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 $\mathbb{Q}$.

  • $\begingroup$ 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. $\endgroup$ Jul 22, 2012 at 5:12
  • $\begingroup$ That's nice. Thanks. $\endgroup$ Jul 22, 2012 at 5:14
  • 3
    $\begingroup$ $\mathbb{C}$ 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".... $\endgroup$
    – user14972
    Jul 22, 2012 at 5:39
  • $\begingroup$ @Hurkyl I'm just asking a proof using only elementary properties of $\mathbb{Q}$. $\endgroup$ Jul 22, 2012 at 12:14
  • 2
    $\begingroup$ 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. $\endgroup$
    – user14972
    Jul 22, 2012 at 14:37

1 Answer 1


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

  • 2
    $\begingroup$ I think you mean "with $a, b \in L$". (Then define $|a|$ by means of the assigned total ordering on $L$.) $\endgroup$
    – user43208
    Aug 23, 2013 at 13:33
  • $\begingroup$ Thanks, you're right of course. I edited to fix that. $\endgroup$ Aug 23, 2013 at 14:40
  • 2
    $\begingroup$ I can imagine there being other proofs which are more satisfying than the one above. For instance, if we write $f(X)=\sum_{i=0}^{l-2} a_i X^i$ with $a_i\in\mathbb{Q}$, then the norm from $\mathbb{Q}(\zeta)$ to $\mathbb{Q}$ of $f(\zeta)$ equals $H(a_0,a_1,\dots,a_{l-2})$ for some $H(x_0,\dots,x_{l-2})\in\mathbb{Z}[x_0,\dots,x_{l-2}]$. The question is to show that this polynomial $H$ takes nonnegative values at all points of $\mathbb{Q}^{l-1}$. Perhaps $H$ is a sum of squares in $\mathbb{Z}[x_0,\dots,x_{l-2}]$? $\endgroup$ Aug 23, 2013 at 15:22
  • $\begingroup$ Dear Michael: can you give some hint on how to solve the exercise in the end? $\endgroup$ Aug 23, 2013 at 20:38
  • $\begingroup$ (Obviously you can prove that $L$ is formally real and that there's an ordering with $(\zeta + \zeta^{-1})^2 < 4$ by using the obvious embedding of $L$ in $\mathbb R$. But it should be clear that this is not the sought for solution.) $\endgroup$ Aug 23, 2013 at 21:24

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .