Let $f=a_0 + a_1 x + .... x^n \in \Bbb{Z}[x]$. If $\overline{f} =\overline{a}_0 + ... + x^n \in \Bbb{Z}_p[x]$ is irreducible for some $p$, then $f$ is irreducible.
I tried searching through google to find a duplicate, but I couldn't really find anything. Here is what I have so far
Suppose that $f = gh$. Then the $\overline{f} = \overline{g} \overline{h}$. Since $\overline{f}$ is irreducible, and $\Bbb{Z}_p[x]$ is an integral domain (because $\Bbb{Z}_p$ is a field), we get that, WLOG, $\overline{g} = \overline{b_0} + ... \overline{b_m} x^m$ is a unit and the coefficients of the nonconstant terms are zero in $\Bbb{Z}_p$. This means that $\overline{b_0}$ is a unit or that $p \not\mid b_0$; and it means that $\overline{b}_i = 0$ for every $i > 0$ or that $b_i = k_ip$ for some $k_i \in \Bbb{Z}$....
Here is the problem. Since $f$ is monic and $f=gh$, this means that $|b_m| = 1$. But then this implies $|k_m|p=1$, which can't be true...so what I am doing wrong? I am trying to conclude that $g$ is a unit in $\Bbb{Z}[x]$, but somehow I procured for myself a contradictory statement. Even if I am able to get around this, I am not sure how to finish, since I really need to show that $b_i = 0$ in $\Bbb{Z}$ for all $i > 0$.