# Absolute values of non-unit algebraic integers at all infinite places

Let $$K$$ be a number field and $$O_K$$ its ring of integers. Given any non-unit element $$\alpha\in O_K$$, does there exist a unit $$u\in O^\times_K$$ such that $$|\sigma(u \alpha)| >1 \quad \text{ for all } \sigma \in \mathrm{Hom}(K, \mathbb{C}).$$

I do not find a counter-example; and I can not see how to use Minkowski theory to find such an element in the ideal lattice $$\alpha O_K \subset \prod_{v\mid \infty} K_v$$.

No. Let $$K=\mathbb Q(\sqrt{6})$$ and let $$\alpha=2+\sqrt{6}\in O_K$$. All units of $$O_K$$ are given by $$u=\pm v^n$$, where $$v=5+2\sqrt{6}$$ is the fundamental unit. The key here is that $$\alpha$$ has relatively small norm while the fundamental unit $$v$$ is relatively large.
Consider $$u\alpha=\pm v^n\alpha$$. If $$n<0$$, then $$|u\alpha|=|v|^n|\alpha|\leq|5-2\sqrt{6}|\cdot |2+\sqrt{6}|<1$$. If $$n\geq 0$$, then $$|u\alpha|=|v|^n|\alpha|\geq|2+\sqrt{6}|>2$$. Since $$\alpha$$, and hence also $$u\alpha$$, has absolute norm $$2$$, this implies that the conjugate of $$u\alpha$$ has absolute value smaller than $$1$$.
• @YaakovBaruch No, I really do mean norm. Small norm implies that $\alpha$ and its conjugate can't simultaneously have too large absolute value. So one abs value larger than 2 implies the other smaller than 1. Commented May 14, 2021 at 9:51