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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.

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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.

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  • $\begingroup$ Thank you, I edited to remove zero case. Ok yeah it was just me being silly and not thinking about products of actual numbers enough then! Sometimes its so easy to get all caught up in norms and conjugates etc. that you forget simple facts that will do the job. Thanks! $\endgroup$
    – marineabcd
    Apr 8, 2018 at 9:00

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