[9up數學] 五次方程沒有解?

係歷史的發展上面, 二次方程的解法係公元前古巴比倫已經有, 三次方程係1530年被Tartaglia解決, 四次方程係1540由Ferrari解決, 但之後就無法突破, 而微積分的發明更加令到當時的數學家覺得有方法搵五次方程的解

(year 1 / AL 程度)微積分練習1:
證明任何五次方程
x^5 + ax^4 + cx^3 + dx^2 + ex + f = 0 (a, b, c, d, e都係real numbers)
都會有一個real root.
More generally, 證明任何單數次方程with real number coefficients都有一個real root.

解三次方程的橋大約係對 x^3 + ax^2 + bx + c 用substitution x = y - a/3, 就可以將條equation轉做 y^3 + py = q 的樣, 然後假設個solution係兩個cube roots的和, 再運算一下就可以計到答案

解四次方程的橋第一步同三次一樣, 都係對 x^4 + ax^3 + bx^2 + cx sub x = y - a/4 就可以變成 y^4 + py^2 + qy = r的樣, 然後用一D極華麗的運算就可以將佢變成一條三次方程, 於是就可以用三次方程公式解決

於是當時的數學家就打算用同一條藥方去解五次方程, 但係都無功而還, 因為最後竟然會令條方程越搞越禍, 即使係一條好簡單的五次方程 x^5 + px + q = 0 都搞到一鑊粥. 最後由Ruffini係1799提出, 由Abel (Abel prize果個Abel) 係1824年比一個完整的證明, 其描述如下:

不存在一條只包括四則運算及有理數指數的五次方程公式解

要注意的係呢條定理唔係否定所有五次方程都無只含有四則運算及有理數指數的根 (例如x^5 = 1大家都知個根係1), 而係指無一條只包含四則運算及有理數指數的萬能公式去解一次五次方程

而呢個搵公式的旅程由Galois終結, Galois係1832年前就發現左Galois theory (係1846年 (佢死後約14年) 先由Liouville代其發表), 其表述如下:

一條多項式方程存在使用四則運算及有理數指數的解當且僅當其Galois group is solvable.

順帶一提, 呢度已經出現左兩個悲劇數學家, Abel同Galois, 係入正題之前可以講下佢地幾悲劇

Abel (1802-1829): 大約18歲的時候死老豆, 一直係窮撚, 寄篇文去拎資助果陣篇文又唔知點解唔見左, 而上邊提到佢的大作Abel-Ruffini又因為佢無錢所以只能夠縮頭到得返六頁, 然後佢寄比Gauss又比Gauss掟埋一邊無理到, 佢有另一篇大作又比Cauchy無視 (究竟Gauss同Cauchy呢兩條茂利害左幾多個數學家 lol) (而呢篇文又係係佢死後先有人係櫃桶底度搵返出黎), 之後佢因為肺癆而死, 而係佢死後兩日Crelle (數學界現時著名期刊Crelle's Journal的創辦人) 幫佢搵到份柏林大學教授職位, 享年26歲

Galois (1811-1832): 佢人生完全係傳奇, 因為佢表達完全係1999 (佢留低的手稿都無乜人明佢想講乜, 可以google望下佢的真跡), 因為政治而坐過兩次監, 岩岩坐完第二次監就 (相傳爭女) 就同一決鬥最後傷重不治, 享年20歲. 但係似乎佢知道呢次決鬥會死, 係決鬥之前三日就瘋狂將佢的數學發現寫低, 仲係咁寫"我無時間啦", 而佢對佢細佬的遺言係"唔好喊呀Alfred, 我需要全部勇氣先可以係20歲死架渣"

題外話, 歷史上有好多英年早逝的數學家, 最著名的有Ramanujan (1887- 1920, 32歲, 病死), Riemann (1826-1866, 40歲, 又係肺癆), Mirzakhani (1977-2017, 40歲, 2014 Fields Medalist, 死於癌症)
其他的有
Pavel Urysohn (1898-1924, 26歲, 游水浸死, 讀Topology一定聽過佢個大名)
Andreas Floer (1956-1991, 34歲, 自殺, 發明左Floer homology, topologist應該都聽過佢個大名)
谷山 豊 (1927-1958, 31歲, 自殺, 其谷山-志村猜想的特例可以用黎證明Fermat's Last Theorem)
Gotthold Eisenstein (1823-1852, 29歲, 死於肺癆, 讀代數果陣必定會讀到Eisenstein's criterion)
Oswald Teichmüller (1913-1943, 30歲, KIA(參加納粹軍))

=====================================================
(待續)

lm

呢個post的重點係簡介下入門級的Galois Theory, 所以有好多details會omit, 想睇details可以睇書 (最後會有references)

回顧返 (方便我講我會用返英文...)

Let f(x) be a polynomial. Then f(x) = 0 can be solved by radicals if and only if its Galois group is solvable.

呢度有好多technical terms要解釋, 我會慢慢講

1. Field

簡單黎講一個field就係可以做加減乘除, 而且加法乘法都符合commutative law (交換律), 亦有distributive law (分配律). 基本例子有 rational numbers (Q), real numbers (R), complex numbers (C).

如果F係一個field, 我地記 F[x] 做所有polynomials with coefficients in F 的集合, 而F(x)則係所有rational functions with coefficients in F.

問題2: 點解integers唔係一個field?

問題3: 點解C(x)係一個field但C[x]唔係一個field?

2. Field Extension

給定一個field, 我地都可以整大個field: 即係加多D"數字"落去原本個field度令佢變成一個新的field, 我地會叫呢個大D的field做細D的field的field extension.

例子有: R係Q 的field extension, C係R的field extension.

而呢條定理保證左我地可以加料成功:
Let F be a field and f(x) be an irreducible polynomial in F[x]. Then there is a field extension E such that f(x) has a root in E.

Irreducible polynomial即係個polynomial 唔可以factorised做兩個lower degree的non-constant polynomial.

例如:
x^2 + 1係irreducible over R, 但reducible over C.
x^4 + 1係irreducible over Q, 但reducible over R (x^4 + 1 = (x^2 + \sqrt{2}x + 1)(x^2 - \sqrt{2}x + 1))

點樣分辦一個polynomial係咪irreducible有好多方法, 最簡單有以下兩種:
1. quadratic / cubic: 無root => irreducible, 有root => reducible.
2. (Eisenstein's criterion) 如果f(x) = a_n x^n + ... + a_0 in Z[x], 而且有一個質數 p 令到: p整除 a_{n - 1}, ..., a_1, a_0 但 p^2 唔整除 a_0, 咁 f(x)會係irreducible over Q.
例如x^3 + 2x + 2係irreducible over Q (p = 2).

問題4: 證明 x^4 + 6x^3 + 9x^2 + 12x + 30 係irreducible over Q.

C有一條幾出名的定理:
(Fundamental theorem of algebra) 任何係 C[x] 的 polynomial 都會有一個root in C.

呢一類的field我地會叫佢地 algebraically closed.

而algebraically closed field係存在的:
Let F be a field, then F has a field extension which is algebraically closed.

而最細的 algebraically closed field containing F 我地叫佢做 algebraic closure of F. 例如R同Q的algebraic closure係C.

如果F係一個field, 而E係F的field extension, a 係 E 的一個數字, 我地會寫F(a)做smallest field extension of F that contains a.
例如
R(i) = C.
Q(\sqrt{2}) = { a + b\sqrt{2} : a, b \in Q} 都係一個field.
Q(2^(1/3)) = { a + b 2^(1/3) + c 4^(1/3) : a, b, c \in Q} 都係一個field.
Q(\zeta_n) 係Q 加上 n-th root of unity (n-th root of unity即係 x^n - 1 的根), 呢一種field extension我地叫cyclotomic extension.

問題5: 估下Q(\sqrt{3})的數字會係咩樣?

==========================================================================

禁錯制

an algebraic closure of a field K is an algebraic extension of K that is algebraically closed.

呢個post的重點係簡介下入門級的Galois Theory, 所以有好多details會omit, 想睇details可以睇書 (最後會有references)

回顧返 (方便我講我會用返英文...)

Let f(x) be a polynomial. Then f(x) = 0 can be solved by radicals if and only if its Galois group is solvable.

呢度有好多technical terms要解釋, 我會慢慢講

1. Field

簡單黎講一個field就係可以做加減乘除, 而且加法乘法都符合commutative law (交換律), 亦有distributive law (分配律). 基本例子有 rational numbers (Q), real numbers (R), complex numbers (C).

如果F係一個field, 我地記 F[x] 做所有polynomials with coefficients in F 的集合, 而F(x)則係所有rational functions with coefficients in F.

問題2:
點解integers (Z)唔係一個field?

問題3:
點解C(x)係一個field但C[x]唔係一個field?

2. Field Extension

給定一個field, 我地都可以整大個field: 即係加多D"數字"落去原本個field度令佢變成一個新的field, 我地會叫呢個大D的field做細D的field的field extension.

例子有:
R係Q 的field extension, C係R的field extension.

而呢條定理保證左我地可以加料成功:

Let F be a field and f(x) be an irreducible polynomial in F[x]. Then there is a field extension E such that f(x) has a root in E.

Irreducible polynomial即係個polynomial 唔可以factorised做兩個lower degree的non-constant polynomial.

例如:
1. x^2 + 1係irreducible over R, 但reducible over C.
2. x^4 + 1係irreducible over Q, 但reducible over R (x^4 + 1 = (x^2 + sqrt(2)x + 1)(x^2 - sqrt(2) x + 1))

點樣分辦一個polynomial係咪irreducible有好多方法, 最簡單有以下兩種:

1. quadratic / cubic: 無root => irreducible, 有root => reducible.
2. (Eisenstein's criterion) Let f(x) = a_n x^n + ... + a_0 in Z[x], if there is a prime number p such that p divides a_{n - 1}, ..., a_1, a_0 BUT p^2 does not divide a_0, then f(x) is irreducible over Q.

例如x^3 + 2x + 2係irreducible over Q (p = 2).

問題4: 證明 x^4 + 6x^3 + 9x^2 + 12x + 30 係irreducible over Q.

C有一條幾出名的定理:

(Fundamental theorem of algebra) Any polynomial in C[x] has a root in C.

呢一類的field我地會叫佢地 algebraically closed.

而algebraically closed field係存在的:

Let F be a field, then F has a field extension which is algebraically closed.

(要用Zorn's lemma呢條19野)

而最細的 algebraically closed field containing F 我地叫佢做 algebraic closure of F.

例如R同Q的algebraic closure係C.

如果F係一個field, 而E係F的field extension, a 係 E 的一個數字, 我地會寫F(a)做smallest field extension of F that contains a.

例如
1. R(i) = C.
2. Q(sqrt(2)) = { a + bsqrt(2) : a, b \in Q} 都係一個field.
3. Q(2^(1/3)) = { a + b 2^(1/3) + c 4^(1/3) : a, b, c \in Q} 都係一個field.
4. Q(ζ_n) 係Q 加上 n-th root of unity (n-th root of unity即係 x^n - 1 的根), 呢一種field extension我地叫cyclotomic extension.

問題5: 估下Q(\sqrt{3})的數字會係咩樣?

================================================

(待續)

推廣數學, 值得推

precise definition係咁

我比人block哂

咁C 唔會係Q嘅algebraic closure？

個post出得一陣，消代都要時間

*化

其實解呢D方程有咩用
永遠都會有再高一個次方

唔係, pi 唔係algebraic over Q

咦我知咩事, 原來有typo

但係你寫 R同Q的algebraic closure係C

留名學嘢

求學
想問下

想知有無四則運算以外既算法既定義> matrix係唔係？

一時手誤
連登好衰無得edit

篇文個定位係想俾數撚定大眾睇
如果想做科普好似落得太多專有名詞同definition