, 2013

Assume that there exist two non-constant polynomials $g(X)$ and $h(X)$ with integer coefficients such that $f(X) = g(X) h(X)$. Because $p = g(0) h(0)$ is prime, we can assume that $|g(0)| = 1$.

Because the modulus of the product of the complex roots of $g(X)$ is equal to $1$, at least one of these roots, say $\omega_{0}$, has modulus less than or equal to $1$. But $f\left(\omega_{0}\right) = 0$. We deduce that
$$\begin{aligned} p & = \left| a_{n} \omega_{0}^{n} + a_{n-1} \omega_{0}^{n-1} + \cdots + a_{1} \omega_{0} \right| \\ & \leq \left| a_{n} \right| \cdot \left| \omega_{0} \right|^{n} + \left| a_{n-1} \right| \cdot \left| \omega_{0} \right|^{n-1} + \cdots + \left| a_{1} \right| \cdot \left| \omega_{0} \right| \\ & \leq \left| a_{n} \right| + \left| a_{n-1} \right| + \cdots + \left| a_{1} \right| \end{aligned}$$
which is a contradiction. Therefore, $f(X)$ is irreducible.