y^5+10y^2-15y+6=0有根式解吗？

God→Osiris

如果有是什么样的？

倪举鹏

4个虚根   一个实根   

God→Osiris

顶

sheng_jianguo

MATLAB计算结果：1实根，4虚根，无根式解：

                                                                                                        -2.5598979159298104066483939051127
0.2068011716547840110098820552133002482165421387114054346771252543*i + 0.7002751914814584568917359046729652481650661880210257645222539392
                                   0.579673766483446746432461047883405954526215331785999670545828127 - 2.0149885768278193263556272458593*i
   2.014988576827819326355627245859351123477912268643719173627322247*i + 0.579673766483446746432461047883403478395856251436273001353565174
                                  0.7002751914814584568917359046729652481650661880210257645222539392 - 0.2068011716547840110098820552133*i

mathe

magma计算Galios群120阶，所以无根式解

mathe

magma.maths.usyd.edu.au/calc/

mathe

P<t>:=PolynomialAlgebra(Rationals()); f:=1*t^5+10*t^2-15*t+6; G:=GaloisGroup(f); print G;

mathematica

mathe 发表于 2014-12-3 11:11
P:=PolynomialAlgebra(Rationals()); f:=1*t^5+10*t^2-15*t+6; G:=GaloisGroup(f); print G;

我有一个疑问：如果一个二次多项式与这个方程相乘，那最后结果是什么样？局部可解？

mathematica

mathe 发表于 2014-12-3 11:11
P:=PolynomialAlgebra(Rationals()); f:=1*t^5+10*t^2-15*t+6; G:=GaloisGroup(f); print G;

1. P<t>:=PolynomialAlgebra(Rationals());
   
2. f:=32*t^5+3349456*t^4-5941616812296*t^3-585145514845851080*t^2+147013447513276833423286*t+15377302441624829616294559439;
   
3. f:=1*t^5+10*t^2-15*t+6;
   
4. f:=f*(t^2+3*t+2);
   
5. G:=GaloisGroup(f);
   
6. print G;
   

复制代码

输出结果:
Permutation group G acting on a set of cardinality 7
Order = 120 = 2^3 * 3 * 5
    (1, 2, 3, 4, 5)
    (1, 2)

看这个结果，分成五次方程与二次的，五次不可解，二次可解

nyy

mathematica 发表于 2021-5-14 11:15
输出结果:
Permutation group G acting on a set of cardinality 7
Order = 120 = 2^3 * 3 * 5

1. Clear["Global`*"];(*Clear all variables*)
   
2. f=y^5+10*y^2-15*y+6(*定义多项式*)
   
3. aa={Factor[f,Modulus->#],#}&/@Prime@Range[25];
   
4. MatrixForm[aa]
   

复制代码

求解结果：
\[\left(
\begin{array}{cc}
y (y+1)^4 & 2 \\
y^2 (y+1)^3 & 3 \\
(y+1)^5 & 5 \\
(y+4) \left(y^4+3 y^3+2 y^2+2 y+5\right) & 7 \\
(y+2) (y+5) \left(y^3+4 y^2+6 y+5\right) & 11 \\
\left(y^2+3 y+7\right) \left(y^3+10 y^2+2 y+12\right) & 13 \\
(y+3) \left(y^4+14 y^3+9 y^2+2\right) & 17 \\
(y+4) (y+14) \left(y^3+y^2+2 y+13\right) & 19 \\
(y+9) \left(y^4+14 y^3+12 y^2+17 y+16\right) & 23 \\
(y+13) (y+24) \left(y^3+21 y^2+13 y+24\right) & 29 \\
(y+23) \left(y^2+13 y+28\right) \left(y^2+26 y+8\right) & 31 \\
(y+7) \left(y^4+30 y^3+12 y^2+22\right) & 37 \\
y^5+10 y^2+26 y+6 & 41 \\
\left(y^2+41 y+37\right) \left(y^3+2 y^2+10 y+42\right) & 43 \\
\left(y^2+35 y+5\right) \left(y^3+12 y^2+45 y+20\right) & 47 \\
(y+23) \left(y^4+30 y^3+52 y^2+33 y+21\right) & 53 \\
y^5+10 y^2+44 y+6 & 59 \\
(y+6) (y+55) \left(y^3+36 y+10\right) & 61 \\
(y+39) \left(y^4+28 y^3+47 y^2+53 y+62\right) & 67 \\
y^5+10 y^2+56 y+6 & 71 \\
(y+7) (y+37) (y+61) \left(y^2+41 y+52\right) & 73 \\
y^5+10 y^2+64 y+6 & 79 \\
\left(y^2+22 y+6\right) \left(y^3+61 y^2+63 y+1\right) & 83 \\
(y+47) \left(y^2+8 y+70\right) \left(y^2+34 y+87\right) & 89 \\
\left(y^2+y+36\right) \left(y^3+96 y^2+62 y+81\right) & 97 \\
\end{array}
\right)\]

分析求解结果
只看求解结果中的\(\left(
\begin{array}{cc}
\left(y^2+3 y+7\right) \left(y^3+10 y^2+2 y+12\right) & 13 \\
\end{array}
\right)\)就知道这个群是A5或者S5，因此无根式解。

再看\(\left(
\begin{array}{cc}
y^5+10 y^2+26 y+6 & 41 \\
\end{array}
\right)\)和\(\left(
\begin{array}{cc}
(y+7) (y+37) (y+61) \left(y^2+41 y+52\right) & 73 \\
\end{array}
\right)\)，从这两个知道，这两个是S5的生成元，因此方程无根式解！