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Question Number 30590 by abdo imad last updated on 23/Feb/18
decompose sur R[x] x^(2n+1) −1.
$${decompose}\:{sur}\:{R}\left[{x}\right]\:\:{x}^{\mathrm{2}{n}+\mathrm{1}} \:−\mathrm{1}. \\ $$
Answered by sma3l2996 last updated on 23/Feb/18
let P(x)=x^(2n+1) −1 α_k =e^((2ikπ)/(2n+1)) are roots of P(x) P(x)=Π_(k=0) ^(2n) (x−e^((2ikπ)/(2n+1)) ) α_1 =e^((2iπ)/(2n+1)) ; α_(2n) =e^((2i(2n)π)/(2n+1)) =e^((2i(2n+1−1)π)/(2n+1)) =e^((−2iπ)/(2n+1)) ×e^(2iπ) =α_1 ^(__) (x−α_1 )(x−α_1 ^(__) )=x^2 −2cos(((2π)/(2n+1)))+1 α_2 =α_(2n−1) ^(__) α_(n−1) =α_(n+2) ^(__) In general α_j =α_(2n−(j−1)) ^(__) and (x−α_j )(x−α_(2n−(j−1)) )=x^2 +2cos(((2jπ)/(2n+1)))+1 so P(x)=(x−1)Π_(k=1) ^n (x^2 +2cos(((2kπ)/(2n+1)))+1)
$${let}\:{P}\left({x}\right)={x}^{\mathrm{2}{n}+\mathrm{1}} −\mathrm{1} \\ $$$$\alpha_{{k}} ={e}^{\frac{\mathrm{2}{ik}\pi}{\mathrm{2}{n}+\mathrm{1}}} \:{are}\:{roots}\:{of}\:{P}\left({x}\right) \\ $$$${P}\left({x}\right)=\prod_{{k}=\mathrm{0}} ^{\mathrm{2}{n}} \left({x}−{e}^{\frac{\mathrm{2}{ik}\pi}{\mathrm{2}{n}+\mathrm{1}}} \right) \\ $$$$\alpha_{\mathrm{1}} ={e}^{\frac{\mathrm{2}{i}\pi}{\mathrm{2}{n}+\mathrm{1}}} \:;\:\alpha_{\mathrm{2}{n}} ={e}^{\frac{\mathrm{2}{i}\left(\mathrm{2}{n}\right)\pi}{\mathrm{2}{n}+\mathrm{1}}} ={e}^{\frac{\mathrm{2}{i}\left(\mathrm{2}{n}+\mathrm{1}−\mathrm{1}\right)\pi}{\mathrm{2}{n}+\mathrm{1}}} ={e}^{\frac{−\mathrm{2}{i}\pi}{\mathrm{2}{n}+\mathrm{1}}} ×{e}^{\mathrm{2}{i}\pi} =\overset{\_\_} {\alpha}_{\mathrm{1}} \\ $$$$\left({x}−\alpha_{\mathrm{1}} \right)\left({x}−\overset{\_\_} {\alpha}_{\mathrm{1}} \right)={x}^{\mathrm{2}} −\mathrm{2}{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)+\mathrm{1} \\ $$$$\alpha_{\mathrm{2}} =\overset{\_\_} {\alpha}_{\mathrm{2}{n}−\mathrm{1}} \\ $$$$\alpha_{{n}−\mathrm{1}} =\overset{\_\_} {\alpha}_{{n}+\mathrm{2}} \\ $$$${In}\:{general}\:\:\alpha_{{j}} =\overset{\_\_} {\alpha}_{\mathrm{2}{n}−\left({j}−\mathrm{1}\right)} \\ $$$${and}\:\:\left({x}−\alpha_{{j}} \right)\left({x}−\alpha_{\mathrm{2}{n}−\left({j}−\mathrm{1}\right)} \right)={x}^{\mathrm{2}} +\mathrm{2}{cos}\left(\frac{\mathrm{2}{j}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)+\mathrm{1} \\ $$$${so}\:\:{P}\left({x}\right)=\left({x}−\mathrm{1}\right)\prod_{{k}=\mathrm{1}} ^{{n}} \left({x}^{\mathrm{2}} +\mathrm{2}{cos}\left(\frac{\mathrm{2}{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)+\mathrm{1}\right) \\ $$

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