多元多项式的带余除法

命题:设$K$是域,$f(x_{1},\cdots,x_{n})\in K[x_{1},\cdots,x_{n}]$是$n$元多项式，
对于任意$K$中$n$个元素$a_{1},\cdots,a_{n}\in K,$存在$f_{i}\in K[x_{1},\cdots,x_{2}](i=1,\cdots,n),c\in K$使得

$$f(x_{1},\cdots,x_{n})=f_{1}\cdot(x_{1}-a_{1})+\cdots+f_{n}\cdot(x_{n}-a_{n})+c.$$

证明: 对$f(x_{1},\cdots,x_{n})$的次数(即$f$中所有单项式的最高次数)归纳.

若$deg(f)=1,$即$f$有形式

$$f(x_{1},\cdots,x_{n})=\sum_{i=1}^{n}b_{i}x_{i}+c_{0},$$

则

$$f(x_{1},\cdots,x_{n})=\sum_{i=1}^{n}b_{i}(x_{i}-a_{i})+(\sum_{i=1}^{n}a_{i}b_{i}+c_{0}).$$

即$n=1$时命题成立.

下面设$n=k$时命题成立。若$deg(f)=k+1,$则$f$首先可以写成形式

$$f=x_{1}f_{1}+\cdots+x_{n}f_{n}+c_{0},$$

其中$deg(f_{i})\leq k,i=1,\cdots,n.$

根据归纳法，存在$f_{i,j}\in K[x_{1},\cdots,x_{n}](i,j=1\cdots,n),c_{i}\in K,$使得$f_{i}=\sum_{j=1}^{n}f_{i,j}\cdot(x_{j}-a_{j})+c_{i}.$

于是

$$f=\sum_{i=1}^{n}x_{i}(\sum_{j=1}^{n}f_{i,j}(x_{j}-a_{j}))+\sum_{i=1}^{n}x_{i}c_{i}+c_{0}$$ $$=\sum_{j=1}^{n}(\sum_{i=1}^{n}x_{i}f_{i,j})(x_{j}-a_{j})+\sum_{j=1}^{n}c_{j}(x_{j}-a_{j})+\sum_{k=1}^{n}a_{k}c_{k}+c_{0}$$ $$=\sum_{j=1}^{n}(\sum_{i=1}^{n}x_{i}f_{i,j}+c_{j})(x_{j}-a_{j})+\sum_{k=1}^{n}a_{k}c_{k}+c_{0}.$$

令

$$g_{j}=\sum_{i=1}^{n}x_{i}f_{i,j}+c_{j}(j=1,\cdots,n),c=\sum_{k=1}^{n}a_{k}c_{k}+c_{0}.$$

则$f=\sum_{j=1}^{n}g_{j}(x_{j}-a_{j})+c.$归纳成立.