Suppose that a polynomial $f(X)\in\mathbb{Z}[X]$ has the form $f(X) = φ(X)^e + pM(X)$, where $p$ is a prime number, $φ(X)$ is an irreducible polynomial modulo $p$, and $M(X)$ is a polynomial relatively prime to $φ(X)$ modulo $p$, with $\deg M<\deg f$. Then $f$ is irreducible over $Q$.
I am trying to get a hang of these criteria, and wonder if there is anyone who have proof to why this is true. I know the Eisenstein criteria , but am not sure how to connect these two
The key fact is that $\mathbb{F}_p[X]$ is a UFD. If $f = f_1f_2$, then $f_1 \equiv \varphi^{k_1} \mod p $ and $f_2 \equiv \varphi^{k_2} \mod p$ for some $k_1$ and $k_2$ satisfying $k_1 + k_2 = e$. Hence $f_1 = \varphi^k_1 + p m_1, f_2 = \varphi^{k_2} + pm_2$ for $m_1,m_2 \in \mathbb Z[X]$. Now
$$\varphi^e + pM = f_1f_2 = \varphi^e + p(m_1\varphi^{k_2} + \varphi^{k_1}m_2) + p^2(m_1m_2).$$ Can you do it from here? Hint: Subtract $\varphi^e$ from both sides. The full solution is below the break.
Solution: Subtract $\varphi^e$ and divide by $p$ to get $$M = m_1\varphi^{k_2} + \varphi^{k_1}m_2 + pm_1m_2,$$ but since $\varphi$ does not divide $m$ modulo $p$, we conclude one of $k_1$ and $k_2$ is zero, say $k_1 = 0$ without loss of generality. But then $\deg f_1 \geq \deg \varphi^e = \deg f$ (since $\deg M < \deg f$), so $f_1 = f$ and $f_2$ is a unit.