求多项式的Galois群

God→Osiris

$x^5+x^4-5x^3-5x^2+4x+3=0$

xiaoshuchong

Galois群为$S_5$， 因而无根式解

nyy

1. P<x>:=PolynomialAlgebra(Rationals());
   
2. f:=x^5+x^4-5*x^3-5*x^2+4*x+3;
   
3. G:=GaloisGroup (f);print G;Order(G);IsSolvable(G);
   
4. TransitiveGroupDescription(G);

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求解结果
http://magma.maths.usyd.edu.au/calc/

1. Symmetric group G acting on a set of cardinality 5
   
2. Order = 120 = 2^3 * 3 * 5
   
3. 120
   
4. false
   
5. S(5)

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群是S5，因此无根式解

1. Clear["Global`*"];(*Clear all variables*)
   
2. f=x^5+x^4-5*x^3-5*x^2+4*x+3 (*定义多项式*)
   
3. aa={Factor[f,Modulus->#],#}&/@Prime@Range[30];
   
4. MatrixForm[aa]
   

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求解结果
\[\left(
\begin{array}{cc}
x^5+x^4+x^3+x^2+1 & 2 \\
x \left(x^4+x^3+x^2+x+1\right) & 3 \\
\left(x^2+3 x+4\right) \left(x^3+3 x^2+2 x+2\right) & 5 \\
\left(x^2+4 x+5\right) \left(x^3+4 x^2+2 x+2\right) & 7 \\
\left(x^2+2 x+2\right) \left(x^3+10 x^2+6 x+7\right) & 11 \\
(x+7) \left(x^4+7 x^3+11 x^2+9 x+6\right) & 13 \\
x^5+x^4+12 x^3+12 x^2+4 x+3 & 17 \\
x^5+x^4+14 x^3+14 x^2+4 x+3 & 19 \\
(x+10) \left(x^4+14 x^3+16 x^2+19 x+21\right) & 23 \\
(x+25) \left(x^4+5 x^3+15 x^2+26 x+21\right) & 29 \\
(x+27) \left(x^2+6 x+19\right) \left(x^2+30 x+2\right) & 31 \\
(x+30) \left(x^4+8 x^3+14 x^2+19 x+26\right) & 37 \\
\left(x^2+10 x+19\right) \left(x^3+32 x^2+25 x+39\right) & 41 \\
\left(x^2+13 x+8\right) \left(x^3+31 x^2+14 x+38\right) & 43 \\
(x+8) (x+34) \left(x^3+6 x^2+35 x+42\right) & 47 \\
(x+50) \left(x^4+4 x^3+7 x^2+16 x+52\right) & 53 \\
(x+17) \left(x^2+15 x+4\right) \left(x^2+28 x+20\right) & 59 \\
x^5+x^4+56 x^3+56 x^2+4 x+3 & 61 \\
(x+30) (x+61) \left(x^3+44 x^2+57 x+48\right) & 67 \\
(x+29) (x+54) \left(x^3+60 x^2+52 x+54\right) & 71 \\
x^5+x^4+68 x^3+68 x^2+4 x+3 & 73 \\
(x+32) \left(x^4+48 x^3+39 x^2+11 x+47\right) & 79 \\
(x+56) \left(x^4+28 x^3+4 x^2+20 x+46\right) & 83 \\
(x+21) \left(x^4+69 x^3+59 x^2+2 x+51\right) & 89 \\
x^5+x^4+92 x^3+92 x^2+4 x+3 & 97 \\
(x+36) \left(x^2+12 x+95\right) \left(x^2+54 x+7\right) & 101 \\
(x+42) (x+51) (x+70) \left(x^2+44 x+33\right) & 103 \\
x^5+x^4+102 x^3+102 x^2+4 x+3 & 107 \\
(x+38) \left(x^4+72 x^3+93 x^2+58 x+89\right) & 109 \\
x^5+x^4+108 x^3+108 x^2+4 x+3 & 113 \\
\end{array}
\right)\]

从上面的计算结果知道伽罗瓦群是S5，因此无根式解！