# Positivity of the norm of an element of a cyclotomic field

Let $l$ be an odd prime number and $\zeta$ be an $l$-th primitive root of unity in $\mathbb{C}$. Let $\mathbb{Q}(\zeta)$ be the cyclotomic field and $\alpha$ be a non-zero element of $\mathbb{Q}(\zeta)$.

There exists a polynomial $f(X) \in \mathbb{Q}[X]$ such that $\alpha = f(\zeta)$. Let $N(\alpha) = f(\zeta)f(\zeta^2)...f(\zeta^{l-1})$.

From $\bar\zeta = \zeta^{-1}$ it follows that $\bar f(\zeta) = f(\zeta^{-1})$. Likewise, $\bar f(\zeta^i) = f(\zeta^{-i})$ for $i = 1,2,\cdots,l - 1$. Since $f(\zeta^i)\bar f(\zeta^i) = |f(\zeta^i)|^2 > 0$, it follows that $N(\alpha) > 0$.

We used the fact that the field of complex numbers $\mathbb{C}$ has an $l$-th primitive root of unity. It seems to me that this fact can only be proved by using some (elementary) analysis. My question is:

Can we prove $N(\alpha) > 0$ purely algebraically?

In other words, can we prove $N(\alpha) > 0$ without using the field of real numbers? Please note that $\mathbb{Q}(\zeta) \cong \mathbb{Q}[X]/(1 + X + ... + X^{l-1})$ can be constructed purely algebraically from $\mathbb{Q}$.

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I edited your question so that it is a little easier to read. Feel free to undo the edit if you prefer it as it was before. – Ragib Zaman Jul 22 '12 at 5:12
That's nice. Thanks. – Makoto Kato Jul 22 '12 at 5:14
$\mathbb{C}$ is a red herring; it has nothing to do with the argument, and was presumably only mentioned out of habit. You could replace $\mathbb{C}$ with $\overline{\mathbb{Q}}$, but it would still be unrelated. But as long as I'm on the unrelated issue, you may be interested in the notion of a "real closed field".... – Hurkyl Jul 22 '12 at 5:39
@Hurkyl I'm just asking a proof using only elementary properties of $\mathbb{Q}$. – Makoto Kato Jul 22 '12 at 12:14
What part of the proof do you think isn't an elementary property of $\mathbb{Q}(\zeta)$? I'm honestly not sure what you think this proof uses that you want to avoid. – Hurkyl Jul 22 '12 at 14:37
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