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[求助] 谁能帮下载伽罗瓦理论天才的激情 [章璞 著] 2013年版.pdf

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发表于 2021-5-13 08:53:15 | 显示全部楼层 |阅读模式

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https://download.csdn.net/download/yulao2001/14992626
伽罗瓦理论天才的激情 [章璞 著] 2013年版.pdf
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
发表于 2021-5-13 08:59:46 | 显示全部楼层
已经上传到百度网盘. 链接: https://pan.baidu.com/s/1KfstnIDkkj7JNZ7mz8fL6g 提取码: yaws

点评

能看懂如何判定是否有根式解吗?  发表于 2021-5-14 09:49
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
 楼主| 发表于 2021-5-13 09:36:00 | 显示全部楼层
liangbch 发表于 2021-5-13 08:59
已经上传到百度网盘. 链接: https://pan.baidu.com/s/1KfstnIDkkj7JNZ7mz8fL6g 提取码: yaws

谢谢了,我已经把这个电子书传到论坛QQ群了
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
 楼主| 发表于 2021-5-13 09:47:31 | 显示全部楼层
https://bbs.emath.ac.cn/thread-3018-1-1.html
这个五次方程怎么解?

求一个可解的五次方程
https://bbs.emath.ac.cn/thread-3024-1-1.html

y^5+10y^2-15y+6=0有根式解吗?
https://bbs.emath.ac.cn/thread-5946-1-1.html

看完这本书,估计这两个问题就能彻底解决了
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
 楼主| 发表于 2021-5-13 11:12:30 | 显示全部楼层
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)? - 知乎
https://www.zhihu.com/question/458961859/answer/1883098711

这个回答我觉得有意思
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
 楼主| 发表于 2021-5-14 10:23:46 | 显示全部楼层
https://kconrad.math.uconn.edu/blurbs/galoistheory/galoisSnAn.pdf
RECOGNIZING GALOIS GROUPS Sn AND An
没怎么看懂
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
 楼主| 发表于 2021-5-14 14:19:55 | 显示全部楼层
本帖最后由 mathematica 于 2021-5-14 14:21 编辑
  1. //变量是t
  2. P<t>:=PolynomialAlgebra(Rationals());
  3. //注意多项式变量是t,下面方程不可以求根式解
  4. f:=1*t^5+10*t^2-15*t+6;
  5. f:=f*(t^2+3*t+2);
  6. //注意多项式变量是t,下面方程可以求根式解
  7. f:=32*t^5+3349456*t^4-5941616812296*t^3-585145514845851080*t^2+147013447513276833423286*t+15377302441624829616294559439;
  8. //注意多项式变量是t,下面方程可以求根式解
  9. f:=t^8-t^7+29*t^2+29;
  10. G:=GaloisGroup(f);
  11. print G;
  12. //看群是否可解
  13. IsSolvable(G);
复制代码


网站:
http://magma.maths.usyd.edu.au/calc/

运行结果:
  1. Permutation group G acting on a set of cardinality 8
  2. Order = 56 = 2^3 * 7
  3.     (1, 4, 6, 2, 5, 3, 8)
  4.     (1, 4)(2, 8)(3, 7)(5, 6)
  5.     (1, 3)(2, 5)(4, 7)(6, 8)
  6.     (1, 5)(2, 3)(4, 6)(7, 8)
  7. true
复制代码


参考资料
Galois group command for Magma online calculator?
https://math.stackexchange.com/q ... ulator?noredirect=1

On the solvable octic x8−x7+29x2+29=0
https://mathoverflow.net/questio ... octic-x8-x729x229-0



毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
 楼主| 发表于 2021-5-16 12:34:52 | 显示全部楼层
Galois Groups
Finding Galois groups (of normal closures) of algebraic fields is a hard problem, in general. All practical currently-used algorithms fall into two groups: The absolute resolvent method [SM85] and the method of Stauduhar [Sta73].

The Magma implementation is based on an extension of the method of Stauduhar by [GK00], [Gei03] and recent work by Klüners and Fieker [FK14], Elsenhans [Els12], [Els14], [Els16] and Sutherland [Sut15].

For polynomials over Z, Q, number fields and global function fields and irreducible polynomials over function fields over Q, Magma is able to compute the Galois group without any a-priori restrictions on the degree. Note, however, that the running time and memory constraints can make computations in degree >50 impossible, although computations in degree >200 have been successful as well. In contrast to the absolute resolvent method, it also provides the explicit action on the roots of the polynomial f which generates the algebraic field. On demand, the older version which is restricted to a maximum degree of 23, is still available.

Roughly speaking, the method of Stauduhar traverses the subgroup lattice of transitive permutation groups of degree n from the symmetric group to the actual Galois group. This is done by using so-called relative resolvents. Resolvents are polynomials whose splitting fields are subfields of the splitting field of the given polynomial which are computed using approximations of the roots of the polynomial f.

If the field (or the field defined by a polynomial) has subfields (i.e. the Galois group is imprimitive) the current implementation changes the starting point of the algorithm in the subgroup lattice, to get as close as possible to the actual Galois group. This is done via computation of subfields of a stem field of f, that is the field extension of Q which we get by adjoining a root of f to Q. Using this knowledge of the subfields, the Galois group is found as a subgroup of the intersection of suitable wreath products which may be easily computed. This intersection is a good starting point for the algorithm.

If the field (or the field defined by a polynomial) does not have subfields (i.e. the Galois group is primitive) we use a combination of the method of Stauduhar and the absolute resolvent method. The Frobenius automorphism of the underlying p-adic field or the complex conjugation, when using complex approximations of the roots of the polynomial f, already determines a subgroup of the Galois group, which is used to speed up computations in the primitive case.

http://magma.maths.usyd.edu.au/magma/handbook/text/419
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
 楼主| 发表于 2021-5-16 14:15:47 | 显示全部楼层
毋因群疑而阻独见  毋任己意而废人言
毋私小惠而伤大体  毋借公论以快私情
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