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.

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