求cos 9°的极小多项式？

hbghlyj

$\zeta=\exp(i\frac{2\pi}{40})$
$\varphi(40)=16$
\begin{array}{|c|c|c|c|}
\hline
1 & 3 & 7 & 9 \\
\hline
11 & 13 & 17 & 19 \\
\hline
21 & 23 & 27 & 29 \\
\hline
31 & 33 & 37 & 39 \\
\hline
\end{array}2cos 9°$=\zeta+\zeta^{39}$
2cos 27°$=\zeta^3+\zeta^{37}$
2cos 63°$=\zeta^7+\zeta^{33}$
2cos 81°$=\zeta^9+\zeta^{31}$
2cos 99°$=\zeta^{11}+\zeta^{29}$
2cos 117°$=\zeta^{13}+\zeta^{27}$
2cos 153°$=\zeta^{17}+\zeta^{23}$
2cos 171°$=\zeta^{19}+\zeta^{21}$
所以cos 9°的极小多项式为
(x-cos 9°)(x-cos 27°)(x-cos 63°)(x-cos 81°)(x-cos 99°)(x-cos 117°)(x-cos 153°)(x-cos 171°)=
$$\frac{1}{256}\left(256 x^8-512 x^6+304 x^4-48 x^2+1\right)$$
wolframalpha.com/input?i=Expand%5B%28x-cos+9%C2%B0%29%28x-cos+27 ... -cos+171%C2%B0%29%5D

hbghlyj

PARI/GP
? polgalois(256*x^8-512*x^6+304*x^4-48*x^2+1)
%1 = [8, 1, 2, "4[x]2"]

C₂×C₄
$\xymatrix { &C_2\times C_4 \\ C_4\ar@[red][ur]&C_2^2\ar@[red][u ]&C_4\ar@[red][ul] \\
C_2\ar@[red][ur]&C_2\ar@[red][ul]\ar@[red][u ]\ar@[red][ur]&C_2\ar@[red][ul] \\
&C_1\ar@[red][u ]\ar@[red][ul]\ar@[red][ur]}$
按照$\Bbb Z^\times_{40}/\{\pm1\}$乘法写出：
$\xymatrix { &\{\{\pm1\},\{\pm3\},\{\pm7\},\{\pm9\},\{\pm11\},\{\pm13\},\{\pm17\},\{\pm19\}\}\\ \{\{\pm1\},\{\pm3\},\{\pm9\},\{\pm13\}\}\ar@[red][ur]&\{\{\pm1\},\{\pm9\},\{\pm11\},\{\pm19\}\}\ar@[red][u ]&\{\{\pm1\},\{\pm7\},\{\pm9\},\{\pm17\}\}\ar@[red][ul] \\
\{\{\pm1\},\{\pm11\}\}\ar@[red][ur]&\{\{\pm1\},\{\pm9\}\}\ar@[red][ul]\ar@[red][u ]\ar@[red][ur]&\{\{\pm1\},\{\pm19\}\}\ar@[red][ul] \\
&\{\{\pm1\}\}\ar@[red][u ]\ar@[red][ul]\ar@[red][ur]}$
写出$\zeta$的幂的和：
如果和为0，用它们的乘积，如$(\zeta+\zeta^{-1})(\zeta^{19}+\zeta^{-19})$
$\xymatrix { &ℚ(\zeta+\zeta^{-1}+\zeta^3+\zeta^{-3}+\zeta^7+\zeta^{-7}+\zeta^9+\zeta^{-9}+\zeta^{11}+\zeta^{-11}+\zeta^{13}+\zeta^{-13}+\zeta^{17}+\zeta^{-17}+\zeta^{19}+\zeta^{-19}) \\ ℚ(\zeta+\zeta^{-1}+\zeta^3+\zeta^{-3}+\zeta^9+\zeta^{-9}+\zeta^{13}+\zeta^{-13})\ar@{<-}@[red][ur]& ℚ((\zeta+\zeta^{-1})(\zeta^9+\zeta^{-9})(\zeta^{11}+\zeta^{-11})(\zeta^{19}+\zeta^{-19}))\ar@{<-}@[red][u ]& ℚ(\zeta+\zeta^{-1}+\zeta^7+\zeta^{-7}+\zeta^9+\zeta^{-9}+\zeta^{17}+\zeta^{-17})\ar@{<-}@[red][ul] \\ ℚ(\zeta+\zeta^{-1}+\zeta^{11}+\zeta^{-11})\ar@{<-}@[red][ur]&ℚ(\zeta+\zeta^{-1}+\zeta^9+\zeta^{-9})\ar@{<-}@[red][ul]\ar@{<-}@[red][u ]\ar@{<-}@[red][ur]&ℚ((\zeta+\zeta^{-1})(\zeta^{19}+\zeta^{-19}))\ar@{<-}@[red][ul]
\\&ℚ(\zeta+\zeta^{-1})\ar@{<-}@[red][ul]\ar@{<-}@[red][u ]\ar@{<-}@[red][ur]}$
代数化简：
$\zeta+\zeta^{-1}+\zeta^3+\zeta^{-3}+\zeta^9+\zeta^{-9}+\zeta^{13}+\zeta^{-13}$
$(\zeta+\zeta^{-1})(\zeta^9+\zeta^{-9})(\zeta^{11}+\zeta^{-11})(\zeta^{19}+\zeta^{-19})$
$\zeta+\zeta^{-1}+\zeta^7+\zeta^{-7}+\zeta^9+\zeta^{-9}+\zeta^{17}+\zeta^{-17}$
$\zeta+\zeta^{-1}+\zeta^{11}+\zeta^{-11}$
$\zeta+\zeta^{-1}+\zeta^9+\zeta^{-9}$
$(\zeta+\zeta^{-1})(\zeta^{19}+\zeta^{-19})$
$\xymatrix { &ℚ \\ ℚ(\sqrt{10})\ar@{<-}@[red][ur]& ℚ(\sqrt5)\ar@{<-}@[red][u ]& ℚ(\sqrt2)\ar@{<-}@[red][ul] \\ ℚ(\sqrt{5-\sqrt{5}})\ar@{<-}@[red][ur]&ℚ(\sqrt{3+\sqrt{5}})\ar@{<-}@[red][ul]\ar@{<-}@[red][u ]\ar@{<-}@[red][ur]&ℚ(\sqrt{\frac{5+\sqrt5}2})\ar@{<-}@[red][ul]
\\&ℚ(\frac{1}{2} \sqrt{2+\sqrt{\frac{5+\sqrt{5}}{2}}})\ar@{<-}@[red][ul]\ar@{<-}@[red][u ]\ar@{<-}@[red][ur]}$
其中$2\sqrt{3+\sqrt5}=\sqrt2+\sqrt{10}$，所以$\Bbb Q(\sqrt{3+\sqrt5})=\Bbb Q(\sqrt2,\sqrt5)=\Bbb Q(\sqrt2,\sqrt{10})=\Bbb Q(\sqrt5,\sqrt{10})$

hbghlyj

cos 9°可以写成以下根式形式：
$$=\frac{1}{4} \left(\sqrt{5-\sqrt{5}}+\sqrt{3+\sqrt{5}}\right)\tag1\label1$$
$$=\frac{1}{8}(2 \sqrt{5-\sqrt{5}}+\sqrt{2}+\sqrt{10})\tag2\label2$$
$$=\frac{1}{2} \sqrt{2+\sqrt{\frac{5+\sqrt{5}}{2}}}\tag3\label3$$

1. Factor[1-48 x^2+304 x^4-512 x^6+256 x^8,Extension->Sqrt[5-Sqrt[5]]]

复制代码

cos 9°在$\Bbb Q(\sqrt{5-\sqrt5})$上的极小多项式为$$8 x^2-4 \sqrt{5-\sqrt{5}} x-\sqrt{5}+1$$
解二次方程得\eqref{1}，化简得\eqref{2}

1. Factor[1-48 x^2+304 x^4-512 x^6+256 x^8,Extension->Sqrt[3+Sqrt[5]]]

复制代码

cos 9°在$\Bbb Q(\sqrt{3+\sqrt5})$上的极小多项式为$$8 x^2-4 \sqrt{3+\sqrt{5}} x+\sqrt{5}-1$$解二次方程得\eqref{1}，化简得\eqref{2}

1. Factor[1-48 x^2+304 x^4-512 x^6+256 x^8,Extension->Sqrt[(5 + Sqrt[5])/2]]

复制代码

cos 9°在$\Bbb Q(\sqrt{5+\sqrt5\over2})$上的极小多项式为$$4x^2-\sqrt{5+\sqrt{5}\over2}-2$$解二次方程得\eqref{3}

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因为\eqref{2}和\eqref{3}是从不同的域分解出的，很难直接从\eqref{3}得到\eqref{2}

Czhang271828

mathoverflow.net/questions/287109/minimal-polynomial-of-cos%CF%80-n