something here?
That paper has some problems. I wouldn't spend much time on it unless you have a specific reason you need to do so.
For instance, 2.7 as written is trivially true, but then the proof is invalid.
And as you've pointed out, 2.8 is false. As a simple counterexample, note that $(x^2 + x + 1)(2x+1)$ is reducible in $\mathbb{Z}_4[x]$ but its image is irreducible in $\mathbb{Z}_2[x]$.
I would say it's possible we're restricting our attention to just monic $f$, but comments near the beginning of chapter 2 make me think that's not even the case.