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 $\mathbb{Q}$.
 A: Yes, this can be proved using "only elementary properties of $\mathbb{Q}$", since the original argument can be rewritten in this manner.  Let $K=\mathbb{Q}[X]/(1+X+X^2+\dots+X^{\ell-1})$, and let $\zeta$ be the image of $X$ under the natural map $\mathbb{Q}[X]\to K$.  Let $L=\mathbb{Q}(\zeta+1/\zeta)$.  All that needs to be done is to define a total ordering on $L$ which makes $L$ into an ordered field, and for which $-2\le\zeta+1/\zeta\le 2$.  For, if this is accomplished then the classical proof works: any element of $K$ is $a+b\zeta$ with $a,b\in L$, and 
$$
N_{K/L}(a+b\zeta)=(a+b\zeta)(a+b/\zeta)=a^2+b^2+ab(\zeta+1/\zeta)\ge a^2+b^2-2ab = (a-b)^2 \ge 0,
$$
so for any $h(X)\in\mathbb{Q}[X]$ it follows that
$$
N_{K/\mathbb{Q}}(h(\zeta)) = \prod_{i=1}^{(l-1)/2} h(\zeta^i) h(1/\zeta^i)
= \prod_{i=1}^{(l-1)/2} N_{K/L}(h(\zeta)) \ge 0.
$$
It remains only to define a total ordering on $L$ which is compatible with the field operations, and for which $-2\le\zeta+1/\zeta\le 2$.  If $\ell=3$ then $\mathbb{Q}(\zeta+1/\zeta)=\mathbb{Q}$, and there is nothing to prove.  So assume $\ell>3$.  Let  $f(X)$ be the minimal polynomial of $\zeta+1/\zeta$ over $\mathbb{Q}$.  For any rational number $\beta$, define $\zeta+1/\zeta$ to be less than $\beta$ if $\beta$ is greater than all rational numbers which are smaller than all positive rational numbers $\gamma$ for which $f(\gamma)\cdot f(0)<0$.  If $\zeta+1/\zeta$ is not less than $\beta$, then define $\zeta+1/\zeta$ to be greater than $\beta$.  Exercise for the OP: verify that this ordering has a unique extension to a total ordering on $L$ which is compatible with the field operations.  QED
