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let $l_1,...,l_n$ be roots of unity. I want to prove that the norm(the product of all conjugates)of $a=l_1+...+l_n$ is not greater than $n$, not smaller than $-n$. how can I do to prove this?

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  • $\begingroup$ Have you tried to calculated a conjugate of $a$ under a field automorphism? In which field are you working? $\endgroup$ – BIS HD Oct 17 '13 at 8:41
  • $\begingroup$ let l_i^(n_i)=1,and n be least common multiple of n_i.then the problem is to prove that the integer ring of Q(¥zeta_n) is Z(¥zeta_n)... $\endgroup$ – Yuma Oct 17 '13 at 9:02
  • $\begingroup$ For a proof of your assertion, I refer to [Washington: Cyclotomic Fields], Theorem 2.6. $\endgroup$ – BIS HD Oct 17 '13 at 9:18
  • $\begingroup$ If $n=2$ and $l_1=l_2=\omega$ is a primitive third root of unity, then the norm of $a=2\omega$ is $4>2$. So it seems the assertion is false, or am I missing something? $\endgroup$ – ladisch Oct 17 '13 at 11:46
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We know that $l_k = e^{(\frac{2\pi i (k-1)}{n})}$, for $k=1,...n$. Then: $$a = \sum_{k=1}^n e^{(\frac{2\pi i (k-1)}{n})}$$ We can evaluate the norm of $a$ and use the triangular inequality: $$|a| = \left|\sum_{k=1}^n e^{(\frac{2\pi i (k-1)}{n})}\right| \leq \sum_{k=1}^n \left|e^{(\frac{2\pi i (k-1)}{n})}\right| = \sum_{k=1}^n 1 = n$$ Then $|a| \leq n$. Anyway, the norm is always a positive (or null) number, so it is always satisfied that it is not smaller than $-n$.

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  • $\begingroup$ is the absolute value of a same to algebraic norm of a? $\endgroup$ – Yuma Oct 17 '13 at 8:56
  • $\begingroup$ yes, by definition $|a|^2 = aa^*$, where $a^*$ is the coniugate of $a$ $\endgroup$ – the_candyman Oct 17 '13 at 9:01
  • $\begingroup$ but, I think that "conjugate" is "algebraic" conjugate. $\endgroup$ – Yuma Oct 17 '13 at 9:07
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    $\begingroup$ the_candyman answered your question if you're working in an algebraic closure of $\mathbb{Q}$, for example $\mathbb{C}$. For an arbitrary field (e.g. $\mathbb{Q}$) the roots of unity are not necessarily elements of that field, so your question just makes sense in a field extension containing $\mathbb{Q[\zeta_1,\dots,\zeta_n]}$ but in this case the algebraic and analytic conjugates are the same. $\endgroup$ – BIS HD Oct 17 '13 at 9:14
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    $\begingroup$ I think OP is asking about the field norm $N(a)=\prod_{\sigma}a^{\sigma}$, where $\sigma$ runs over all field automorphisms of $\mathbb{Q}(a)$. $\endgroup$ – ladisch Oct 17 '13 at 11:32
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You can use the triangle inequality for that, i.e. $|a+b|\leq |a| + |b|$, then the result follows immediately by induction. (BTW: That the norm is larger $-n$ should be obvious, since a norm is always non-negative).

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  • $\begingroup$ when i want to try |a|(absolute value) <= n ,this problem is easy, but i want to know the algebraic norm of a.both is the same? $\endgroup$ – Yuma Oct 17 '13 at 8:58
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I don't know why you might think the field norm would be at most $n$ in absolute value. For example, if $x=e^{2\pi i/11}$ then I calculate that the norm from ${\bf Q}(x)$ to $\bf Q$ of $x+x^2+x^4$ is 23.

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