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 $1$.

  • 1
    $\begingroup$ The first occurrence of "norm" should be "absolute value"? $\endgroup$ May 14 at 4:54
  • $\begingroup$ @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. $\endgroup$
    – Wojowu
    May 14 at 9:51

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