【定理】

设

$$A(x)=\sum _{i=0}^m{a}_i{x}^{i}B(x)=\sum _{i=0}^n{b}_i{x}^i\in \mathcal{R}[x]$$

这里$m=\mathrm{deg}(A,x)、n=\mathrm{deg}(B,x)$，而$\mathcal{R}$为一个代数闭域。设${\alpha }_i(i=1,\dots ,m)$和${\beta }_j(j=1,\dots ,n)$分别是$A$和$B$的代数闭域上的零点，那么有恒等式：

$$res(A,B,x)=a_m^n\prod _{i=1}^{m}B\left({\alpha }_i\right)=(-1{)}^{mn}{b}_n^m\prod _{j=1}^{n}A\left({\beta }_i\right)=a_m^n{b}_n^m\prod _{i=1}^m\prod _{j=1}^n\left({\alpha }_i-{\beta }_j\right)$$

【证明思路】

利用数学归纳法来证明，

$$res(A,B,x)=a_m^n\prod _{i=1}^{m}B\left({\alpha }_i\right)$$

以$m=6、n=5$为例子证明。

1.当$k=m=1$时，有

$$res(A,B,x)=\left|\begin{array}{cccccc}{a}_1& a_0& 0& 0& 0& 0\\ 0& a_1& a_0& 0& 0& 0\\ 0& 0& a_1& a_0& 0& 0\\ 0& 0& 0& a_1& a_0& 0\\ 0& 0& 0& 0& a_1& a_0\\ b_5& b_4& b_3& b_2& b_1& b_0\end{array}\right|=\left|\begin{array}{cccccc}{a}_1& a_0& 0& 0& 0& x^{4}A(x)\\ 0& a_1& a_0& 0& 0& x^{3}A(x)\\ 0& 0& a_1& a_0& 0& x^{2}A(x)\\ 0& 0& 0& a_1& a_0& x^{1}A(x)\\ 0& 0& 0& 0& a_1& A(x)\\ b_5& b_4& b_3& b_2& b_1& B(x)\end{array}\right|$$

可以让$x={\alpha }_1$，那么

$$res(A,B,x)=\left|\begin{array}{cccccc}{a}_1& a_0& 0& 0& 0& 0\\ 0& a_1& a_0& 0& 0& 0\\ 0& 0& a_1& a_0& 0& 0\\ 0& 0& 0& a_1& a_0& 0\\ 0& 0& 0& 0& a_1& 0\\ b_5& b_4& b_3& b_2& b_1& B\left({\alpha }_1\right)\end{array}\right|=B\left({\alpha }_1\right)\left|\begin{array}{ccccc}{a}_1& a_0& 0& 0& 0\\ 0& a_1& a_0& 0& 0\\ 0& 0& a_1& a_0& 0\\ 0& 0& 0& a_1& a_0\\ 0& 0& 0& 0& a_1\end{array}\right|=a_1^{5}B\left({\alpha }_1\right)$$

2.当$k=m-1$时成立，即

$$res\left(A_{(5)},B,x\right)=a_5^5\prod _{i=1}^{5}B\left({\alpha }_i\right)$$

下面证明令$k=m$时成立

$$res(A,B,x)=\left|\begin{array}{ccccccccccc}{a}_6& a_5& a_4& a_3& a_2& a_1& a_0& 0& 0& 0& 0\\ 0& a_6& a_5& a_4& a_3& a_2& a_1& a_0& 0& 0& 0\\ 0& 0& a_6& a_5& a_4& a_3& a_2& a_1& a_0& 0& 0\\ 0& 0& 0& a_6& a_5& a_4& a_3& a_2& a_1& a_0& 0\\ 0& 0& 0& 0& a_6& a_5& a_4& a_3& a_2& a_1& a_0\\ b_5& b_4& b_3& b_2& b_1& b_0& 0& 0& 0& 0& 0\\ 0& b_5& b_4& b_3& b_2& b_1& b_0& 0& 0& 0& 0\\ 0& 0& b_5& b_4& b_3& b_2& b_1& b_0& 0& 0& 0\\ 0& 0& 0& b_5& b_4& b_3& b_2& b_1& b_0& 0& 0\\ 0& 0& 0& 0& b_5& b_4& b_3& b_2& b_1& b_0& 0\\ 0& 0& 0& 0& 0& b_5& b_4& b_3& b_2& b_1& b_0\end{array}\right|$$

需要注意，$\mathbf{A}\left(\mathbf{x}\right)$的${\mathbf{a}}_6$和$\mathbf{B}\left(\mathbf{x}\right)$的${\mathbf{b}}_0$正好都处于行列式对角线上。

将第1列乘以${\alpha }_1$加到第2行，然后将第2列乘以${\alpha }_1$加到第3行、第3列乘以${\alpha }_1$加到第4行，重复进行下去，将倒数第二列乘以${\alpha }_1$加到最后一列。

设

$$a_6^{{}^{\prime }}=a_6$$
$$a_5^{{}^{\prime }}=a_5+{\alpha }_1{a}_6^{{}^{\prime }}$$
$$a_4^{{}^{\prime }}=a_4+{\alpha }_1{a}_5^{{}^{\prime }}$$
$$a_3^{{}^{\prime }}=a_3+{\alpha }_1{a}_4^{{}^{\prime }}$$
$$a_2^{{}^{\prime }}=a_2+{\alpha }_1{a}_3^{{}^{\prime }}$$
$$a_1^{{}^{\prime }}=a_1+{\alpha }_1{a}_2^{{}^{\prime }}$$
$$a_0^{{}^{\prime }}=a_0+{\alpha }_1{a}_1^{{}^{\prime }}\equiv 0$$​

$$b_5^{{}^{\prime }}=b_5^{{}^{\prime }}$$
$$b_4^{{}^{\prime }}=b_4+{\alpha }_1{b}_5^{{}^{\prime }}$$
$$b_3^{{}^{\prime }}=b_3+{\alpha }_1{b}_4^{{}^{\prime }}$$
$$b_2^{{}^{\prime }}=b_2+{\alpha }_1{b}_3^{{}^{\prime }}$$
$$b_1^{{}^{\prime }}=b_1+{\alpha }_1{b}_2^{{}^{\prime }}$$
$$b_0^{{}^{\prime }}=b_0+{\alpha }_1{b}_1^{{}^{\prime }}$$
可得

$$res(A,B,x)=\left|\begin{array}{ccccccccccc}{a}_6^{{}^{\prime }}& a_5^{{}^{\prime }}& a_4^{{}^{\prime }}& a_3^{{}^{\prime }}& a_2^{{}^{\prime }}& a_1^{{}^{\prime }}& a_0^{{}^{\prime }}& {\alpha }_1{a}_0^{{}^{\prime }}& {\alpha }_1^2{a}_0^{{}^{\prime }}& {\alpha }_1^3{a}_0^{{}^{\prime }}& {\alpha }_1^4{a}_0^{{}^{\prime }}\\ 0& a_6^{{}^{\prime }}& a_5^{{}^{\prime }}& a_4^{{}^{\prime }}& a_3^{{}^{\prime }}& a_2^{{}^{\prime }}& a_1^{{}^{\prime }}& a_0^{{}^{\prime }}& {\alpha }_1{a}_0^{{}^{\prime }}& {\alpha }_1^2{a}_0^{{}^{\prime }}& {\alpha }_1^3{a}_0^{{}^{\prime }}\\ 0& 0& a_6^{{}^{\prime }}& a_5^{{}^{\prime }}& a_4^{{}^{\prime }}& a_3^{{}^{\prime }}& a_2^{{}^{\prime }}& a_1^{{}^{\prime }}& a_0^{{}^{\prime }}& {\alpha }_1{a}_0^{{}^{\prime }}& {\alpha }_1^2{a}_0^{{}^{\prime }}\\ 0& 0& 0& a_6^{{}^{\prime }}& a_5^{{}^{\prime }}& a_4^{{}^{\prime }}& a_3^{{}^{\prime }}& a_2^{{}^{\prime }}& a_1^{{}^{\prime }}& a_0^{{}^{\prime }}& {\alpha }_1{a}_0^{{}^{\prime }}\\ 0& 0& 0& 0& a_6^{{}^{\prime }}& a_5^{{}^{\prime }}& a_4^{{}^{\prime }}& a_3^{{}^{\prime }}& a_2^{{}^{\prime }}& a_1^{{}^{\prime }}& a_0^{{}^{\prime }}\\ b_5^{{}^{\prime }}& b_4^{{}^{\prime }}& b_3^{{}^{\prime }}& b_2^{{}^{\prime }}& b_1^{{}^{\prime }}& b_0^{{}^{\prime }}& {\alpha }_1{b}_0^{{}^{\prime }}& {\alpha }_1^2{b}_0^{{}^{\prime }}& {\alpha }_1^3{b}_0^{{}^{\prime }}& {\alpha }_1^4{b}_0^{{}^{\prime }}& {\alpha }_1^5{b}_0^{{}^{\prime }}\\ 0& b_5^{{}^{\prime }}& b_4^{{}^{\prime }}& b_3^{{}^{\prime }}& b_2^{{}^{\prime }}& b_1^{{}^{\prime }}& b_0^{{}^{\prime }}& {\alpha }_1{b}_0^{{}^{\prime }}& {\alpha }_1^2{b}_0^{{}^{\prime }}& {\alpha }_1^3{b}_0^{{}^{\prime }}& {\alpha }_1^4{b}_0^{{}^{\prime }}\\ 0& 0& b_5^{{}^{\prime }}& b_4^{{}^{\prime }}& b_3^{{}^{\prime }}& b_2^{{}^{\prime }}& b_1^{{}^{\prime }}& b_0^{{}^{\prime }}& {\alpha }_1{b}_0^{{}^{\prime }}& {\alpha }_1^2{b}_0^{{}^{\prime }}& {\alpha }_1^3{b}_0^{{}^{\prime }}\\ 0& 0& 0& b_5^{{}^{\prime }}& b_4^{{}^{\prime }}& b_3^{{}^{\prime }}& b_2^{{}^{\prime }}& b_1^{{}^{\prime }}& b_0^{{}^{\prime }}& {\alpha }_1{b}_0^{{}^{\prime }}& {\alpha }_1^2{b}_0^{{}^{\prime }}\\ 0& 0& 0& 0& b_5^{{}^{\prime }}& b_4^{{}^{\prime }}& b_3^{{}^{\prime }}& b_2^{{}^{\prime }}& b_1^{{}^{\prime }}& b_0^{{}^{\prime }}& {\alpha }_1{b}_0^{{}^{\prime }}\\ 0& 0& 0& 0& 0& b_5^{{}^{\prime }}& b_4^{{}^{\prime }}& b_3^{{}^{\prime }}& b_2^{{}^{\prime }}& b_1^{{}^{\prime }}& b_0^{{}^{\prime }}\end{array}\right|$$

从$b_i^{{}^{\prime }}$所在几行（第6行）开始，第6行减去"第7行乘以${\alpha }_1$“，第7行减去"第8行乘以${\alpha }_1$”，重复进行直至最后第二行减去"最后一行乘以${\alpha }_1$"，可得

$$res(A,B,x)=\left|\begin{array}{ccccccccccc}{a}_6^{{}^{\prime }}& a_5^{{}^{\prime }}& a_4^{{}^{\prime }}& a_3^{{}^{\prime }}& a_2^{{}^{\prime }}& a_1^{{}^{\prime }}& a_0^{{}^{\prime }}& {\alpha }_1{a}_0^{{}^{\prime }}& {\alpha }_1^2{a}_0^{{}^{\prime }}& {\alpha }_1^3{a}_0^{{}^{\prime }}& {\alpha }_1^4{a}_0^{{}^{\prime }}\\ 0& a_6^{{}^{\prime }}& a_5^{{}^{\prime }}& a_4^{{}^{\prime }}& a_3^{{}^{\prime }}& a_2^{{}^{\prime }}& a_1^{{}^{\prime }}& a_0^{{}^{\prime }}& {\alpha }_1{a}_0^{{}^{\prime }}& {\alpha }_1^2{a}_0^{{}^{\prime }}& {\alpha }_1^3{a}_0^{{}^{\prime }}\\ 0& 0& a_6^{{}^{\prime }}& a_5^{{}^{\prime }}& a_4^{{}^{\prime }}& a_3^{{}^{\prime }}& a_2^{{}^{\prime }}& a_1^{{}^{\prime }}& a_0^{{}^{\prime }}& {\alpha }_1{a}_0^{{}^{\prime }}& {\alpha }_1^2{a}_0^{{}^{\prime }}\\ 0& 0& 0& a_6^{{}^{\prime }}& a_5^{{}^{\prime }}& a_4^{{}^{\prime }}& a_3^{{}^{\prime }}& a_2^{{}^{\prime }}& a_1^{{}^{\prime }}& a_0^{{}^{\prime }}& {\alpha }_1{a}_0^{{}^{\prime }}\\ 0& 0& 0& 0& a_6^{{}^{\prime }}& a_5^{{}^{\prime }}& a_4^{{}^{\prime }}& a_3^{{}^{\prime }}& a_2^{{}^{\prime }}& a_1^{{}^{\prime }}& a_0^{{}^{\prime }}\\ b_5& b_4& b_3& b_2& b_1& b_0& 0& 0& 0& 0& 0\\ 0& b_5& b_4& b_3& b_2& b_1& b_0& 0& 0& 0& 0\\ 0& 0& b_5& b_4& b_3& b_2& b_1& b_0& 0& 0& 0\\ 0& 0& 0& b_5& b_4& b_3& b_2& b_1& b_0& 0& 0\\ 0& 0& 0& 0& b_5& b_4& b_3& b_2& b_1& b_0& 0\\ 0& 0& 0& 0& 0& b_5& b_4& b_3& b_2& b_1& b_0^{{}^{\prime }}\end{array}\right|$$

由于$a_0^{{}^{\prime }}=a_0+{\alpha }_1{a}_1^{{}^{\prime }}=A\left({\alpha }_1\right)=0、b_0^{{}^{\prime }}=B\left({\alpha }_1\right)$，所以

$$res(A,B,x)=\left|\begin{array}{ccccccccccc}{a}_6^{{}^{\prime }}& a_5^{{}^{\prime }}& a_4^{{}^{\prime }}& a_3^{{}^{\prime }}& a_2^{{}^{\prime }}& a_1^{{}^{\prime }}& 0& 0& 0& 0& 0\\ 0& a_6^{{}^{\prime }}& a_5^{{}^{\prime }}& a_4^{{}^{\prime }}& a_3^{{}^{\prime }}& a_2^{{}^{\prime }}& a_1^{{}^{\prime }}& 0& 0& 0& 0\\ 0& 0& a_6^{{}^{\prime }}& a_5^{{}^{\prime }}& a_4^{{}^{\prime }}& a_3^{{}^{\prime }}& a_2^{{}^{\prime }}& a_1^{{}^{\prime }}& 0& 0& 0\\ 0& 0& 0& a_6^{{}^{\prime }}& a_5^{{}^{\prime }}& a_4^{{}^{\prime }}& a_3^{{}^{\prime }}& a_2^{{}^{\prime }}& a_1^{{}^{\prime }}& 0& 0\\ 0& 0& 0& 0& a_6^{{}^{\prime }}& a_5^{{}^{\prime }}& a_4^{{}^{\prime }}& a_3^{{}^{\prime }}& a_2^{{}^{\prime }}& a_1^{{}^{\prime }}& 0\\ b_5& b_4& b_3& b_2& b_1& b_0& 0& 0& 0& 0& 0\\ 0& b_5& b_4& b_3& b_2& b_1& b_0& 0& 0& 0& 0\\ 0& 0& b_5& b_4& b_3& b_2& b_1& b_0& 0& 0& 0\\ 0& 0& 0& b_5& b_4& b_3& b_2& b_1& b_0& 0& 0\\ 0& 0& 0& 0& b_5& b_4& b_3& b_2& b_1& b_0& 0\\ 0& 0& 0& 0& 0& b_5& b_4& b_3& b_2& b_1& B\left({\alpha }_1\right)\end{array}\right|$$

设

$$A^{{}^{\prime }}(x)=a_6\left(x-{\alpha }_2\right)\left(x-{\alpha }_3\right)\left(x-{\alpha }_4\right)\left(x-{\alpha }_5\right)\left(x-{\alpha }_6\right)$$

那么有

$$A^{{}^{\prime }}(x)=a_6^{{}^{\prime }}{x}^5+a_5^{{}^{\prime }}{x}^5+a_4^{{}^{\prime }}{x}^5+a_3^{{}^{\prime }}{x}^5+a_2^{{}^{\prime }}{x}^5+a_1^{{}^{\prime }}$$

所以

$$res(A,B,x)=B\left({\alpha }_1\right)res\left(A^{{}^{\prime }},B,x\right)=B\left({\alpha }_1\right)a_5^5\prod _{i=1}^{5}B\left({\alpha }_i^{{}^{\prime }}\right)=a_5^5\prod _{i=1}^{6}B\left({\alpha }_i\right)$$

也就是$k=m$时成立。

3.同理可证

$$res(A,B,x)=(-1{)}^{mn}{b}_n^m\prod _{j=1}^{n}A\left({\beta }_i\right)$$

注意需要先将$B(x)$的系数移动到行列式最前面几行。

4.因为

$$B(x)=b_n^m\prod _{i=1}^m\left(x-{\beta }_j\right)$$

代入

$$res(A,B,x)=a_m^n\prod _{i=1}^{m}B\left({\alpha }_i\right)$$

可得

$$res(A,B,x)=a_m^n{b}_n^m\prod _{i=1}^m\prod _{j=1}^n\left({\alpha }_i-{\beta }_j\right)$$