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业余的业余
Posted 2023-10-23 01:10
Last edited by hbghlyj 2025-5-16 03:45现在考虑 $q k+1$ 其中 $q$ 不是质数。
$o_p(a)=q \Longrightarrow p \equiv 1\pmod q$ 继续成立。其难点在于找到多项式 $\Phi_q(x)$ 使其根模 $p$ 的阶正好是 $q$, 而不是 $q$ 的某个真约数。可以试一下, 在 $\mathbb{Z}[x]$ 中找这个多项式有些棘手。办法是跳出整数甚至实数的限制, 在复数域中, $x^q-1$ 可因式分解为:
$$
x^q-1=\prod_{k=1}^q\left(x-\zeta_q^k\right)
$$
其中 $\zeta_q=e^{2 \pi i / q}$, 是单位数(1)的第 $q$ 个原始根。 $\zeta_q^{k}$ 均匀地分布在单位圆上, 这大概是分圆多项式得名的原因。这个跳出 $\mathbb{Z}$ 进到 $\mathbb{C}$ 的做法和后面的 跳出 $\mathbb{F}_p[x]$ 进到某个 $\mathbb{F}_{p^n}[x]$ 的处理如出一辄。两个对照着看更容易理解。
我们来看看 $\zeta_q^k$ 的阶, 即满足 $\zeta_q^{k d}=1$ 的最小正整数, 记为 $o\left(\zeta_q^k\right)$. 我们有
$$
\zeta_q^{k d}=1 \Leftrightarrow q\mid k d \Leftrightarrow \frac{q}{\operatorname{gcd}(q, k)}\mid \frac{k}{\operatorname{gcd}(q, k)} d \Leftrightarrow \frac{q}{\operatorname{gcd}(q, k)} \mid d
$$
故有 $o\left(\zeta_q^k\right)=q / \gcd(q, k)$. 当且仅当 $\operatorname{gcd}(q, k)=1$ 时, 这个阶为 $q$. 于是有如下定义
第 $q$ 分圆多项式
$$
\Phi_q(x)=\prod_{\substack{1 \leq k \leq q \\ \operatorname{gcd}(q, k)=1}}\left(x-\zeta_q^k\right)
$$
A 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].
Peter Borwein - Computational Excursions in Analysis and Number Theory (2002).pdf page55 |
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hbghlyj
Posted 2023-10-21 01:09
abababa
Posted 2023-11-6 19:46
