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Last edited by hbghlyj at 2023-2-4 16:40:00A primitive $n$th root of unity is a complex number $\omega$ that satisfies $\omega^{n}=1$ and $\omega^{k} \neq 1$ for any positive $k< n$. Let $\zeta_{n}:=\exp (2 \pi i / n)$; then $\zeta_{n}$ is a primitive $n$th root of unity. The $\phi(n)$ primitive $n$th roots of unity are $\left\{\zeta_{n}^{m}: \operatorname{gcd}(m, n)=1\right\}$.
The $n$th cyclotomic polynomial $\Phi_{n}$ is the minimal polynomial of any primitive $n$th root of unity. This is an irreducible polynomial of degree $\phi(n)$ given by
$$
\Phi_{n}(z)=\prod_{\substack{1 \leq j \leq n \\ \operatorname{gcd}(j, n)=1}}\left(z-\exp \left(j {2 \pi i \over n}\right)\right)
$$
Show that
$$
z^{n}-1=\prod_{d \mid n} \Phi_{n}(z)
$$
Show that
$$
\Phi_{p^{n}}(z)=\Phi_{p}\left(z^{p^{n-1}}\right),
$$
and more generally, if every prime that divides $m$ also divides $n$, then
$$
\Phi_{m n}(z)=\Phi_{n}\left(z^{m}\right)
$$
Show that for odd $n$,
$$
\Phi_{n}(-z)=\Phi_{2 n}(z)
$$
Show that if $p$ is a prime not dividing an integer $n$, then
$$
\Phi_{p n}(z)=\Phi_{n}\left(z^{p}\right) / \Phi_{n}(z)
$$
Show, with $\mu$ is Mobius function, that
$$
\Phi_{n}(z)=\prod_{d \mid n}\left(z^{d}-1\right)^{\mu(n / d)}
$$
Show that
$$
\Phi_{p^{k}}(1)=p
$$
if $p$ is a prime and that $\Phi_{n}(1)=1$ if $n$ is not a power of a prime. Also show that
$$
\Phi_{2 p^{k}}(-1)=p
$$
if $p$ is a prime, $\Phi_{1}(-1)=-2, \Phi_{2}(-1)=0$, and that $\Phi_{n}(-1)=1$ otherwise.
If $p$ is a prime not dividing $n$, then $\Phi_{n}$ factors in $\mathbb{Z}_{p}[z]$ into $\phi(n) / d$ irreducible factors, each of degree $d$, where $d$ is the smallest positive integer solution of $p^{d} \equiv 1\pmod n$. See Lidl and Niederreiter [1983].
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业余的业余
Posted at 2023-10-23 01:08:10
