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Let $F$ be a field and $n \in \mathbb{N}$. 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 $X^n - 1 \in F[X]$. If I denote by $W_n$ the set of the $n$-th roots of unity, then it is a cyclic group and after I define the $n$-th cyclotomic polynomial by $$ {\phi}_n(X) = \prod_{\varepsilon \in W_n^*} (X - \varepsilon)\mbox{,} $$ where $W_n^*$ is the set of the elements $\varepsilon \in W_n$ such that $\langle \varepsilon \rangle = W_n$ (it is the set of the $n$-th primitive roots of unity). And now I can make my question: is the formula $$ X^n - 1 = \prod_{d \mid n} {\Phi}_d(X) $$ valid for each field $F$ or I need to consider $F = \mathbb{C}$?

mathcounterexamples.net
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joseabp91
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    Your set $W_n^*$ is NOT what is usually defined as "the set of $n$th primitive roots of unity", if the characteristic of $F$ divides $n$. – Eric Wofsey Aug 19 '18 at 19:35

1 Answers1

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There are subtleties if $n\cdot1_F=0_F$. If $F$ has characteristic $p$, then we have $$X^p-1=(X-1)^p.$$ This implies $W_{p}^*=W_{p^2}^*=\{1_F\}$, and the formula fails.

Other than this, it does hold. See a relatively recent answer by yours truly.

Jyrki Lahtonen
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  • Why does the formula fail? Is it not possible to obtain ${\Phi}_p(X) = {(X - 1)}^{p - 1}$? – joseabp91 Aug 19 '18 at 19:44
  • This is uncomfortably close to being a dupe, actually. Switching to CW for I should not try and get paid twice for the same work. – Jyrki Lahtonen Aug 19 '18 at 19:44
  • @joseabp91 Because $W_p^*$ has a single element $1$ it follows from your definition that $\Phi_p(X)=X-1$. Similarly $\Phi_{p^2}(X)=X-1$. Therefore the formula fails when $n=p$ and when $n=p^2$. It actually fails for a similar reason whenever $p\mid n$. – Jyrki Lahtonen Aug 19 '18 at 19:45
  • What happens if $W_p^* = {1}$? – joseabp91 Aug 19 '18 at 19:47
  • $W_p={1}$, so $W_p^*={1}$. Therefore $\Phi_p(X)$ has degree $1$. – Jyrki Lahtonen Aug 19 '18 at 19:50
  • Oh I see, the contradiction follows from the equality $X - 1 = {(X - 1)}^{p - 1}$, right? They should have the same degree and we have not it for $p > 2$. – joseabp91 Aug 19 '18 at 19:52
  • Correct @joseabp91. $p=2$ is, indeed, a special case. I knew there was a reason why I brought up $p^2$!!! In characteristic two your definition also leads to $\Phi_4(X)=X-1$ :-) – Jyrki Lahtonen Aug 19 '18 at 20:02
  • Can I state now that the next statement in an arbitrary field is false? This: if $\varepsilon$ is a $n$-th root of unity and $m$-th root of unity, where $n , m \in \mathbb{N}$ satisfy $\gcd(n , m) = 1$, then $\varepsilon = 1$. – joseabp91 Aug 19 '18 at 20:50