多元多项式的带余除法

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

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

若 $d e g (f) = 1,$ 即 $f$ 有形式

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

则

f (x_{1}, \dots, 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 $d e g (f) = k + 1,$ 则 $f$ 首先可以写成形式

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

其中 $d e g (f_{i}) \leq k, i = 1, \dots, n .$

根据归纳法，存在 $f_{i, j} \in K [x_{1}, \dots, x_{n}] (i, j = 1 \dots, n), c_{i} \in K,$ 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, \dots, 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.