[數撚圍爐區] 將一個波分成五份, 然後砌兩個同原本一樣的波的鬼故事 (108)

唔係我都唔係為教學 又唔係數學系 點教你

Lol oh its an open invite 🙈
I so low level當興趣睇d undergrad stuff
有問題都too trivial dare not post here ask 🤦🏻‍♀️
just try ha my luck ~

依家ai消息下下都震撼彈咁
有啲想知多啲

開始明點解要證如果呢個係cyclic extension of degree n/cyclic extension with degree p，咁就會有root解x^p - a同x^p -x -a，一切都係要為solvable by radical做準備

Extension field就會有

已經睇完solvable <=> solvable by radical（core argument都係Galois correspondence）
我咁估：
剩返落嚟就係要喺F中Construct 一個quintic polynomial，只要佢Galois group is of order 120，咁佢就必然要係S5，又或者話佢Galois group係A5，咁佢就係not solvable，換言之唔可以透過自身同加插啲根式元素span出嚟(減同除係+ inverse同* inverse），換言之就係無general solution
問題只係剩番要搵irreducible quintic polynomial有order 120
當初俾五次方程式以上無公式解水咗入嚟，後來先發現原來Galois correspondence先係重點

終於完

但係我依家咁睇，其實佢係話唔可以透過根式base field元素同base field自身元素搵出嚟，冇話冇其他方法喎。雖然我諗唔到除咗加減乘除同根式，同類似newton-rhapson之外，人仲有無其他方法可以搵根

三年了 終於完咗數學其中一樣

Module theory同category theory有需要睇先再睇，可以完啦

Proof: 任何有三隻real root同兩隻conjugate root嘅irreducible polynomial都係S5
因為如果該group中係用一隻5-cycle同2 cycle generate嘅就係S5，所以存在兩隻conjugate root互換
［F(alpha1, …, alpha 5): F]，因為irr就係min polynomial and by tower law，hence 5| [F(alpha1, F] ，hence by correspondence, and by Cauchy，至少有唔係e嘅element of order 5 generate 5次返到e(ie, 5 cycle)
QED

呢五隻唔可以重覆

F做Q

Real root => Real/Rational

Proof: 任何有三隻R/Q root同兩隻conjugate complex root in C/R 嘅irreducible separable polynomial都係S5
因為如果該group中係用一隻5-cycle同2 cycle generate嘅就係S5，所以存在兩隻conjugate root互換
［Q(alpha1, …, alpha 5): Q]，因為irr就係min polynomial and by tower law，hence 5| [Q(alpha1, Q] ，hence by correspondence, and by Cauchy，至少有唔係e嘅element of order 5 generate 5次返到e(ie, 5 cycle)
QED

Polynomial over Q

5|［Q(alpha1, …, alpha 5): Q]

不過咁講，Galois點解知道要將搵extension of field變咗做根map根嘅automorphism嘅group，可能佢見到以前2-4次方程都牽涉開根號係類似循環群，定係從root of unity諗到abelian group？從而諗到嘅？點諗？

睇完Galois可以同ext tor開戰

真係怕

我已經唔記得哂

如果俾我諗，我一世都associate唔到兩樣野有乜關係？

我見到後面仲有啲咩representation theory，唔好搞啦返analysis算

要做返啲exercise，我心滿意足