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 $\alpha O\_K \subset \prod\_{v\mid \infty} K\_v$.



Absolute values of non-unit algebraic integers at all infinite places
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 $1$.


