n(
)次一般代数方程的根式不可解性
根式扩张
定义：若)
,且
使
,即![E=F(\sqrt[n]{a}),a\in F](https://math.jianshu.com/math?formula=E%3DF(%5Csqrt%5Bn%5D%7Ba%7D)%2Ca%5Cin%20F)
,则称
为根式扩张
根式可解
定义：设F为域,![f(x)\in F[x]](https://math.jianshu.com/math?formula=f(x)%5Cin%20F%5Bx%5D)
且为首1多项式,
,E为)
在F上的分裂域,方程%3D0)
称为在域F上根式可解,指存在域的根式扩张序列
满足：
1.)
均是根式扩张,即![F_{r+i}=F_i(\sqrt[n_i]{a_i}),a_i\in F_i](https://math.jianshu.com/math?formula=F_%7Br%2Bi%7D%3DF_i(%5Csqrt%5Bn_i%5D%7Ba_i%7D)%2Ca_i%5Cin%20F_i)
2.
注：
1.从域F扩张成K是由有限次添加元的根号得到的
2.f(x)的全部根都在K中
3.f(x)=0的所有根可通过F中元的四则运算和开根号运算表达
可解群
定义：设有限群G有正规子群列
,其中
表示
是
的正规子群,若商群)
均是交换群,则称G为可解群
定理：设域F的特征为0,且为首1多项式,
,则在F上根式可解当且仅当在F上的伽罗瓦群为可解群
定理：设
,
为n个不定元,F是特征为0的域,则一般方程%3Dx%5En-t_1x%5E%7Bn-1%7D%2Bt_2x%5E%7Bn-2%7D%2B%5Ccdots%2B(-1)%5Ent_n%3D0)
在F上的伽罗瓦群为对称群
证明：
%E7%9A%84%E6%A0%B9%E4%B8%BAy_1%2Cy_2%2C%5Ccdots%2Cy_n)
%3D(x-y_1)(x-y_2)%5Ccdots(x-y_n))
%5Ent_n)
%E5%9C%A8F(t_1%2Ct_2%2C%5Ccdots%2Ct_n)%E4%B8%8A%E7%9A%84%E5%88%86%E8%A3%82%E5%9F%9F%E4%B8%BA)
%3DF(y_1%2Cy_2%2C%5Ccdots%2Cy_n))
%3D(x-x_1)(x-x_2)%5Ccdots(x-x_n))
%5En%5Crho_n)
![\therefore 存在环同态\sigma:F[t_1,t_2,\cdots,t_n]\to F[\rho_1,\rho_2,\cdots,rho_n]](https://math.jianshu.com/math?formula=%5Ctherefore%20%E5%AD%98%E5%9C%A8%E7%8E%AF%E5%90%8C%E6%80%81%5Csigma%3AF%5Bt_1%2Ct_2%2C%5Ccdots%2Ct_n%5D%5Cto%20F%5B%5Crho_1%2C%5Crho_2%2C%5Ccdots%2Crho_n%5D)
%3D%5Crho_i(1%5Cle%20i%5Cle%20n)%2C%5Csigma%7C_F%3D1)
![\therefore 有环同态\tau:F[x_1,x_2,\cdots,x_n]\to F[y_1,y_2,\cdots,y_n]](https://math.jianshu.com/math?formula=%5Ctherefore%20%E6%9C%89%E7%8E%AF%E5%90%8C%E6%80%81%5Ctau%3AF%5Bx_1%2Cx_2%2C%5Ccdots%2Cx_n%5D%5Cto%20F%5By_1%2Cy_2%2C%5Ccdots%2Cy_n%5D)
%3Dy_i(1%5Cle%20i%5Cle%20n)%2C%5Ctau%7C_F%3D1%E2%80%8B)
%3D%5Ctau(%5Crho_i)%3D%5Ctau(x_%7Bj_1%7Dx_%7Bj_2%7D%5Ccdots%20x_%7Bj_i%7D)%E2%80%8B)
%5CRightarrow%20(h)%3D0%5CRightarrow%20%5Ctau%5Csigma(h)%3D0%5CRightarrow%20h%3D0)
![\therefore \sigma是F[t_1,t_2,\cdots,t_n]与F[\rho_0,\rho_2,\cdots,\rho_n]的环同构](https://math.jianshu.com/math?formula=%5Ctherefore%20%5Csigma%E6%98%AFF%5Bt_1%2Ct_2%2C%5Ccdots%2Ct_n%5D%E4%B8%8EF%5B%5Crho_0%2C%5Crho_2%2C%5Ccdots%2C%5Crho_n%5D%E7%9A%84%E7%8E%AF%E5%90%8C%E6%9E%84)
%5Ccong%20F(%5Crho_1%2C%5Crho_2%2C%5Ccdots%2C%5Crho_n))
![\sigma:F(t_1,t_2,\cdots,t_n)[x]\cong F(\rho_1,\rho_2,\cdots,\rho_n)[x]](https://math.jianshu.com/math?formula=%5Csigma%3AF(t_1%2Ct_2%2C%5Ccdots%2Ct_n)%5Bx%5D%5Ccong%20F(%5Crho_1%2C%5Crho_2%2C%5Ccdots%2C%5Crho_n)%5Bx%5D)
%3Dx)
)%3D%5Csigma(x%5En-t_1x%5E%7Bn-1%7D%2B%5Ccdots%2B(-1)%5Ent_n))
%5En%5Crho_n%3Dg(x))
)%3DF(%5Crho_1%2C%5Crho_2%2C%5Ccdots%2C%5Crho_n))
%2FF(t_1%2Ct_2%2C%5Ccdots%2Ct_n)))
%2FF(%5Crho_1%2C%5Crho_2%2C%5Ccdots%2C%5Crho_n)))
%E7%9B%B8%E5%AF%B9F(%5Crho_1%2C%5Crho_2%2C%5Ccdots%2C%5Crho_n)%E7%9A%84%E8%87%AA%E5%90%8C%E6%9E%84)
%2FF(%5Crho_1%2C%5Crho_2%2C%5Ccdots%2C%5Crho_n))%E6%98%AF%E5%AF%B9%E7%A7%B0%E7%BE%A4S_n)
%E5%9C%A8F(t_1%2Ct_2%2C%5Ccdots%2Ct_n)%E4%B8%8A%E7%9A%84%E4%BC%BD%E7%BD%97%E7%93%A6%E7%BE%A4%E6%98%AFS_n%5Cqquad%5Cmathcal%7BQ.E.D%7D)
当
时,是不可解群
推论：当时,n次一般代数方程是根式不可解的
时,一般代数方程是根式可解的
时,
的解为
时,
中
代入
得关于y的三次方程,且无
项
故考虑方程
令![z_1=\sqrt[3]{-{27\over 2}q+\sqrt{{27\over 4}(4p^3+27q^2)}}](https://math.jianshu.com/math?formula=z_1%3D%5Csqrt%5B3%5D%7B-%7B27%5Cover%202%7Dq%2B%5Csqrt%7B%7B27%5Cover%204%7D(4p%5E3%2B27q%5E2)%7D%7D)
![z_2=\sqrt[3]{-{27\over 2}q-\sqrt{27\over 4}(4p^3+27q^2)}](https://math.jianshu.com/math?formula=z_2%3D%5Csqrt%5B3%5D%7B-%7B27%5Cover%202%7Dq-%5Csqrt%7B27%5Cover%204%7D(4p%5E3%2B27q%5E2)%7D)
使
此时方程的三个解为
)
)
)
其中
为3次单位根
时类似,考虑方程
解出三次方程%5Ctheta%2Bq%5E2%3D0)
的三个根
令
,且
此时方程的四个解为
)
)
%E2%80%8B)
%E2%80%8B)
