nyy 发表于 2024-1-15 13:45:15

五次方程、伽罗瓦群、群论等相关问题

已知首一系数的五次方程的五个系数,能确定方程的伽罗瓦群吗? - 酱紫君的回答 - 知乎
https://www.zhihu.com/question/64900539/answer/2476505448



补充内容 (2024-1-18 11:16):
这个回答下有不少错误

补充内容 (2024-1-18 14:13):
A5 = 11111, 122, 23, 5 S5 = 11111, 122, 14, 23, 5类型错误

补充内容 (2024-1-18 14:13):
还有应该强调不可约多项式

nyy 发表于 2024-1-15 14:40:20

本帖最后由 nyy 于 2024-1-15 14:45 编辑

可否建立数学模型并应用程序对任意魔方进行还原? - luikore的回答 - 知乎
https://www.zhihu.com/question/29716151/answer/403913780

如何求解五次方程 x^5 + 20x + 32 = 0,要精确的根式解? - 笑横野的回答 - 知乎
https://www.zhihu.com/question/457898479/answer/2254543107


已知首一系数的五次方程的五个系数,能确定方程的伽罗瓦群吗? - 酱紫君的回答 - 知乎
https://www.zhihu.com/question/64900539/answer/2476505448
整系数的多项式方程有个常用的手算方法叫模 p 约化(Mod P Reducing),

请列出一个最难相信但确实有根式解的一元五次方程? - 笑横野的回答 - 知乎
https://www.zhihu.com/question/423626791/answer/2253853548

请列出一个最难相信但确实有根式解的一元五次方程? - 失落之城的回答 - 知乎
https://www.zhihu.com/question/423626791/answer/2275679199


对于可解的高次方程有没有普遍的求根方法? - 笑横野的回答 - 知乎
https://www.zhihu.com/question/617548797/answer/3169537194

数学史上你认为最丑陋的公式是什么? - 酱紫君的回答 - 知乎
https://www.zhihu.com/question/21400777/answer/2475732494
介绍一下五次方程存在根式解得简单判定方法, 记得拉到下面做一下课后习题看下自己学废了没有.
Dummit 判别式

五次方程也有求根公式? - Achatinidae的文章 - 知乎
https://zhuanlan.zhihu.com/p/434045872
从上面的例子我们可以看出,仅仅通过观察一个五次方程,我们似乎很难判断它到底有没有根式解。然而办法总是有的,在本文中,我们将给出一种方法,它能判断任意一个系数均为有理数的五次方程是否有根式解,如果有的话,也可以给出它的根式解。由于这种方法是由Dummit在1991年的一篇论文中提出的,我称其为Dummit法。

五次方程 x^5+px+q 根式求解 - 酱紫君的文章 - 知乎
https://zhuanlan.zhihu.com/p/108460142

借助Mathematica和Magma体验伽罗瓦理论 - 核桃的文章 - 知乎
https://zhuanlan.zhihu.com/p/26436790

为什么课本上计算 Galois 群都是先求出根再计算的(不可解的除外)? - 王纪翔的回答 - 知乎
https://www.zhihu.com/question/458961859/answer/1882411563

高斯作出正 17 边形的依据是什么? - 缺心眼的香菜的回答 - 知乎
https://www.zhihu.com/question/26096850/answer/2625164779
这个很有意思,转载过来

为什么课本上计算 Galois 群都是先求出根再计算的(不可解的除外)? - 包遵信的回答 - 知乎
https://www.zhihu.com/question/458961859/answer/1883003440
https://math.stackexchange.com/questions/2223818/computing-a-galois-group-by-reducing-mod-p



nyy 发表于 2024-1-16 09:45:31

https://arxiv.org/pdf/2010.01281v2.pdf

CONSTRUCTIONS USING GALOIS THEORY
CLAUS FIEKER AND NICOLE SUTHERLAND
Abstract. We describe algorithms to compute fixed fields, minimal degree splitting fields and towers of radical extensions without constructing
any field containing the splitting field, instead extending Galois group
computations for this task. We also describe the computation of geometric Galois groups and their use in computing absolute factorizations.

nyy 发表于 2024-1-16 14:13:07

对称群S5的所有子群
https://zhuanlan.zhihu.com/p/668083184

Perform(AllSubgroups(SymmetricGroup(5)), function(x) Print(x, " => ", StructureDescription(x), " ", Order(x), "\n"); end);

nyy 发表于 2024-1-16 15:12:22

本帖最后由 nyy 于 2024-1-16 15:15 编辑

已知首一系数的五次方程的五个系数,能确定方程的伽罗瓦群吗? - 酱紫君的回答 - 知乎
https://www.zhihu.com/question/64900539/answer/2476505448

这个回答里
五次方程所有情况就五种

Z5 = 11111, 5
D5 = 11111, 122, 5
F20 = 11111, 14, 122, 5
A5 = 11111, 122, 23, 5
S5 = 11111, 122, 14, 23, 5

S5里还可能出现113这种情况,
比如多项式
f = x^5 + x + 10在mod 29的情况下,
出现了,
\[\left(
\begin{array}{cc}
(x+6) (x+16) \left(x^3+7 x^2+11 x+14\right) & 29 \\
\end{array}
\right)\]

真相不到他会出这种错误!


根据他的回答,只要根据23(只有A5 S5里面包含23这种情况)就能判别五次方程是否可以根式解(如果不含有23就能根式解,含有23就不能根式解),但是
f=(x^2+1)(x^3-x^2+x+1)=0
这个方程,在mod 3的情况下,就是23这种情况,但是这个方程很明显能根式解。

nyy 发表于 2024-1-16 15:28:50

本帖最后由 nyy 于 2024-1-16 15:31 编辑

nyy 发表于 2024-1-16 15:12
已知首一系数的五次方程的五个系数,能确定方程的伽罗瓦群吗? - 酱紫君的回答 - 知乎
https://www.zhihu.c ...

(*根据A5的生成元,查看所有元素*)
Clear["Global`*"];(*Clear all variables*)
gg=PermutationGroup[{Cycles[{{1,4},{2,3}}],Cycles[{{2,5,4}}]}]
hh=GroupElements
MatrixForm@Transpose@{hh}


生成所有的A5的所有元素,均没有发现14这种情况
我把A5的所有都列出来。
Cycles[{}]
Cycles[{{3,4,5}}]
Cycles[{{3,5,4}}]
Cycles[{{1,2,5}}]
Cycles[{{2,3,4}}]
Cycles[{{2,3,5}}]
Cycles[{{2,4,3}}]
Cycles[{{2,4,5}}]
Cycles[{{1,4,2}}]
Cycles[{{2,5,3}}]
Cycles[{{2,5,4}}]
Cycles[{{1,3,2}}]
Cycles[{{1,2,4}}]
Cycles[{{1,3,5}}]
Cycles[{{1,3,4}}]
Cycles[{{1,2,3}}]
Cycles[{{1,4,3}}]
Cycles[{{1,4,5}}]
Cycles[{{1,5,4}}]
Cycles[{{1,5,2}}]
Cycles[{{1,5,3}}]
Cycles[{{1,5,2,4,3}}]
Cycles[{{1,2,3,4,5}}]
Cycles[{{1,2,3,5,4}}]
Cycles[{{1,2,4,5,3}}]
Cycles[{{1,2,4,3,5}}]
Cycles[{{1,2,5,4,3}}]
Cycles[{{1,3,4,5,2}}]
Cycles[{{1,3,5,4,2}}]
Cycles[{{1,5,3,2,4}}]
Cycles[{{1,2,5,3,4}}]
Cycles[{{1,3,2,4,5}}]
Cycles[{{1,3,5,2,4}}]
Cycles[{{1,3,2,5,4}}]
Cycles[{{1,3,4,2,5}}]
Cycles[{{1,4,5,3,2}}]
Cycles[{{1,4,3,5,2}}]
Cycles[{{1,4,5,2,3}}]
Cycles[{{1,4,2,3,5}}]
Cycles[{{1,4,2,5,3}}]
Cycles[{{1,4,3,2,5}}]
Cycles[{{1,5,4,3,2}}]
Cycles[{{1,5,3,4,2}}]
Cycles[{{1,5,4,2,3}}]
Cycles[{{1,5,2,3,4}}]
Cycles[{{1,2} , {3,5}}]
Cycles[{{1,3} , {4,5}}]
Cycles[{{1,4} , {2,5}}]
Cycles[{{1,4} , {3,5}}]
Cycles[{{1,2} , {4,5}}]
Cycles[{{1,3} , {2,4}}]
Cycles[{{1,2} , {3,4}}]
Cycles[{{2,5} , {3,4}}]
Cycles[{{1,5} , {3,4}}]
Cycles[{{1,4} , {2,3}}]
Cycles[{{1,5} , {2,3}}]
Cycles[{{1,3} , {2,5}}]
Cycles[{{2,3} , {4,5}}]
Cycles[{{2,4} , {3,5}}]
Cycles[{{1,5} , {2,4}}]

从上面可以看出
11111、113、5、122这三种情况,而不是
A5 = 11111, 122, 23, 5
也就是A5包含113,不包含23这种情况!

A5生成元见https://zhuanlan.zhihu.com/p/668083184

nyy 发表于 2024-1-16 15:37:29

nyy 发表于 2024-1-16 15:12
已知首一系数的五次方程的五个系数,能确定方程的伽罗瓦群吗? - 酱紫君的回答 - 知乎
https://www.zhihu.c ...

S5=11111、1112、113、14、5、122、23
而不是他说的
S5 = 11111, 122, 14, 23, 5
他漏掉了1112、113这两种情况。
(*根据S5的生成元,查看所有元素*)
Clear["Global`*"];(*Clear all variables*)
gg=PermutationGroup[{Cycles[{{1,2,3,4,5}}],Cycles[{{1,2}}]}]
hh=GroupElements
MatrixForm@Transpose@{hh}


所有结果
Cycles[{}]
Cycles[{{4,5}}]
Cycles[{{3,4}}]
Cycles[{{3,5}}]
Cycles[{{2,3}}]
Cycles[{{2,4}}]
Cycles[{{1,4}}]
Cycles[{{2,5}}]
Cycles[{{1,2}}]
Cycles[{{1,5}}]
Cycles[{{1,3}}]
Cycles[{{3,4,5}}]
Cycles[{{3,5,4}}]
Cycles[{{1,2,5}}]
Cycles[{{2,3,4}}]
Cycles[{{1,5,2}}]
Cycles[{{1,5,4}}]
Cycles[{{2,3,5}}]
Cycles[{{2,4,3}}]
Cycles[{{1,3,2}}]
Cycles[{{2,4,5}}]
Cycles[{{1,4,5}}]
Cycles[{{1,3,4}}]
Cycles[{{2,5,3}}]
Cycles[{{2,5,4}}]
Cycles[{{1,2,4}}]
Cycles[{{1,4,3}}]
Cycles[{{1,3,5}}]
Cycles[{{1,5,3}}]
Cycles[{{1,4,2}}]
Cycles[{{1,2,3}}]
Cycles[{{1,3,2,4}}]
Cycles[{{2,3,4,5}}]
Cycles[{{2,3,5,4}}]
Cycles[{{2,4,5,3}}]
Cycles[{{1,3,2,5}}]
Cycles[{{2,4,3,5}}]
Cycles[{{2,5,4,3}}]
Cycles[{{2,5,3,4}}]
Cycles[{{1,4,3,2}}]
Cycles[{{1,4,5,3}}]
Cycles[{{1,4,3,5}}]
Cycles[{{1,4,2,3}}]
Cycles[{{1,4,2,5}}]
Cycles[{{1,5,3,2}}]
Cycles[{{1,5,4,2}}]
Cycles[{{1,2,3,4}}]
Cycles[{{1,5,4,3}}]
Cycles[{{1,5,3,4}}]
Cycles[{{1,2,3,5}}]
Cycles[{{1,2,4,3}}]
Cycles[{{1,5,2,3}}]
Cycles[{{1,2,4,5}}]
Cycles[{{1,5,2,4}}]
Cycles[{{1,2,5,3}}]
Cycles[{{1,2,5,4}}]
Cycles[{{1,3,4,2}}]
Cycles[{{1,3,5,2}}]
Cycles[{{1,3,4,5}}]
Cycles[{{1,3,5,4}}]
Cycles[{{1,4,5,2}}]
Cycles[{{1,4,5,2,3}}]
Cycles[{{1,3,2,4,5}}]
Cycles[{{1,5,2,4,3}}]
Cycles[{{1,5,3,2,4}}]
Cycles[{{1,4,5,3,2}}]
Cycles[{{1,5,2,3,4}}]
Cycles[{{1,4,3,2,5}}]
Cycles[{{1,3,2,5,4}}]
Cycles[{{1,5,4,3,2}}]
Cycles[{{1,2,3,4,5}}]
Cycles[{{1,2,3,5,4}}]
Cycles[{{1,2,4,5,3}}]
Cycles[{{1,2,5,3,4}}]
Cycles[{{1,5,3,4,2}}]
Cycles[{{1,2,5,4,3}}]
Cycles[{{1,4,2,3,5}}]
Cycles[{{1,4,3,5,2}}]
Cycles[{{1,3,5,4,2}}]
Cycles[{{1,3,5,2,4}}]
Cycles[{{1,3,4,5,2}}]
Cycles[{{1,2,4,3,5}}]
Cycles[{{1,3,4,2,5}}]
Cycles[{{1,4,2,5,3}}]
Cycles[{{1,5,4,2,3}}]
Cycles[{{2,3} , {4,5}}]
Cycles[{{2,4} , {3,5}}]
Cycles[{{2,5} , {3,4}}]
Cycles[{{1,2} , {4,5}}]
Cycles[{{1,2} , {3,4}}]
Cycles[{{1,4} , {2,3}}]
Cycles[{{1,3} , {2,5}}]
Cycles[{{1,5} , {2,4}}]
Cycles[{{1,5} , {3,4}}]
Cycles[{{1,3} , {4,5}}]
Cycles[{{1,3} , {2,4}}]
Cycles[{{1,4} , {2,5}}]
Cycles[{{1,5} , {2,3}}]
Cycles[{{1,4} , {3,5}}]
Cycles[{{1,2} , {3,5}}]
Cycles[{{1,2} , {3,4,5}}]
Cycles[{{1,2} , {3,5,4}}]
Cycles[{{1,2,3} , {4,5}}]
Cycles[{{1,2,4} , {3,5}}]
Cycles[{{1,4,2} , {3,5}}]
Cycles[{{1,3,2} , {4,5}}]
Cycles[{{1,3} , {2,5,4}}]
Cycles[{{1,4,5} , {2,3}}]
Cycles[{{1,4,3} , {2,5}}]
Cycles[{{1,4} , {2,5,3}}]
Cycles[{{1,5,4} , {2,3}}]
Cycles[{{1,3,5} , {2,4}}]
Cycles[{{1,3} , {2,4,5}}]
Cycles[{{1,5} , {2,3,4}}]
Cycles[{{1,5,3} , {2,4}}]
Cycles[{{1,5,2} , {3,4}}]
Cycles[{{1,3,4} , {2,5}}]
Cycles[{{1,5} , {2,4,3}}]
Cycles[{{1,2,5} , {3,4}}]
Cycles[{{1,4} , {2,3,5}}]


真没想到他的科普回答居然这么多错误!

nyy 发表于 2024-1-16 15:43:40

https://zhuanlan.zhihu.com/p/668083184
从这边的子群可以看出
只要包含轮换长度=3且轮换长度=5的情况,那么群的阶必然是15的倍数,那么群的阶只可能是15、30、45、60等这些情况
而S5的可解群最高也就是24阶情况,且S5没有15阶的子群,因此只要群的阶是15的倍数,那么必然是A5或者S5,这时候也就不可解了!

nyy 发表于 2024-1-16 15:51:51

nyy 发表于 2024-1-16 15:43
https://zhuanlan.zhihu.com/p/668083184
从这边的子群可以看出
只要包含轮换长度=3且轮换长度=5的情况,那 ...

S5=11111、1112、113、14、5、122、23
A5=11111、113、5、122
两者的区别是1112、14、23,都是奇数个偶数(1112中的2是偶数,14中的4是偶数,23中的2是偶数)

nyy 发表于 2024-1-16 15:59:48

本帖最后由 nyy 于 2024-1-16 16:21 编辑

nyy 发表于 2024-1-16 15:51
S5=11111、1112、113、14、5、122、23
A5=11111、113、5、122
两者的区别是1112、14、23,都是奇数个偶数 ...

RECOGNIZING GALOIS GROUPS Sn AND An

https://kconrad.math.uconn.edu/blurbs/galoistheory/galoisSnAn.pdf

https://kconrad.math.uconn.edu/blurbs/

https://math.stackexchange.com/questions/2223818/computing-a-galois-group-by-reducing-mod-p

If you now give me an equation that you have chosen at will, and you wish
to know whether or not it is solvable by radicals, I will have nothing to do
other than to indicate to you the way to respond to your question, without
wishing to charge either myself or anyone else with doing it. In a word,
the calculations are impractical. [...] But most of the time in applications
[...] one is led to equations all of whose properties one knows beforehand:
properties by means of which it will always be easy to answer the question
by the rules which we shall expound. E. Galois .

GALOIS GROUPS OVER Q AND FACTORIZATIONS MOD p
https://kconrad.math.uconn.edu/blurbs/gradnumthy/galois-Q-factor-mod-p.pdf

https://kconrad.math.uconn.edu/



页: [1] 2
查看完整版本: 五次方程、伽罗瓦群、群论等相关问题