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Let $F$ be a field and $n , m \in \mathbb{N}$. Then an element $\varepsilon \in \overline{F}$, being $\overline{F}$ an algebraic closure of $F$, is called a $n$-th root of unity if it is a root of the polynomial $X^n - 1 \in F[X]$. Now suppose that $n$ and $m$ are coprime numbers ($\gcd(n , m) = 1$) and $\varepsilon$ is $n$-th and $m$-th root of the unity. Can I state that $\varepsilon = 1$? If $F = \mathbb{C}$, then it is trivial because we have the formula $$ X^n - 1 = \prod_{d\big|_n} {\Phi}_d(X)\mbox{,} $$ being ${\Phi}_d$ the $d$-th cyclotomic polynomial in $\mathbb{C}[X]$. But this formula fails in an abstract field $F$, so is it false in general?

joseabp91
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  • Why do you say the formula fails in an abstract field? It is true in $\mathbb{Z}[X]$, so it is true in any commutative ring. – xarles Aug 19 '18 at 21:39
  • If we take $n = 3$ and $F = {\mathbb{Z}}_3$, we would obtain ${(X - 1)}^2 = X - 1$. Look at this:https://math.stackexchange.com/questions/2888050/cyclotomic-polynomial-formula-is-it-valid-in-an-arbitrary-field – joseabp91 Aug 19 '18 at 21:41
  • What you say in not true. In $\mathbb{F}_3$, the field with three elements, $(X-1)^3=X^3-1$. But $X^3-1=(X-1)(X^2+X+1)$ also, and $\Phi_3(X)=X^2+X+1$. What it is not true is that $\Phi_3(X)$ over $\mathbb{F}_3$ can be obtained as $(X-\xi)(X-\xi^2)$ for $\xi$ a primitive $3$-root of unity over $\mathbb{F}_3$. – xarles Aug 19 '18 at 21:49
  • Then how can you get ${\Phi}_3(X)$? I do not understand what you mean – joseabp91 Aug 19 '18 at 21:59
  • $$\Phi_n(x):=\frac{x^{n}-1}{\prod_{\stackrel{d|n}{{}{d<n}}}\Phi{d}(x)}$$ by definition. – xarles Aug 19 '18 at 22:06
  • My definition of cyclotomic polynomial is ${\Phi}n(X) = \prod{\varepsilon} (X - \varepsilon)$, where $\varepsilon$ is $n$-th primitive root of polynomial. In that case, you obtain that ${\Phi}_3(X) = X^2 + X + 1 = {(X - 1)}^2$, so there is only one cubic primitive root of unity, which is $1$. Then you obtain $X^2 + X + 1 = X - 1$. Where I am wrong? – joseabp91 Aug 19 '18 at 22:19
  • Let us continue this discussion in chat. – xarles Aug 19 '18 at 22:19

1 Answers1

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To prove that statement in general, here is the problem that you need to solve (by induction on $\max (n,m)$):

$$\gcd(x^n-1, x^m-1)= x^{\gcd(n,m)}-1$$

This holds over every field! Then your question follows immediately.

A. Pongrácz
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  • Oh I see. Thank you very much! – joseabp91 Aug 19 '18 at 21:47
  • You are welcome! So can you accept my answer? Thank you. – A. Pongrácz Aug 19 '18 at 21:48
  • Yes, of course. And do you know if this statement is generally false? $\varepsilon$ is $n$-th root of unity if and only if there exists $d \in \mathbb{N}$, with $d\big|_n$, such that $\varepsilon$ is $d$-th primitive root of unity. – joseabp91 Aug 19 '18 at 21:51
  • I think this also follows from my answer. The smallest $d$ must divide all good exponents, it seems to me. It should also follow from basic group theory. – A. Pongrácz Aug 19 '18 at 21:55
  • How can I make the inductive step to show that formula? – joseabp91 Aug 20 '18 at 10:41
  • Show that if $n\geq m$, then $\gcd(x^n-1, x^m-1)= \gcd(x^{n-m}-1, x^m-1)$. This should kick in the Euclidean algorithm, or you can refer to the induction hypothesis, if you follow my proof strategy. – A. Pongrácz Aug 20 '18 at 10:45
  • I refer to the induction hypothesis. The case $N = \max(n , m) = 1$ is easy to check. How can I complete the proof now? – joseabp91 Aug 20 '18 at 10:50
  • The way I just suggested! – A. Pongrácz Aug 20 '18 at 10:51
  • Sorry but I do not understand your argument. I am customed to work with only one variable when I apply induction method. – joseabp91 Aug 20 '18 at 10:55
  • The only argument is $\max(n,m)$, so there is no problem with that. The assertion clearly holds when $n=m$. So assume that $n> m$: then $\max(n,m)> \max(n-m,m)$, and $\gcd(n,m) = \gcd(n-m,m)$. So you can refer to the induction hypothesis. – A. Pongrácz Aug 20 '18 at 10:58
  • If $n > m$, then $\max(n , m) > \max(n - m , m)$. It is clear. But why $\gcd(n , m) = \gcd(n - m , m)$? And what do I do after this? – joseabp91 Aug 20 '18 at 11:05
  • Please check the basic number theoretical properties of the greatest common divisor. You do not do anything after this, YOU ARE DONE! (I gave you the complete proof now, you just have to understand it.) Please think about it... (It would be nice if you could accept the answer now.) – A. Pongrácz Aug 20 '18 at 11:07