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I think it's true that in a field $F$ of characteristic zero there are exactly $n$ distinct $n$th roots of unity (in some algebraic closure $\bar{F}$), that is, roots of the polynomial $x^n-1$.

I know there can be at most $n$ roots (since $x^n-1$ is of degree $n$), but how can we show that all $n$ roots are distinct?

I also know that in fields of characteristic zero all irreducible polynomials have zeros of multiplicity 1. But $x^n-1$ is reducible, so I think there must be some additional arguments that are escaping me.

user26857
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WillG
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    If $x^n-1$ had repeated roots then they would have to be common roots of $x^n-1$ and its derivative... – Daniel Schepler Sep 14 '18 at 23:13
  • Does using derivatives in this context require invoking theorems from calculus/analysis? Because don't those presuppose that we are working in $\mathbb{R}$ as opposed to generic fields? – WillG Sep 14 '18 at 23:57
  • Or is there a way to use formal derivatives for polynomials to prove this without relying on analysis concepts? – WillG Sep 14 '18 at 23:58
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    Yes, use the formal derivative $Df$ of a polynomial $f$. You can prove that $f$ and $Df$ have a common root $\alpha$ iff $f$ has a repeated root at $\alpha$ (that is, $(x-\alpha)^2$ divides $f$). – Daniel Mroz Sep 15 '18 at 00:00
  • Right, the formal derivative works, i.e. it satisfies the Leibniz product rule, so if you had $f(x) = g(x) (x-a)^n$ then $f'(x) = g'(x) (x-a)^n + g(x) \cdot n (x-a)^{n-1} = (x-a)^{n-1} (g'(x) (x-a) + n g(x))$. – Daniel Schepler Sep 15 '18 at 00:01
  • Somewhat related: https://en.wikipedia.org/wiki/K%C3%A4hler_differential or https://en.wikipedia.org/wiki/Differential_algebra - so $F[t]$ is a differential $F$-algebra, or $\Omega^1_{F[t] / F} \simeq F[t] dt$. – Daniel Schepler Sep 15 '18 at 00:06

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Suppose we have a primitive $n$th root of unity, $\omega$. Then for all $1 < a \le n$, $(\omega^a)^n = (\omega^n)^a = 1^a = 1$. If $1 \le a < b \le n$ then $\omega^a = \omega^b \implies \omega^{b-a} = 1$, contradicting the assumption that $\omega$ is a primitive $n$th root of unity. Therefore the existence of a primitive $n$th root of unity implies that there are $n$ distinct $n$th roots of unity, independently of the characteristic.

Now, let $\omega$ be a repeated root of $x^n - 1$ in the splitting field of $x^n - 1$ over $F$. It must be a primitive $m$th root of unity for some $m$ which is a proper factor of $n$. Let $k = \frac nm$. Then $$\gcd\left(\frac{x^{km}-1}{x^m-1}, x^m - 1\right) \neq 1$$ Now, $\gcd(p(x), q(x)) = \gcd(p(x) - q(x)r(x), q(x))$. Taking $r(x) = \sum_{i=0}^{k-2} (k-1-i)x^{im}$ we have $$\frac{x^{km} - 1}{x^m-1} - (x^m-1) \sum_{i=0}^{k-2} (k-1-i)x^{im} = \frac{(x^{km} - 1) - (x^m-1)^2 \sum_{i=0}^{k-2} (k-1-i)x^{im}}{x^m-1}$$ But that $r$ was carefully chosen to telescope: $$(x^m-1)^2 \sum_{i=0}^{k-2} (k-1-i)x^{im} = x^{km} -k x^{m} + (k-1)$$ so that $$\gcd\left(\frac{x^{km}-1}{x^m-1}, x^m - 1\right) = \gcd(k, x^m - 1)$$ This is non-constant iff $k$ is a multiple of the characteristic of $F$.

Peter Taylor
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