## If all the conjugates of an algebraic integer in some algebraic number field have modulus less than 1, then the integer is 0

Let $\alpha$ be an algebraic integer in some algebraic field extension $K$ of $\mathbb{Q}$. Prove that if all of the conjugates of $\alpha$ have modulus strictly less than 1 then $\alpha = 0$.

Recall that the norm of $\alpha$ is the product of its conjugates $\alpha_i$. By Lemma 7.1 in TAN, $N(\alpha)$ is a rational integer. Letting bars denote taking the complex modulus, we also have $|N(\alpha)| = \prod |\alpha_i| < 1$. So $N(\alpha) = 0$, and thus $\alpha = 0$.