Habicht定理中有关子结式命题3.4.6的证明

在这里插入图片描述

个人认为红色区域有问题，因为 $\deg{\left( \phi\left( S_{j} \right) \right) = r}$ ，当 $\geq r$ 时， $subres_{i}\left( \phi(S_{j + 1}),\phi\left( S_{j} \right) \right)$ 的定义不存在！！！

下面是我的证明过程：

【证明】

(a)

因为 ${\phi\left( R_{j + 1}^{2(j - i)}S_{i} \right) }{= \phi\left( subres_{i}\left( S_{j + 1},S_{j} \right) \right) }{= detpol\left( x^{j - i - 1}\phi\left( S_{j + 1} \right),\ldots,\phi\left( S_{j + 1} \right),x^{j - i}\phi\left( S_{j} \right),\ldots,\phi\left( S_{j} \right) \right)}$

而当 $\leq i \leq j - 1$ 时，有

$\deg\left( \phi\left( S_{j + 1} \right) \right) = j + 1 > \deg\left( x^{j - i}\phi\left( S_{j} \right) \right) + 1 = j - (r + 1) + r + 1 = j$

此时有 $detpol\left( x^{j - i - 1}\phi\left( S_{j + 1} \right),\ldots,\phi\left( S_{j + 1} \right),x^{j - i}\phi\left( S_{j} \right),\ldots,\phi\left( S_{j} \right) \right) = 0$ ，也就是 $\phi\left( R_{j + 1}^{2(j - i)}S_{i} \right) = 0$ ，即 $\phi\left( S_{j - 1} \right) = \phi\left( S_{j - 2} \right) = \ldots = \phi\left( S_{r + 1} \right) = 0$ 。

(b)

若 $j = n$ ，有

$\phi\left( S_{r} \right) = detpol\left( x^{n - r - 1}\phi\left( S_{n + 1} \right),\ldots,\phi\left( S_{n + 1} \right),x^{n - r}\phi\left( S_{n} \right),\ldots,\phi\left( S_{n} \right) \right)$

由于 $\deg{\left( \phi\left( S_{n + 1} \right) \right) = n + 1 = \deg\left( x^{n - r}\phi\left( S_{n} \right) \right) + 1 = n - r + r + 1 = n + 1}$ ，所以

$\phi\left( S_{r} \right) = \left\lbrack lc\left( \phi\left( S_{n + 1} \right),x \right)lc\left( \phi\left( S_{n} \right),x \right) \right\rbrack^{n - r}\phi\left( S_{n} \right)$

若 $j < n$ ，则

$\phi\left( R_{j + 1}^{2(j - r)}S_{r} \right) = detpol\left( x^{j - r - 1}\phi\left( S_{j + 1} \right),\ldots,\phi\left( S_{j + 1} \right),x^{j - r}\phi\left( S_{j} \right),\ldots,\phi\left( S_{j} \right) \right)$

由于

$\deg{\left( \phi\left( S_{j + 1} \right) \right) = j + 1 = \deg\left( x^{j - r}\phi\left( S_{n} \right) \right) + 1 = j - r + r + 1 = j + 1}$

所以

$\phi\left( R_{j + 1}^{2(j - r)}S_{r} \right) = \left\lbrack \phi\left( R_{j + 1} \right)lc\left( \phi\left( S_{n} \right),x \right) \right\rbrack^{j - r}\phi\left( S_{j} \right)$

即

$\phi\left( R_{j + 1}^{j - r}S_{r} \right) = {lc\left( \phi\left( S_{n} \right),x \right)}^{j - r}\phi\left( S_{j} \right)\$

(c)

若 $j = n$ ，有

${\phi\left( S_{r - 1} \right) = detpol\left( x^{n - r}\phi\left( S_{n + 1} \right),\ldots,\phi\left( S_{n + 1} \right),x^{n - r + 1}\phi\left( S_{n} \right),\ldots,x\phi\left( S_{n} \right),\phi\left( S_{n} \right) \right) }{= ( - 1)^{n - r + 2}detpol\left( x^{n - r}\phi\left( S_{n + 1} \right),\ldots,x\phi\left( S_{n + 1} \right),x^{n - r + 1}\phi\left( S_{n} \right),\ldots,x\phi\left( S_{n} \right),\phi\left( S_{n} \right),\phi\left( S_{n + 1} \right) \right)}$

由于

$\deg{\left( x\phi\left( S_{n + 1} \right) \right) = n + 2 = \deg\left( x^{n - r + 1}\phi\left( S_{n} \right) \right) + 1 = n - r + 1 + r + 1 = n + 2}$ ，

$\deg{\left( \phi\left( S_{n + 1} \right) \right) = n + 1 = \deg\left( x^{n - r + 1}\phi\left( S_{n} \right) \right)}$

所以

$\phi\left( S_{r - 1} \right) = \left\lbrack - lc\left( \phi\left( S_{n + 1} \right),x \right) \right\rbrack^{n - r}detpol\left( x^{n - r + 1}\phi\left( S_{n} \right),\ldots,x\phi\left( S_{n} \right),\phi\left( S_{n} \right),\phi\left( S_{n + 1} \right) \right) = \left\lbrack - lc\left( \phi\left( S_{n + 1} \right),x \right) \right\rbrack^{n - r}prem\left( \phi\left( S_{n + 1} \right),\phi\left( S_{n} \right),x \right)$

若 $j < n$ ，有

${\phi\left( R_{j + 1}^{2(j - r + 1)}S_{r - 1} \right) = detpol\left( x^{j - r}\phi\left( S_{j + 1} \right),\ldots,\phi\left( S_{j + 1} \right),x^{j - r + 1}\phi\left( S_{j} \right),\ldots,x\phi\left( S_{j} \right),\phi\left( S_{j} \right) \right) }{= ( - 1)^{j - r + 2}detpol\left( x^{j - r}\phi\left( S_{j + 1} \right),\ldots,x\phi\left( S_{j + 1} \right),x^{j - r + 1}\phi\left( S_{j} \right),\ldots,x\phi\left( S_{j} \right),\phi\left( S_{j} \right),\phi\left( S_{j + 1} \right) \right)}$

由于

${\deg\left( x\phi\left( S_{j + 1} \right) \right) = j + 2 }{= \deg\left( x^{j - r + 1}\phi\left( S_{n} \right) \right) + 1 = j - r + r + 1 + 1 = j + 2 }{= \deg{\left( \left( S_{j + 1} \right) \right) + 1} = j + 1 + 1 = j + 2}$

所以

${\phi\left( R_{j + 1}^{2(j - r + 1)}S_{r-1} \right) }{= ( - 1)^{j - r + 2}\left\lbrack \phi\left( R_{j + 1} \right) \right\rbrack^{j - r}detpol\left( x^{j - r + 1}\phi\left( S_{j} \right),\ldots,x\phi\left( S_{j} \right),\phi\left( S_{j} \right),\phi\left( S_{j + 1} \right) \right) }$