个人认为红色区域有问题，因为
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subres_{i}\left( \phi(S_{j + 1}),\phi\left( S_{j} \right) \right)
subresi(ϕ(Sj+1),ϕ(Sj))的定义不存在！！！
下面是我的证明过程：
【证明】
(a)
因为 ϕ ( R j + 1 2 ( j − i ) S i ) = ϕ ( s u b r e s i ( S j + 1 , S j ) ) = d e t p o l ( x j − i − 1 ϕ ( S j + 1 ) , … , ϕ ( S j + 1 ) , x j − i ϕ ( S j ) , … , ϕ ( S j ) ) {\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)} ϕ(Rj+12(j−i)Si)=ϕ(subresi(Sj+1,Sj))=detpol(xj−i−1ϕ(Sj+1),…,ϕ(Sj+1),xj−iϕ(Sj),…,ϕ(Sj))
而当 r + 1 ≤ i ≤ j − 1 r + 1 \leq i \leq j - 1 r+1≤i≤j−1时，有
deg ( ϕ ( S j + 1 ) ) = j + 1 > deg ( x j − i ϕ ( S j ) ) + 1 = j − ( r + 1 ) + r + 1 = j \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 deg(ϕ(Sj+1))=j+1>deg(xj−iϕ(Sj))+1=j−(r+1)+r+1=j
此时有 d e t p o l ( x j − i − 1 ϕ ( S j + 1 ) , … , ϕ ( S j + 1 ) , x j − i ϕ ( S j ) , … , ϕ ( S j ) ) = 0 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 detpol(xj−i−1ϕ(Sj+1),…,ϕ(Sj+1),xj−iϕ(Sj),…,ϕ(Sj))=0，也就是 ϕ ( R j + 1 2 ( j − i ) S i ) = 0 \phi\left( R_{j + 1}^{2(j - i)}S_{i} \right) = 0 ϕ(Rj+12(j−i)Si)=0，即 ϕ ( S j − 1 ) = ϕ ( S j − 2 ) = … = ϕ ( S r + 1 ) = 0 \phi\left( S_{j - 1} \right) = \phi\left( S_{j - 2} \right) = \ldots = \phi\left( S_{r + 1} \right) = 0 ϕ(Sj−1)=ϕ(Sj−2)=…=ϕ(Sr+1)=0。
(b)
- 若 j = n j = n j=n，有
ϕ ( S r ) = d e t p o l ( x n − r − 1 ϕ ( S n + 1 ) , … , ϕ ( S n + 1 ) , x n − r ϕ ( S n ) , … , ϕ ( S n ) ) \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) ϕ(Sr)=detpol(xn−r−1ϕ(Sn+1),…,ϕ(Sn+1),xn−rϕ(Sn),…,ϕ(Sn))
由于 deg ( ϕ ( S n + 1 ) ) = n + 1 = deg ( x n − r ϕ ( S n ) ) + 1 = n − r + r + 1 = n + 1 \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} deg(ϕ(Sn+1))=n+1=deg(xn−rϕ(Sn))+1=n−r+r+1=n+1，所以
ϕ ( S r ) = [ l c ( ϕ ( S n + 1 ) , x ) l c ( ϕ ( S n ) , x ) ] n − r ϕ ( S n ) \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) ϕ(Sr)=[lc(ϕ(Sn+1),x)lc(ϕ(Sn),x)]n−rϕ(Sn)
- 若 j < n j < n j<n，则
ϕ ( R j + 1 2 ( j − r ) S r ) = d e t p o l ( x j − r − 1 ϕ ( S j + 1 ) , … , ϕ ( S j + 1 ) , x j − r ϕ ( S j ) , … , ϕ ( S j ) ) \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) ϕ(Rj+12(j−r)Sr)=detpol(xj−r−1ϕ(Sj+1),…,ϕ(Sj+1),xj−rϕ(Sj),…,ϕ(Sj))
由于
deg ( ϕ ( S j + 1 ) ) = j + 1 = deg ( x j − r ϕ ( S n ) ) + 1 = j − r + r + 1 = j + 1 \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} deg(ϕ(Sj+1))=j+1=deg(xj−rϕ(Sn))+1=j−r+r+1=j+1
所以
ϕ ( R j + 1 2 ( j − r ) S r ) = [ ϕ ( R j + 1 ) l c ( ϕ ( S n ) , x ) ] j − r ϕ ( S j ) \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) ϕ(Rj+12(j−r)Sr)=[ϕ(Rj+1)lc(ϕ(Sn),x)]j−rϕ(Sj)
即
ϕ ( R j + 1 j − r S r ) = l c ( ϕ ( S n ) , x ) j − r ϕ ( S j ) \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)\ ϕ(Rj+1j−rSr)=lc(ϕ(Sn),x)j−rϕ(Sj)
(c)
- 若 j = n j = n j=n，有
ϕ ( S r − 1 ) = d e t p o l ( x n − r ϕ ( S n + 1 ) , … , ϕ ( S n + 1 ) , x n − r + 1 ϕ ( S n ) , … , x ϕ ( S n ) , ϕ ( S n ) ) = ( − 1 ) n − r + 2 d e t p o l ( x n − r ϕ ( S n + 1 ) , … , x ϕ ( S n + 1 ) , x n − r + 1 ϕ ( S n ) , … , x ϕ ( S n ) , ϕ ( S n ) , ϕ ( S n + 1 ) ) {\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)} ϕ(Sr−1)=detpol(xn−rϕ(Sn+1),…,ϕ(Sn+1),xn−r+1ϕ(Sn),…,xϕ(Sn),ϕ(Sn))=(−1)n−r+2detpol(xn−rϕ(Sn+1),…,xϕ(Sn+1),xn−r+1ϕ(Sn),…,xϕ(Sn),ϕ(Sn),ϕ(Sn+1))
由于
deg ( x ϕ ( S n + 1 ) ) = n + 2 = deg ( x n − r + 1 ϕ ( S n ) ) + 1 = n − r + 1 + r + 1 = n + 2 \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(xϕ(Sn+1))=n+2=deg(xn−r+1ϕ(Sn))+1=n−r+1+r+1=n+2，
deg ( ϕ ( S n + 1 ) ) = n + 1 = deg ( x n − r + 1 ϕ ( S n ) ) \deg{\left( \phi\left( S_{n + 1} \right) \right) = n + 1 = \deg\left( x^{n - r + 1}\phi\left( S_{n} \right) \right)} deg(ϕ(Sn+1))=n+1=deg(xn−r+1ϕ(Sn))
所以
ϕ ( S r − 1 ) = [ − l c ( ϕ ( S n + 1 ) , x ) ] n − r d e t p o l ( x n − r + 1 ϕ ( S n ) , … , x ϕ ( S n ) , ϕ ( S n ) , ϕ ( S n + 1 ) ) = [ − l c ( ϕ ( S n + 1 ) , x ) ] n − r p r e m ( ϕ ( S n + 1 ) , ϕ ( S n ) , x ) \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) ϕ(Sr−1)=[−lc(ϕ(Sn+1),x)]n−rdetpol(xn−r+1ϕ(Sn),…,xϕ(Sn),ϕ(Sn),ϕ(Sn+1))=[−lc(ϕ(Sn+1),x)]n−rprem(ϕ(Sn+1),ϕ(Sn),x)
- 若 j < n j < n j<n，有
ϕ ( R j + 1 2 ( j − r + 1 ) S r − 1 ) = d e t p o l ( x j − r ϕ ( S j + 1 ) , … , ϕ ( S j + 1 ) , x j − r + 1 ϕ ( S j ) , … , x ϕ ( S j ) , ϕ ( S j ) ) = ( − 1 ) j − r + 2 d e t p o l ( x j − r ϕ ( S j + 1 ) , … , x ϕ ( S j + 1 ) , x j − r + 1 ϕ ( S j ) , … , x ϕ ( S j ) , ϕ ( S j ) , ϕ ( S j + 1 ) ) {\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)} ϕ(Rj+12(j−r+1)Sr−1)=detpol(xj−rϕ(Sj+1),…,ϕ(Sj+1),xj−r+1ϕ(Sj),…,xϕ(Sj),ϕ(Sj))=(−1)j−r+2detpol(xj−rϕ(Sj+1),…,xϕ(Sj+1),xj−r+1ϕ(Sj),…,xϕ(Sj),ϕ(Sj),ϕ(Sj+1))
由于
deg ( x ϕ ( S j + 1 ) ) = j + 2 = deg ( x j − r + 1 ϕ ( S n ) ) + 1 = j − r + r + 1 + 1 = j + 2 = deg ( ( S j + 1 ) ) + 1 = j + 1 + 1 = j + 2 {\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} deg(xϕ(Sj+1))=j+2=deg(xj−r+1ϕ(Sn))+1=j−r+r+1+1=j+2=deg((Sj+1))+1=j+1+1=j+2
所以
ϕ ( R j + 1 2 ( j − r + 1 ) S r − 1 ) = ( − 1 ) j − r + 2 [ ϕ ( R j + 1 ) ] j − r d e t p o l ( x j − r + 1 ϕ ( S j ) , … , x ϕ ( S j ) , ϕ ( S j ) , ϕ ( S j + 1 ) ) {\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) } ϕ(Rj+12(j−r+1)Sr−1)=(−1)j−r+2[ϕ(Rj+1)]j−rdetpol(xj−r+1ϕ(Sj),…,xϕ(Sj),ϕ(Sj),ϕ(Sj+1))
即
ϕ ( − R j + 1 j − r + 2 ) ϕ ( S r − 1 ) = p r e m ( ϕ ( S j ) , ϕ ( S j + 1 ) ) \phi\left( - R_{j + 1}^{j - r + 2} \right)\phi\left( S_{r-1} \right) = prem\left( \phi\left( S_{j} \right),\phi\left( S_{j + 1} \right) \right) ϕ(−Rj+1j−r+2)ϕ(Sr−1)=prem(ϕ(Sj),ϕ(Sj+1))
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