Geometry of number

This question seems not very hard, but it is starting to embarrass me. So I thought I can use your ideas to solve it, and I would be thankful in advance.

Let $K/\mathbb{Q}$ be a number field of degree $n$ with the discriminant $D$. For $\alpha\in\mathcal{O}_K$ define $$||\alpha||:=\max\{|\sigma(\alpha)|:\mbox{over all euclidean place} \}$$

I want to show that there exists an $0\neq\alpha\in\mathcal{O}_K$ such that $Tr_{K/\mathbb{Q}}(\alpha)=0$ and $||\alpha||\ll|D|^{\frac{1}{2(n-1)}}$, where the implied constant depends only on $n$.

I was trying to use geometry of number, then it is easy to find an $0\neq\alpha\in\mathcal{O}_K$ such that and $||\alpha||\ll|D|^{\frac{1}{2(n-1)}}$ but I could not prove this $\alpha$ can be choosen so that $Tr(\alpha)=0$.

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If $\beta\ne\alpha$ is a conjugate of $\alpha$ then $\alpha-\beta$ has trace zero. Can you get a bound on $\|\alpha-\beta\|$ from one on $\|\alpha\|$? –  Gerry Myerson Jul 9 '11 at 5:01
Seems working, since $||\alpha-\beta||\leq ||\alpha||+||\beta||\ll |D|^{\frac{1}{2(n-1)}}$. –  m.b Jul 9 '11 at 5:29
OK, I'll promote my comment to an answer. –  Gerry Myerson Jul 9 '11 at 7:49

1 Answer

If $\beta\ne\alpha$ is a conjugate of $\alpha$ then $\alpha-\beta$ has trace zero. If as you say $\|\alpha-\beta\|\le\|\alpha\|+\|\beta\|$, then you're done.

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it seems there is just one simple problem, to show $\alpha\neq 0$ we need to show $\alpha\not\in \mathbb{Z}$ since then $\alpha-\beta=0$ –  m.b Jul 9 '11 at 14:02
Good point. Your proof that you can find $\alpha$ satisfying the bound - can you incorporate something in that proof to guarantee that the $\alpha$ you wind up with isn't an integer? –  Gerry Myerson Jul 10 '11 at 11:31
I have fixed it, i.e I can proof one can choose $\alpha\not\in\mathbb{Z}$, so your idea is working. –  m.b Jul 10 '11 at 12:21