[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)
膠智的kai龍
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Word + excel
用開md
Algebra姐係y=x^2 個d
係arithmetic 
竟然會有funamental theorem of algebra
analytic proof
只見過d Alembert one同Cauchy integral formula嘅corollary
但Algebra proof真係第一次見，一定要睇
analytic proof
只見過d Alembert one同Cauchy integral formula嘅corollary但Algebra proof真係第一次見，一定要睇
當年d Alembert同Gauss無topology 呢啲工具但都好犀利
又交返個波俾我
證p-group has all its p subgroup, order p, p^2,…p^k-1
證p-group has all its p subgroup, order p, p^2,…p^k-1
唔係sylow d朋友?
係，首先假設R[i]對上仲有個finite extension E，而K就係嗰個finite extension用automorphism集合，R[i]-E-K
做做下就證到有個entire Galois group有2-Sylow subgroup存在，咁Gal(K/R) 就會係2^k，但仲有odd part。不過odd當中irreducible polynomial嘅degree有1(by IVT, all odd degree has one real root, hence only possible degree of irreducible polynomial is 1，但1，其實就等於喺R，所以其實係冇)。
證埋呢度(by induction, center, correspondnece)就可以話存在quadratic extension F，但extend嘅元素其實都係屬於R[i]，因為all positive real number has root and complex number has root as well，所以F = R[i]，一路用呢個argument，K=R[i]，咁頭先R[i]-E-K=R[i]就要令E都係R[i]，i.e. 所有arbitrary finite extension都係佢自己，咁R[i]就係algebraic closure of R，every polynomial of R has a root in R[i]，i.e. C
做做下就證到有個entire Galois group有2-Sylow subgroup存在，咁Gal(K/R) 就會係2^k，但仲有odd part。不過odd當中irreducible polynomial嘅degree有1(by IVT, all odd degree has one real root, hence only possible degree of irreducible polynomial is 1，但1，其實就等於喺R，所以其實係冇)。
證埋呢度(by induction, center, correspondnece)就可以話存在quadratic extension F，但extend嘅元素其實都係屬於R[i]，因為all positive real number has root and complex number has root as well，所以F = R[i]，一路用呢個argument，K=R[i]，咁頭先R[i]-E-K=R[i]就要令E都係R[i]，i.e. 所有arbitrary finite extension都係佢自己，咁R[i]就係algebraic closure of R，every polynomial of R has a root in R[i]，i.e. C
所以用到少少analysis，但係呢個proof第一次見
