May I verify if my proof to this problem is correct?
Let $p \in \mathbb{P}.$ For $x \in \mathbb{Z},$ let $\overline{x}$ be remainder of $x$ when divided by $p.$ Let $f(X)= \sum^{n}_{i=0}a_iX^i \in\mathbb{Z}[X]$ such that $p$$ \not|a_n$ in $\mathbb{Z}.$ If $g(X)=\sum^{n}_{i=0}\overline{a}_iX^i $ is irreducible in $\mathbb{Z}_p[X],$ then $f(X)$ is irreducible in $\mathbb{Q}[X]. $
Proof: $\psi:\mathbb{Z} \to \mathbb{Z}_p, x \mapsto\overline{x},$ is a unital ring homomorphism.
So $\Phi:\mathbb{Z}[X] \to \mathbb{Z}_p[X], \ \sum a_iX^i \mapsto \sum\overline{a}_iX^i, \ $ is a ring homomorphism. If $f(X)=c,$ for some $c \in \mathbb{Z},$ then $g(X)= \overline{c} \neq \overline{0},$ since $p$$\not | c$ in $\mathbb{Z}.$ Since $\mathbb{Z}_p$ is a field, $g(X) \in U(\mathbb{Z}_p[X]) =U(\mathbb{Z}_p),$ contradicting that $g(X)$ is irreducible in $\mathbb{Z}_p[X].$
Now, suppose $\text{deg}(f(X)) \geq 1$ and assume, to the contrary, $f$ is not irreducible in $\mathbb{Q}[X].$ This means $f$ is reducible in $\mathbb{Q}[X]. $ Since $\mathbb{Z}$ is $\text{UFD}, \ f$ is not irreducible in $\mathbb{Z}[X].$ Clearly, $f \not\in U(\mathbb{Z}[X]) \cup \{0\},$ so $f$ is reducible in $\mathbb{Z}[X].$
Let $f(X) =h(X)k(X)$, where $h(X)= \sum^{m}_{i=0}b_iX^i, \ k(X)= \sum^{k}_{i=0}c_iX^i \in \mathbb{Z}[X]$ and $\text{deg}(h), \text{deg}(k) \geq 1.$ Hence$, \ \Phi(f(X))=g(X)= \Phi(h(X))\Phi(k(X))=\sum^{m}_{i=0}\overline{b}_iX^i \sum^{k}_{i=0} \overline{c}_{i}X^{i}.$
Since $p$$\not |a_n, $ we have $ p$$\not| b_m$ and $p$$\not | c_k,$ so $\overline{b}_m \neq \overline{0}, \overline{c}_k \neq \overline{0}.$ Hence $\Phi(h(X))$ and $ \Phi(k(X))$ are non-constant polynomials and hence are not units of $\mathbb{Z}_p[X].$ This contradicts that $g(X)$ is irreducible in $\mathbb{Z}_p[X].$
p.s. $U(R)$ denotes set of units of ring $R.$
Thank you for your attention.
It seems correct, but possibly too involved.
The projection $a\mapsto \bar{a}$ extends uniquely to a ring homomorphism $\def\Z{\mathbb{Z}}\varphi\colon\Z[X]\to\Z_p[X]$ by sending $X$ to $X$.
If $f\in\Z[X]$ is reducible, then $f=gh$ with $g$ and $h$ having degree at least $1$
$\varphi(f)=\varphi(g)\varphi(h)$ is, by hypothesis, irreducible, which means that at least one among $\varphi(g)$ or $\varphi(h)$ is constant (and nonzero)
Assume $\varphi(g)$ is constant: then $g=pg_1(X)+a$, where $a\in\Z$ is not divisible by $p$
In particular, the leading coefficient of $g$ is divisible by $p$, contradicting the fact that $a_n$ is not divisible by $p$
Therefore $f$ is irreducible in $\Z[X]$
Irreducibility in $\Z[X]$ implies irreducibility in $\mathbb{Q}[X]$ (Gauss' lemma)