复数域内的不可约多项式在四元数域内可约吗？

zeroieme

比如\(x^2+y^2-1\)、\(x^3+y^3-1\)

zeroieme

顶起，求@mathe @wayne 解惑

mathe

复数域内不可约多项式只有一次多项式

zeroieme

mathe 发表于 2019-9-5 06:08
复数域内不可约多项式只有一次多项式

复数域内，一元高次多项式是肯定可约的。在多元情形不一定。

https://en.wikipedia.org/wiki/Irreducible_polynomial

A polynomial that is irreducible over any field containing the coefficients is absolutely irreducible. By the fundamental theorem of algebra, a univariate polynomial is absolutely irreducible if and only if its degree is one. On the other hand, with several indeterminates, there are absolutely irreducible polynomials of any degree, such as \({\displaystyle x^{2}+y^{n}-1,} {\displaystyle x^{2}+y^{n}-1,}\) for any positive integer n.

zeroieme

建立参数方程
\(\left(\left(a_r+a_i I+a_j J+a_k K\right) \left(x_r+x_i I+x_j J+x_k K\right)+\left(b_r+b_i I+b_j J+b_k K\right) \left(y_r+y_i I+y_j J+y_k K\right)+\left(c_r+c_i I+c_j J+c_k K\right)\right) \left(\left(d_r+d_i I+d_j J+d_k K\right) \left(x_r+x_i I+x_j J+x_k K\right)+\left(e_r+e_i I+e_j J+e_k K\right) \left(y_r+y_i I+y_j J+y_k K\right)+\left(f_r+f_i I+f_j J+f_k K\right)\right)=\left(x_r+x_i I+x_j J+x_k K\right){}^2+\left(y_r+y_i I+y_j J+y_k K\right){}^2-1\)

zeroieme

方程组先线性简化有
\(\left\{a_i d_i=0,a_j d_i=0,a_k d_i=0,a_r d_i=0,b_i d_i=0,b_j d_i=0,b_k d_i=0,b_r d_i=0,a_i d_j=0,a_j d_j=0,a_k d_j=0,a_r d_j=0,b_i d_j=0,b_j d_j=0,b_k d_j=0,b_r d_j=0,a_i d_k=0,a_j d_k=0,a_k d_k=0,a_r d_k=0,b_i d_k=0,b_j d_k=0,b_k d_k=0,b_r d_k=0,a_i d_r=0,a_j d_r=0,a_k d_r=0,a_r d_r-1=0,b_i d_r=0,b_j d_r=0,b_k d_r=0,b_r d_r=0,a_i e_i=0,a_j e_i=0,a_k e_i=0,a_r e_i=0,b_i e_i=0,b_j e_i=0,b_k e_i=0,b_r e_i=0,a_i e_j=0,a_j e_j=0,a_k e_j=0,a_r e_j=0,b_i e_j=0,b_j e_j=0,b_k e_j=0,b_r e_j=0,a_i e_k=0,a_j e_k=0,a_k e_k=0,a_r e_k=0,b_i e_k=0,b_j e_k=0,b_k e_k=0,b_r e_k=0,a_i e_r=0,a_j e_r=0,a_k e_r=0,a_r e_r=0,b_i e_r=0,b_j e_r=0,b_k e_r=0,b_r e_r-1=0,a_i f_i=0,a_j f_i=0,a_k f_i=0,a_r f_i=0,b_i f_i=0,b_j f_i=0,b_k f_i=0,b_r f_i=0,a_i f_j=0,a_j f_j=0,a_k f_j=0,a_r f_j=0,b_i f_j=0,b_j f_j=0,b_k f_j=0,b_r f_j=0,a_i f_k=0,a_j f_k=0,a_k f_k=0,a_r f_k=0,b_i f_k=0,b_j f_k=0,b_k f_k=0,b_r f_k=0,c_r f_i-c_k f_j+c_j f_k+c_i f_r=0,c_k f_i+c_r f_j-c_i f_k+c_j f_r=0,c_j f_i-c_i f_j-c_r f_k-c_k f_r=0,c_i f_i+c_j f_j+c_k f_k-c_r f_r-1=0,c_r d_i-c_k d_j+c_j d_k+c_i d_r+a_i f_r=0,c_k d_i+c_r d_j-c_i d_k+c_j d_r+a_j f_r=0,c_j d_i-c_i d_j-c_r d_k-c_k d_r-a_k f_r=0,c_i d_i+c_j d_j+c_k d_k-c_r d_r-a_r f_r=0,c_r e_i-c_k e_j+c_j e_k+c_i e_r+b_i f_r=0,c_k e_i+c_r e_j-c_i e_k+c_j e_r+b_j f_r=0,c_j e_i-c_i e_j-c_r e_k-c_k e_r-b_k f_r=0,c_i e_i+c_j e_j+c_k e_k-c_r e_r-b_r f_r=0\right\}\)

zeroieme

无解

zeroieme

能否证明对于任何正整数n都不存在以下关系：
(A X+B Y+C)(D X+E Y+F)=X^2+Y^2-I
A,B,C,D,E,F,X,Y均为n阶方阵，I为n阶单位矩阵。