# One of the conjugates of an algebraic integer must have absolute value $\geq$ 1

Let $0\neq\omega \in \mathcal{O}_K$ be an algebraic integer. Prove that one of its conjugates has absolute value $\geq$ 1.

My Thoughts: We know that the norm and trace of $\omega$ are in $\mathbb{Z}$, surely this is relevant.

Call the conjugates $\omega = \omega_1,\dots,\omega_n$ all roots of $\mu_\omega \in \mathbb{Z}[x]$ its minimal polynomial. I think we then know in $\mathbb{Q}(\omega)$, we have embeddings $\sigma_i : \mathbb{Q}(\omega)\rightarrow \mathbb{C}$ where $\sigma_i(\omega)=\omega_i$ and so Norm($\omega$) = $\omega_1\cdots\omega_n$ and Trace($\omega$) = $\omega_1 + \cdots + \omega_n$. (But is this norm over $K$ or $\mathbb{Q}(\omega)$ I'm a bit confused here too?)

I feel close but couldn't quite pull it all together. Any thoughts appreciated.

This is not true if $\omega=0$. Otherwise it is true. Suppose $\omega$ has degree $n$. Let $\omega_1=\omega,\ldots,\omega_n$ be its conjugates. Then $\omega_1\omega_2\cdots\omega_n$ is the norm of $\omega$, and so a nonzero integer. Not all $|\omega_j|<1$ for then their product has absolute value less than one.