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[数论] $\sum_{k=1}^{\text{ord}_n(2)}\zeta_n^{2^k}$的次数?

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hbghlyj Posted at 2024-5-10 04:09:44 |Read mode
$n\gt2$为奇数,$\zeta_n=\mathrm{e}^{i\frac{2\pi}n}$,$μ(n)$是Möbius function
  • $\mu(n)=0$,则$\sum_{k=1}^{\text{ord}_n(2)}\zeta_n^{2^k}=0$
  • $\mu(n)\ne0$,则$\sum_{k=1}^{\text{ord}_n(2)}\zeta_n^{2^k}$的极小多项式的次数为$\varphi(n)\over\text{ord}_n(2)$

Sum[Exp[I*2Pi/n*2^k],{k,1,MultiplicativeOrder[2,n]}] for n=3$=-1$的极小多项式$x + 1$的次数为1
Sum[Exp[I*2Pi/n*2^k],{k,1,MultiplicativeOrder[2,n]}] for n=5$=-1$的极小多项式$x + 1$的次数为1
Sum[Exp[I*2Pi/n*2^k],{k,1,MultiplicativeOrder[2,n]}] for n=7$=\frac{1}{2}(-1+i \sqrt{7})$的极小多项式$x^2 + x + 2$的次数为2
Sum[Exp[I*2Pi/n*2^k],{k,1,MultiplicativeOrder[2,n]}] for n=15$=\frac{1}{2}(1+i \sqrt{15})$的极小多项式$x^2 - x + 4$的次数为2

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 Author| hbghlyj Posted at 2024-5-10 04:44:53
发到MSE试试:https://math.stackexchange.com/q ... -textord-n2-zeta-n2
有回覆了。


证明见上面的帖子的回答中的链接linear independence of characters的Theorem 3.8

character真的很好用

相关:
Section 9.13 (0CKK): Linear independence of characters

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