玉溪师范学院数学系，云南 玉溪

E

收稿日期：2020年8月17日；录用日期：2020年9月8日；发布日期：2020年9月15日

摘要

最大公因式是高等代数的重要内容之一。本文推广了关于最大公因式的一个命题。

关键词

最大公因式，命题，推广

A Generalization of a Proposition of the Greatest Common Factor

Jiming Yang

Department of Mathematics, Yuxi Normal University, Yuxi Yunnan

Received: Aug. 17^th, 2020; accepted: Sep. 8^th, 2020; published: Sep. 15^th, 2020

ABSTRACT

The greatest common factor is an important part of advanced algebra. This paper generalizes a proposition of the greatest common factor.

Keywords:The Greatest Common Factor, Proposition, Generalization

Copyright © 2020 by author(s) and Hans Publishers Inc.

This work is licensed under the Creative Commons Attribution International License (CC BY 4.0).

http://creativecommons.org/licenses/by/4.0/

1. 引言

文 [1] 中有这样一个定理：若 $a,b,c$ 是三个整数，且 $\left(a,c\right)=1$，$b,c$ 至少有一个不为零，则 $\left(ab,c\right)=\left(b,c\right)$。文 [2] 把这个定理作了推广。通过类比，本文对高等代数多项式理论中相应结果进行讨论。

2. 主要结果及应用

本文中讨论的多项式都是数域P上的多项式。在多项式理论中有这样的一个命题：

命题 若 $f\left(x\right),g\left(x\right),h\left(x\right)$ 是三个多项式，且 $\left(f\left(x\right),h\left(x\right)\right)=1$，$g\left(x\right),h\left(x\right)$ 至少有一个不为零多项式，则 $\left(f\left(x\right)g\left(x\right),h\left(x\right)\right)=\left(g\left(x\right),h\left(x\right)\right)$。

该命题的证明可参阅文 [3]。

引理 [3] 设 $f\left(x\right),g\left(x\right)$ 不全为零多项式，多项式 $h\left(x\right)$ 的首项系数为1，则

$\left(f\left(x\right)h\left(x\right),g\left(x\right)h\left(x\right)\right)=\left(f\left(x\right),g\left(x\right)\right)h\left(x\right).$ (1)

文 [3] 给出该引理的一个证明，下面再给出一个证明。

证易知 $\left(f\left(x\right),g\left(x\right)\right)h\left(x\right)|f\left(x\right)h\left(x\right),\left(f\left(x\right),g\left(x\right)\right)h\left(x\right)|g\left(x\right)h\left(x\right)$，故

$\left(f\left(x\right),g\left(x\right)\right)h\left(x\right)|\left(f\left(x\right)h\left(x\right),g\left(x\right)h\left(x\right)\right).$

另一方面，存在多项式 $u\left(x\right),v\left(x\right)$ 使得

$u\left(x\right)f\left(x\right)+v\left(x\right)g\left(x\right)=\left(f\left(x\right),g\left(x\right)\right).$

故

$\left(f\left(x\right),g\left(x\right)\right)h\left(x\right)=u\left(x\right)f\left(x\right)h\left(x\right)+v\left(x\right)g\left(x\right)h\left(x\right).$

于是 $\left(f\left(x\right)h\left(x\right),g\left(x\right)h\left(x\right)\right)|\left(f\left(x\right),g\left(x\right)\right)h\left(x\right)$。因此，(1)式成立。

把以上这个命题进行推广，就可以得到如下几个定理。

定理1 设 $f\left(x\right),g\left(x\right),h\left(x\right)$ 是三个多项式， $f\left(x\right),h\left(x\right)$ 不全为零多项式， $g\left(x\right),h\left(x\right)$ 也不全为零多项式，且 $\left(\left(f\left(x\right),h\left(x\right)\right),\left(g\left(x\right),h\left(x\right)\right)\right)=1$，则

$\left(f\left(x\right)g\left(x\right),h\left(x\right)\right)=\left(f\left(x\right),h\left(x\right)\right)\left(g\left(x\right),h\left(x\right)\right).$ (2)

证 因 $f\left(x\right),h\left(x\right)$ 不全为零多项式， $g\left(x\right),h\left(x\right)$ 也不全为零多项式，故 $f\left(x\right)g\left(x\right),h\left(x\right)$ 不全为零多项式，于是 $\left(f\left(x\right)g\left(x\right),h\left(x\right)\right),\left(f\left(x\right),h\left(x\right)\right)$ 与 $\left(g\left(x\right),h\left(x\right)\right)$ 都存在。下面首先证明

$\left(f\left(x\right),h\left(x\right)\right)\left(g\left(x\right),h\left(x\right)\right)|\left(f\left(x\right)g\left(x\right),h\left(x\right)\right).$ (3)

因 $\left(f\left(x\right),h\left(x\right)\right)|f\left(x\right)g\left(x\right),\left(f\left(x\right),h\left(x\right)\right)|h\left(x\right)$，故 $\left(f\left(x\right),h\left(x\right)\right)|\left(f\left(x\right)g\left(x\right),h\left(x\right)\right)$。同理， $\left(g\left(x\right),h\left(x\right)\right)|\left(f\left(x\right)g\left(x\right),h\left(x\right)\right)$。又因 $\left(\left(f\left(x\right),h\left(x\right)\right),\left(g\left(x\right),h\left(x\right)\right)\right)=1$，故(3)式成立。

再证

$\left(f\left(x\right)g\left(x\right),h\left(x\right)\right)|\left(f\left(x\right),h\left(x\right)\right)\left(g\left(x\right),h\left(x\right)\right).$ (4)

由引理得

$\left(f\left(x\right),h\left(x\right)\right)\left(g\left(x\right),h\left(x\right)\right)=\left(f\left(x\right)\left(g\left(x\right),h\left(x\right)\right),h\left(x\right)\left(g\left(x\right),h\left(x\right)\right)\right),$

故要证(4)成立，只需证明

$\left(f\left(x\right)g\left(x\right),h\left(x\right)\right)|\left(f\left(x\right)\left(g\left(x\right),h\left(x\right)\right),h\left(x\right)\left(g\left(x\right),h\left(x\right)\right)\right).$ (5)

下面先证

$\left(f\left(x\right)g\left(x\right),h\left(x\right)\right)|f\left(x\right)\left(g\left(x\right),h\left(x\right)\right).$ (6)

当 $f\left(x\right)$ 为零多项式时，结论显然正确。下面设 $f\left(x\right)$ 不为零多项式，a为多项式 $f\left(x\right)$ 的首项系数，

$f_1\left(x\right)=\frac{1}{a}f\left(x\right)$。因为 $\left(f\left(x\right)g\left(x\right),h\left(x\right)\right)|f\left(x\right)g\left(x\right),\left(f\left(x\right)g\left(x\right),h\left(x\right)\right)|f\left(x\right)h\left(x\right)$，所以由引理得

$\left(f\left(x\right)g\left(x\right),h\left(x\right)\right)|\left(f\left(x\right)g\left(x\right),f\left(x\right)h\left(x\right)\right)=f_1\left(x\right)\left(ag\left(x\right),ah\left(x\right)\right)=f_1\left(x\right)\left(g\left(x\right),h\left(x\right)\right).$

于是，(6)式成立。又因 $\left(f\left(x\right)g\left(x\right),h\left(x\right)\right)|h\left(x\right)\left(g\left(x\right),h\left(x\right)\right)$，故(5)式成立。

由(3)和(4)两式，即知(2)式成立。

定理2 设 $f_1\left(x\right),f_2\left(x\right),\cdots ,f_n\left(x\right),g\left(x\right)$ 是 $n+1$ 个多项式， $f_i\left(x\right)$ 和 $g\left(x\right)$ 不全为零多项式， $i=1,2,\cdots ,n$，且

$\left(\left(f_i\left(x\right),g\left(x\right)\right),\left(f_j\left(x\right),g\left(x\right)\right)\right)=1,\left(i,j=1,2,\cdots ,n,但i\ne j\right),$

则

$\left(f_1\left(x\right)f_2\left(x\right)\cdots f_n\left(x\right),g\left(x\right)\right)=\left(f_1\left(x\right),g\left(x\right)\right)\left(f_2\left(x\right),g\left(x\right)\right)\cdots \left(f_n\left(x\right),g\left(x\right)\right).$

证 对n作数学归纳法证明。

当 $n=1$ 时，根据定理1，结论正确。

假设定理2的结论对 $n-1\left(n\ge 2\right)$ 正确，下面由此可推出定理2的结论对n也正确。因为 $f_i\left(x\right)$ 和 $g\left(x\right)$ 不全为零多项式， $i=1,2,\cdots ,n$，所以 $f_1\left(x\right)f_2\left(x\right)\cdots f_{n-1}\left(x\right),g\left(x\right)$ 不全为零多项式， $f_n\left(x\right),g\left(x\right)$ 不全为零多项式。设

$\left(\left(f_1\left(x\right)f_2\left(x\right)\cdots f_{n-1}\left(x\right),g\left(x\right)\right),\left(f_n\left(x\right),g\left(x\right)\right)\right)=d\left(x\right).$

若 $d\left(x\right)$ 的次数 $\partial \left(d\left(x\right)\right)>0$，则 $d\left(x\right)$ 存在不可约因式 $p\left(x\right)$，从而

$p\left(x\right)|\left(f_1\left(x\right)f_2\left(x\right)\cdots f_{n-1}\left(x\right),g\left(x\right)\right),p\left(x\right)|\left(f_n\left(x\right),g\left(x\right)\right).$

于是， $p\left(x\right)|f_1\left(x\right)f_2\left(x\right)\cdots f_{n-1}\left(x\right),p\left(x\right)|g\left(x\right)$。由 $p\left(x\right)|f_1\left(x\right)f_2\left(x\right)\cdots f_{n-1}\left(x\right)$ 得， $p\left(x\right)|$ 某个 $f_r\left(x\right)$ $\left(1\le r\le n-1\right)$。从而 $p\left(x\right)|\left(f_r\left(x\right),g\left(x\right)\right)$，故 $p\left(x\right)|\left(f_r\left(x\right),g\left(x\right)\right),p\left(x\right)|\left(f_n\left(x\right),g\left(x\right)\right)$，这与 $\left(\left(f_r\left(x\right),g\left(x\right)\right),\left(f_n\left(x\right),g\left(x\right)\right)\right)=1$ 矛盾。故

$\left(\left(f_1\left(x\right)f_2\left(x\right)\cdots f_{n-1}\left(x\right),g\left(x\right)\right),\left(f_n\left(x\right),g\left(x\right)\right)\right)=1.$

因上式成立，故由定理1及归纳假设得

$\begin{array}{l}\left(f_1\left(x\right)f_2\left(x\right)\cdots f_n\left(x\right),g\left(x\right)\right)=\left(f_1\left(x\right)f_2\left(x\right)\cdots f_{n-1}\left(x\right),g\left(x\right)\right)\left(f_n\left(x\right),g\left(x\right)\right)\\ =\left(f_1\left(x\right),g\left(x\right)\right)\left(f_2\left(x\right),g\left(x\right)\right)\cdots \left(f_n\left(x\right),g\left(x\right)\right).\end{array}$

推论1 设非零多项式 $f_1\left(x\right),f_2\left(x\right),\cdots ,f_n\left(x\right)$ 两两互素， $g\left(x\right)$ 为任意多项式，则

$\left(f_1\left(x\right)f_2\left(x\right)\cdots f_n\left(x\right),g\left(x\right)\right)=\left(f_1\left(x\right),g\left(x\right)\right)\left(f_2\left(x\right),g\left(x\right)\right)\cdots \left(f_n\left(x\right),g\left(x\right)\right).$

推论2 设非零多项式 $f_1\left(x\right),f_2\left(x\right),\cdots ,f_n\left(x\right)$ 两两互素， $g\left(x\right)$ 为任意多项式，若

$f_1\left(x\right)|g\left(x\right),f_2\left(x\right)|g\left(x\right),\cdots ,f_n\left(x\right)|g\left(x\right),$

则

$f_1\left(x\right)f_2\left(x\right)\cdots f_n\left(x\right)|g\left(x\right).$

证 设多项式 $f_i\left(x\right)$ 的首项系数为 $a_i,i=1,2,\cdots ,n$。因

$f_1\left(x\right)|g\left(x\right),f_2\left(x\right)|g\left(x\right),\cdots ,f_n\left(x\right)|g\left(x\right),$

故 $\left(f_1\left(x\right),g\left(x\right)\right)=\frac{1}{a_1}{f}_1\left(x\right),\left(f_2\left(x\right),g\left(x\right)\right)=\frac{1}{a_2}{f}_2\left(x\right),\cdots ,\left(f_n\left(x\right),g\left(x\right)\right)=\frac{1}{a_n}{f}_n\left(x\right)$。 又因 $f_1\left(x\right),f_2\left(x\right),\cdots ,f_n\left(x\right)$ 两两互素，故由推论1得

$\left(f_1\left(x\right)f_2\left(x\right)\cdots f_n\left(x\right),g\left(x\right)\right)=\left(f_1\left(x\right),g\left(x\right)\right)\left(f_2\left(x\right),g\left(x\right)\right)\cdots \left(f_n\left(x\right),g(\; x\; )\right)$

$=\frac{1}{a_1}{f}_1\left(x\right)\cdot \frac{1}{a_2}{f}_2\left(x\right)\cdot \cdots \cdot \frac{1}{a_n}{f}_n\left(x\right)=\frac{1}{a_1{a}_2\cdots a_n}{f}_1\left(x\right)f_2\left(x\right)\cdots f_n\left(x\right),$

所以 $f_1\left(x\right)f_2\left(x\right)\cdots f_n\left(x\right)|g\left(x\right)$。

定理3 设 $f_1\left(x\right),f_2\left(x\right),\cdots ,f_n\left(x\right)$ 及 $g_1\left(x\right),g_2\left(x\right),\cdots ,g_m\left(x\right)$ 是任意两组多项式，且 $f_i\left(x\right),g_j\left(x\right)$ 至少有一个不全为零多项式， $i=1,2,\cdots ,n;j=1,2,\cdots ,m$。若

$\left(\left(f_i\left(x\right),g_j\left(x\right)\right),\left(f_k\left(x\right),g_l\left(x\right)\right)\right)=1$，$i,k=1,2,\cdots ,n;j,l=1,2,\cdots ,m$，但有序数对 $\left(i,j\right)\ne$ 有序数对 $\left(k,l\right)$，这里规定当且仅当 $i=j$ 且 $k=l$ 时有序数对 $\left(i,j\right)=$ 有序数对 $\left(m,l\right)$，则

$\left(f_1\left(x\right)f_2\left(x\right)\cdots f_n\left(x\right),g_1\left(x\right)g_2\left(x\right)\cdots g_m\left(x\right)\right)={\displaystyle \underset{i=1}{\overset{n}{\prod }}{\displaystyle \underset{j=1}{\overset{m}{\prod }}\left(f_i\left(x\right),g_j\left(x\right)\right)}}.$

证 对m作数学归纳法。

当 $m=1$ 时，由定理2得定理3的结论正确。假设定理3的结论对 $m-1\left(m\ge 2\right)$ 正确，我们将由此推出定理3的结论对m也正确。因为 $f_i\left(x\right),g_j\left(x\right)$ 至少有一个不为零多项式， $i=1,2,\cdots ,n;j=1,2,\cdots ,m$，故 $f_1\left(x\right)f_2\left(x\right)\cdots f_n\left(x\right),g_1\left(x\right)g_2\left(x\right)\cdots g_{m-1}\left(x\right)$ 不全为零多项式， $f_1\left(x\right)f_2\left(x\right)\cdots f_n\left(x\right),g_m\left(x\right)$ 不全为零多项式。设

$\left(\left(f_1\left(x\right)f_2\left(x\right)\cdots f_n\left(x\right),g_1\left(x\right)g_2\left(x\right)\cdots g_{m-1}\left(x\right)\right),\left(f_1\left(x\right)f_2\left(x\right)\cdots f_n\left(x\right),g_m\left(x\right)\right)\right)=d\left(x\right).$

若多项式 $d\left(x\right)$ 的次数 $\partial \left(d\left(x\right)\right)\ge 1$，则 $d\left(x\right)$ 存在不可约因式 $p\left(x\right)$。从而

$p\left(x\right)|\left(f_1\left(x\right)f_2\left(x\right)\cdots f_n\left(x\right),g_1\left(x\right)g_2\left(x\right)\cdots g_{m-1}\left(x\right)\right),$

$p\left(x\right)|\left(f_1\left(x\right)f_2\left(x\right)\cdots f_n\left(x\right),g_m\left(x\right)\right).$

于是， $p\left(x\right)|f_1\left(x\right)f_2\left(x\right)\cdots f_n\left(x\right),p\left(x\right)|g_1\left(x\right)g_2\left(x\right)\cdots g_{m-1}\left(x\right),p\left(x\right)|g_m\left(x\right)$。但 $p\left(x\right)$ 为不可约多项式，故 $p\left(x\right)|$ 某个 $f_r\left(x\right)\left(1\le r\le n\right)$，$p\left(x\right)|$ 某个 $g_s\left(x\right)\left(1\le s\le m-1\right)$，于是 $p\left(x\right)|\left(f_r\left(x\right),g_s\left(x\right)\right)$，$p\left(x\right)|\left(f_r\left(x\right),g_m\left(x\right)\right)$，这与已知条件 $\left(\left(f_r\left(x\right),g_s\left(x\right)\right),\left(f_r\left(x\right),g_m\left(x\right)\right)\right)=1$ 矛盾。故

$\left(\left(f_1\left(x\right)f_2\left(x\right)\cdots f_n\left(x\right),g_1\left(x\right)g_2\left(x\right)\cdots g_{m-1}\left(x\right)\right),\left(f_1\left(x\right)f_2\left(x\right)\cdots f_n\left(x\right),g_m\left(x\right)\right)\right)=1$。于是，由定理1，定理2及归纳假设得

$\begin{array}{l}\left(f_1\left(x\right)f_2\left(x\right)\cdots f_n\left(x\right),g_1\left(x\right)g_2\left(x\right)\cdots g_m\left(x\right)\right)\\ =\left(f_1\left(x\right)f_2\left(x\right)\cdots f_n\left(x\right),g_1\left(x\right)g_2\left(x\right)\cdots g_{m-1}\left(x\right)\right)\left(f_1\left(x\right)f_2\left(x\right)\cdots f_n\left(x\right),g_m\left(x\right)\right)\\ =\left[{\displaystyle \underset{i=1}{\overset{n}{\prod }}{\displaystyle \underset{j=1}{\overset{m-1}{\prod }}\left(f_i\left(x\right),g_j\left(x\right)\right)}}\right]{\displaystyle \underset{i=1}{\overset{n}{\prod }}\left(f_i\left(x\right),g_m\left(x\right)\right)}\\ ={\displaystyle \underset{i=1}{\overset{n}{\prod }}{\displaystyle \underset{j=1}{\overset{m}{\prod }}\left(f_i\left(x\right),g_j\left(x\right)\right)}}.\end{array}$

推论1 设 $f_1\left(x\right),f_2\left(x\right),\cdots ,f_n\left(x\right)$ 及 $g_1\left(x\right),g_2\left(x\right),\cdots ,g_m\left(x\right)$ 是任意两组多项式，若前一组多项式中任一多项式与后一组多项式中任一多项式互素，则 $f_1\left(x\right)f_2\left(x\right)\cdots f_n\left(x\right)$ 与 $g_1\left(x\right)g_2\left(x\right)\cdots g_m\left(x\right)$ 互素。

推论2 设 $f_1\left(x\right),f_2\left(x\right),\cdots ,f_n\left(x\right)$ 及 $g_1\left(x\right),g_2\left(x\right),\cdots ,g_m\left(x\right)$ 是任意两组多项式，且多项式 $f_1\left(x\right),f_2\left(x\right),\cdots ,f_n\left(x\right)$ 两两互素，多项式 $g_1\left(x\right),g_2\left(x\right),\cdots ,g_m\left(x\right)$ 两两互素，则

$\left(f_1\left(x\right)f_2\left(x\right)\cdots f_n\left(x\right),g_1\left(x\right)g_2\left(x\right)\cdots g_m\left(x\right)\right)={\displaystyle \underset{i=1}{\overset{n}{\prod }}{\displaystyle \underset{j=1}{\overset{m}{\prod }}\left(f_i\left(x\right),g_j\left(x\right)\right)}}.$

推论3 设 $f\left(x\right),g\left(x\right)$ 是任意两个非零多项式，其首项系数分别为 $a,b$，

$f\left(x\right)=a{p}_1^{{\alpha }_1}\left(x\right)p_2^{{\alpha }_2}\left(x\right)\cdots p_k^{{\alpha }_k}\left(x\right)$，整数 ${\alpha }_i\ge 0,i=1,2,\cdots ,k$，

$g\left(x\right)=b{p}_1^{{\beta }_1}\left(x\right)p_2^{{\beta }_2}\left(x\right)\cdots p_k^{{\beta }_k}\left(x\right)$，整数 ${\beta }_i\ge 0,i=1,2,\cdots ,k$，

其中 $p_1\left(x\right),p_2\left(x\right),\cdots ,p_k\left(x\right)$ 是互不相同的首项系数为1的不可约多项式，则

$\left(f\left(x\right),g\left(x\right)\right)=p_1^{\min \left({\alpha }_1,{\beta }_1\right)}\left(x\right)p_2^{\min \left({\alpha }_2,{\beta }_2\right)}\left(x\right)\cdots p_k^{\min \left({\alpha }_k,{\beta }_k\right)}\left(x\right).$

证 由推论2得，

$\begin{array}{c}\left(f\left(x\right),g\left(x\right)\right)=\left(p_1^{{\alpha }_1}\left(x\right)p_2^{{\alpha }_2}\left(x\right)\cdots p_k^{{\alpha }_k}\left(x\right),p_1^{{\beta }_1}\left(x\right)p_2^{{\beta }_2}\left(x\right)\cdots p_k^{{\beta }_k}\left(x\right)\right)\\ ={\displaystyle \underset{k=1}{\overset{k}{\prod }}{\displaystyle \underset{j=1}{\overset{k}{\prod }}\left(p_i^{{\alpha }_i}\left(x\right),p_j^{{\beta }_j}\left(x\right)\right)}}={\displaystyle \underset{i=1}{\overset{k}{\prod }}\left(p_i^{{\alpha }_i}\left(x\right),p_i^{{\beta }_i}\left(x\right)\right)}={\displaystyle \underset{i=1}{\overset{k}{\prod }}{p}_i^{\min \left({\alpha }_i,{\beta }_i\right)}\left(x\right)}.\end{array}$

文章引用

杨继明. 关于最大公因式的一个命题的推广
A Generalization of a Proposition of the Greatest Common Factor[J]. 理论数学, 2020, 10(09): 847-851. https://doi.org/10.12677/PM.2020.109098

参考文献