I've been given the following definition for irreducibility of polynomials:
Definition: Given a field $K$ and a polynomial $f \in K[x]$, we say that $f$ is irreducible over $K[x]$ if for any $g, h \in K[x]$ such that $f = gh$, we have that $\delta(g) \ge 0$ or $\delta(h) \ge 0$.
I have to prove the following statement:
Theorem: Let $p \in \mathbb{Z}$ be a prime and we consider the homomorphism $\bar{\bullet}: \mathbb{Z} \longrightarrow \mathbb{Z}_p$ such that $n \mapsto n + p\mathbb{Z}$, which extends naturally to a homomorphism between $\mathbb{Z}[x]$ and $\mathbb{Z}_p[x]$. If $f \in \mathbb{Z}[x]$ is a monic polynomial such that $\bar{f} \in \mathbb{Z}_p[x]$ is irreducible, then $f$ is also irreducible.
I have a proof of the statement considering the definition given above; however, this definition assumes $K$ is a field but, as we know, $\mathbb{Z}$ isn't a field. I've searched for a more general definition of irreducibility which involves one of the factors of the decomposition being a unit of the field. According to this more general definition, is the statement true?
I have demonstrated that the theorem still holds for the "right" definition of irreducibility by leveraging the property that if $f \in R[x]$, where $R$ is a domain such as $\mathbb{Z}$, is monic and $f = gh$ for some $g, h \in R[x]$, then both $g$ and $h$ must be monic. Additionally, I employed the observation that in this scenario, if $f \in \mathbb{Z}[x]$ is monic, then $\delta(f) = \delta(\bar{f})$.
Let's consider the assumption that $f = gh$ for some $g, h \in \mathbb{Z}[x]$. Utilizing the fact that $\bar{\bullet}$ is a homomorphism, we deduce that $\bar{f} = \bar{g} \bar{h}$. As $\bar{f}$ is monic and irreducible, it follows that both $\bar{g}$ and $\bar{h}$ must be monic, and one of them must have degree 0 (due to the fact that $\mathbb{Z}_p$ is a field). Without loss of generality, let's assume that $\delta(g) = 0$.
Recall that $g$ is monic, and since $\delta(g) = \delta(\bar{g}) = 0$, we conclude that $g$ must be $1$, which is a unit in $\mathbb{Z}$. This completes the proof that $f$ is irreducible.