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\begin_body

\begin_layout Title
结式
\end_layout

\begin_layout Standard
结式 (resultant) 是代数学术语, 指由两个多项式的系数所构成的一种行列式, 或称 Sylvester 行列式, 结式可判断两个多项式是否有公根、是否
互素, 以及判断多项式是否有重根.
 
\end_layout

\begin_layout Standard
设 
\begin_inset Formula 
\[
\begin{aligned}f(x) & =a_{0}x^{n}+a_{1}x^{n-1}+\ldots+a_{n},\\
g(x) & =b_{0}x^{m}+b_{1}x^{m-1}+\ldots+b_{m}.
\end{aligned}
\]

\end_inset

定义下列 
\begin_inset Formula $m+n$
\end_inset

 阶行列式 
\begin_inset Formula 
\[
\mathrm{Res}(f,g)=\begin{vmatrix}a_{0} & a_{1} & a_{2} & \cdots & \cdots & a_{n} & 0 & \cdots & 0\\
0 & a_{0} & a_{1} & \cdots & \cdots & a_{n-1} & a_{n} & \cdots & 0\\
0 & 0 & a_{0} & \cdots & \cdots & a_{n-2} & a_{n-1} & \cdots & 0\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots\\
0 & 0 & \cdots & 0 & a_{0} & \cdots & \cdots & \cdots & a_{n}\\
b_{0} & b_{1} & b_{2} & \cdots & \cdots & \cdots & b_{m} & \cdots & 0\\
0 & b_{0} & b_{1} & \cdots & \cdots & \cdots & b_{m-1} & b_{m} & \cdots\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots\\
0 & \cdots & 0 & b_{0} & b_{1} & \cdots & \cdots & \cdots & b_{m}
\end{vmatrix}
\]

\end_inset

为 
\begin_inset Formula $f(x)$
\end_inset

 与 
\begin_inset Formula $g(x)$
\end_inset

 的结式或称 Sylvester 行列式.
 
\end_layout

\begin_layout Standard
根据以上定义我们可以有下列判断两个多项式存在公共根的定理.
 
\end_layout

\begin_layout Theorem
\begin_inset CommandInset label
LatexCommand label
name "thm:sylvester"

\end_inset

 多项式 
\begin_inset Formula $f(x)$
\end_inset

 与 
\begin_inset Formula $g(x)$
\end_inset

 有公共根 (在复数域中) 的充分必要条件是它们的结式 
\begin_inset Formula $\mathrm{Res}(f,g)=0$
\end_inset

.
 
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Proof
由条件知, 多项式 
\begin_inset Formula $f(x)$
\end_inset

 与 
\begin_inset Formula $g(x)$
\end_inset

 有公共根 
\begin_inset Formula $x=t_{0}$
\end_inset

, 不妨设 
\begin_inset Formula $a_{n}$
\end_inset

 与 
\begin_inset Formula $b_{m}$
\end_inset

 不同时零, (否则结式中最后一列为零, 结论显然), 因此 
\begin_inset Formula $x_{0}\ne0$
\end_inset

.
\end_layout

\begin_layout Proof
考虑到
\begin_inset Formula 
\[
\begin{cases}
a_{0}t_{0}^{m+n-1}+a_{1}t_{0}^{m+n-2}+\ldots+a_{n}t_{0}^{m-1}=0\\
\cdots\\
a_{0}t_{0}^{n+1}+a_{1}t_{0}^{n}+\ldots+a_{n}t_{0}=0\\
a_{0}t_{0}^{n}+a_{1}t_{0}^{n-1}+\ldots+a_{n}=0
\end{cases}\quad\begin{cases}
b_{0}t_{0}^{m+n-1}+b_{1}t_{0}^{m+n-2}+\ldots+b_{m}t_{0}^{n-1}=0\\
\cdots\\
b_{0}t_{0}^{m+1}+b_{1}t_{0}^{m}+\ldots+b_{m}t_{0}=0\\
b_{0}t_{0}^{m}+b_{1}t_{0}^{m-1}+\ldots+b_{m}=0
\end{cases}
\]

\end_inset

现在令 
\begin_inset Formula $x_{r}=t_{0}^{r}$
\end_inset

, (
\begin_inset Formula $r=0,1,2,\cdots,m+n-1$
\end_inset

), 则 
\begin_inset Formula $(m+n)$
\end_inset

 元一次方程组
\begin_inset Formula 
\[
\begin{cases}
a_{0}x_{m+n-1}+a_{1}x_{m+n-2}+\ldots+a_{n}x_{m-1}=0,\\
\cdots\\
a_{0}x_{n}+a_{1}x_{n-1}+\ldots+a_{n}x_{0}=0\\
b_{0}x_{m+n-1}+b_{1}x_{m+n-2}+\ldots+b_{m}x_{n-1}=0\\
\cdots\\
b_{0}x_{m}+b_{1}x_{m-1}+\ldots+b_{m}x_{0}=0
\end{cases}
\]

\end_inset

有非零解 
\begin_inset Formula $(x_{0},x_{1},x_{2},\cdots,x_{m+n-1})=(1,t_{0},t_{0}^{2},\cdots,t_{0}^{m+n-1})$
\end_inset

, 由齐次线性方程组的克莱姆法则的逆定理, 上式对应的系数行列式为零.
 也即
\begin_inset Formula 
\[
\mathrm{Res}(f,g)=\begin{vmatrix}a_{0} & a_{1} & a_{2} & \cdots & \cdots & a_{n} & 0 & \cdots & 0\\
0 & a_{0} & a_{1} & \cdots & \cdots & a_{n-1} & a_{n} & \cdots & 0\\
0 & 0 & a_{0} & \cdots & \cdots & a_{n-2} & a_{n-1} & \cdots & 0\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots\\
0 & 0 & \cdots & 0 & a_{0} & \cdots & \cdots & \cdots & a_{n}\\
b_{0} & b_{1} & b_{2} & \cdots & \cdots & \cdots & b_{m} & \cdots & 0\\
0 & b_{0} & b_{1} & \cdots & \cdots & \cdots & b_{m-1} & b_{m} & \cdots\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots\\
0 & \cdots & 0 & b_{0} & b_{1} & \cdots & \cdots & \cdots & b_{m}
\end{vmatrix}=0.
\]

\end_inset


\end_layout

\begin_layout Corollary
对 
\begin_inset Formula $f(x),g(x)\in\mathbb{C}[X]$
\end_inset

, 方程组 
\begin_inset Formula $\begin{cases}
f(x)=0\\
g(x)=0
\end{cases}$
\end_inset

 有复数解当且仅当结式 
\begin_inset Formula $\mathrm{Res}(f,g)=0$
\end_inset

.
 
\end_layout

\begin_layout Standard
这就把一元方程组是否有解归纳为一个常数 
\begin_inset Formula $\mathrm{Res}(f,g)$
\end_inset

 是否为 
\begin_inset Formula $0$
\end_inset

.
 
\end_layout

\begin_layout Standard
现设 
\begin_inset Formula $f(x,y),g(x,y)\in F[X,Y]$
\end_inset

 为二元多项式, 现在解方程 
\begin_inset Formula 
\[
\begin{cases}
f(x,y)=0,\\
g(x,y)=0.
\end{cases}
\]

\end_inset

可以把 
\begin_inset Formula $f$
\end_inset

 和 
\begin_inset Formula $g$
\end_inset

 看作不定元 
\begin_inset Formula $x$
\end_inset

 的 (以 
\begin_inset Formula $y$
\end_inset

 的多项式为系数的) 多项式, 即 
\begin_inset Formula 
\[
\begin{cases}
f(x,y)=a_{0}(y)x^{n}+\cdots+a_{n}(y)\in F[y][x],\\
g(x,y)=b_{0}(y)x^{m}+\cdots+b_{m}(y)\in F[y][x].
\end{cases}
\]

\end_inset


\end_layout

\begin_layout Standard
于是由定理 
\begin_inset CommandInset ref
LatexCommand ref
reference "thm:sylvester"
plural "false"
caps "false"
noprefix "false"

\end_inset

 (其中 
\begin_inset Formula $a_{k}(y)$
\end_inset

 与 
\begin_inset Formula $b_{k}(y)$
\end_inset

 为变量 
\begin_inset Formula $y$
\end_inset

 的多项式), 如果上述方程有解 
\begin_inset Formula $(x,y)=(x_{0},y_{0})$
\end_inset

 使得
\begin_inset Formula 
\[
\begin{cases}
f(x_{0},y_{0})=0,\\
g(x_{0},y_{0})=0.
\end{cases}
\]

\end_inset

则必有
\begin_inset Formula 
\[
\mathrm{Res}(f,g,x)\coloneqq\begin{vmatrix}a_{0} & a_{1} & a_{2} & \cdots & \cdots & a_{n} & 0 & \cdots & 0\\
0 & a_{0} & a_{1} & \cdots & \cdots & a_{n-1} & a_{n} & \cdots & 0\\
0 & 0 & a_{0} & \cdots & \cdots & a_{n-2} & a_{n-1} & \cdots & 0\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots\\
0 & 0 & \cdots & 0 & a_{0} & \cdots & \cdots & \cdots & a_{n}\\
b_{0} & b_{1} & b_{2} & \cdots & \cdots & \cdots & b_{m} & \cdots & 0\\
0 & b_{0} & b_{1} & \cdots & \cdots & \cdots & b_{m-1} & b_{m} & \cdots\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots\\
0 & \cdots & 0 & b_{0} & b_{1} & \cdots & \cdots & \cdots & b_{m}
\end{vmatrix}(y)=0.
\]

\end_inset


\end_layout

\begin_layout Corollary
多项式 
\begin_inset Formula $f(x)$
\end_inset

 与 
\begin_inset Formula $g(x)$
\end_inset

 互素的充分必要条件是它们的结式 
\begin_inset Formula $\mathrm{Res}(f,g)\neq0$
\end_inset

.
\end_layout

\begin_layout Theorem
设 
\begin_inset Formula 
\[
\begin{aligned}f(x) & =a_{0}x^{n}+a_{1}x^{n-1}+\ldots+a_{n},\\
g(x) & =b_{0}x^{m}+b_{1}x^{m-1}+\ldots+b_{m}.
\end{aligned}
\]

\end_inset


\begin_inset Formula $f(x)$
\end_inset

 的根为 
\begin_inset Formula $x_{1},x_{2},\ldots,x_{n}$
\end_inset

, 
\begin_inset Formula $g(x)$
\end_inset

 的根为 
\begin_inset Formula $y_{1},y_{2},\ldots,y_{m}$
\end_inset

, 则 
\begin_inset Formula $f(x)$
\end_inset

 与 
\begin_inset Formula $g(x)$
\end_inset

 的结式为 
\begin_inset Formula 
\[
\mathrm{Res}(f,g)=a_{0}^{m}b_{0}^{n}\prod_{j=1}^{m}\prod_{i=1}^{n}\left(x_{i}-y_{j}\right).
\]

\end_inset


\end_layout

\begin_layout Definition
利用结式, 可定义多项式的判别式如下.
 多项式 
\begin_inset Formula 
\[
f(x)=a_{0}x^{n}+a_{1}x^{n-1}+\ldots+a_{n}
\]

\end_inset

的判别式定义为 
\begin_inset Formula 
\[
\Delta(f)=(-1)^{\frac{1}{2}n(n-1)}a_{0}^{-1}\mathrm{Res}\left(f,f^{\prime}\right).
\]

\end_inset


\end_layout

\begin_layout Theorem
多项式 
\begin_inset Formula 
\[
f(x)=a_{0}x^{n}+a_{1}x^{n-1}+\ldots+a_{n}
\]

\end_inset

的判别式等于 
\begin_inset Formula 
\[
\Delta(f)=a_{0}^{2n-2}\prod_{1\leq i<j\leq n}\left(x_{i}-x_{j}\right)^{2},
\]

\end_inset

其中 
\begin_inset Formula $x_{1},x_{2},\ldots,x_{n}$
\end_inset

 为 
\begin_inset Formula $f(x)$
\end_inset

 的根.
\end_layout

\begin_layout Corollary
多项式 
\begin_inset Formula $f(x)$
\end_inset

 有重根的充分必要条件是它的判别式 
\begin_inset Formula $\Delta(f)=0$
\end_inset

.
\end_layout

\end_body
\end_document