厦门大学高等代数教案网站IP地址:59.7.1.116;域名: gdjpkc. xmu.edu. cn §5.3最大公因式 教学目的和要求熟练掌握最大公因式的概念,性质与结论;熟练掌握互素的 概念和充要条件;了解中国剩余定理的内容和思想方法. 最大公因式 定义设∫(x),g(x),l(x)∈K[],称d(x)是f(x),g(x)的最大公因式,如果 (1)d(a)lf(a), d(a)lg(a) (2)若h(x)f(x),h(x)l9(x),则h(x)d(x) 注设d(x)是f(x)与g(x)的最大公因式,0≠c∈K,则cd(x)也是f(x)和 g(x)的最大公因式.设d(x),d1(x)是f(x)与g(x)的最大公因式,则存在c∈K, 使得d(x)=cd1(x).首项系数为1的最大公因式唯一,记为(f(x),9(x) 注∫(x)是∫(x)与0的最大公因式;8是6和7的最大公因式(0,2x2)=x2; (0,0)=0 引理设∫(x),9(x)∈K[x],则(f(x),g(x)=(f(x)-l(x)9(x),g(x) 证法1设(f(x),9(x)=d(x),要证(f(x)-l(x)9(x),9(x)=d(x) 1)如果d(x)f(x)且d(x)lg(x),则d(x)|g(x)且d(x)|(f(x)-lx)9(x) (2)若h(x)lg(x)且h(x川|(f(x)-l(x)g(x),则h(r川f(x)且h(x)9(x) 所以h(x)d(x) 证法2设d(x)=(f(x),g(x),dl1(x)=(f(x)-l(x)g(x),9(x).则d(x川f(x)且 d(x)lg(x),故d(x)(∫(x)-l(x)9(x)且d(x)g(x),所以d(x)ld1(x).另一方面,若 d1(x)(f(x)-l(x)9(x)且d1(x)lg(x),则d1(x)∫(x)且d1(x)lg(x),故d1(x)d(x) 又因为d(x),d1(x)首项系数为1,所以d(x)=d1(x).口 定理设∫(x),g(x)∈K[],则存在f(x)与9(x)的最大公因式d(x),且存在 u(x),v(x)∈K[x],使得d(x)=u(x)f(x)+v(x)9(x) 证明若∫(x)=9(x)=0,取d(x)=u(x)=t(x)=0即可,若f(x)=0, g(x)≠0,则(f(x),g(x)=cg(x),其中c是g(x)的首项系数的倒数;同理,若 g(x)=0,f(x)≠0时,结论成立S0#s |% IP $+ 59.77.1.116; W gdjpkc.xmu.edu.cn §5.3 R7;KF >HC8=ID qK&￾33m!/Z￾ -APqK&￾:u! /Z8Æz?MB.6j%H!Yb8t,+

33m 9J i f(x), g(x), d(x) ∈ K[x], Æ d(x) n f(x), g(x) !33m￾ 7 (1) d(x)|f(x), d(x)|g(x); (2) e h(x)|f(x), h(x)|g(x), # h(x)|d(x). Q i d(x) n f(x)  g(x) !33m￾ 0 6= c ∈ K, # cd(x) n f(x) 8 g(x) !33m
i d(x), d1(x) n f(x)  g(x) !33m￾#" c ∈ K, l d(x) = cd1(x). p
r~ 1 !33m}￾>~ (f(x), g(x)). Q f(x) n f(x)  0 !33m8 n 6 8 7 !33m
(0, 2x 2 ) = x 2 ; (0, 0) = 0. L? i f(x), g(x) ∈ K[x], # (f(x), g(x)) = (f(x) − l(x)g(x), g(x)). N: 1 i (f(x), g(x)) = d(x), Æ) (f(x) − l(x)g(x), g(x)) = d(x). (1) 7 d(x)|f(x) \ d(x)|g(x), # d(x)|g(x) \ d(x)|(f(x) − l(x)g(x)); (2) e h(x)|g(x) \ h(x)|(f(x) − l(x)g(x)), # h(x)|f(x) \ h(x)|g(x), w h(x)|d(x). N: 2 i d(x) = (f(x), g(x)), d1(x) = (f(x) − l(x)g(x), g(x)). # d(x)|f(x) \ d(x)|g(x), 4 d(x)|(f(x) − l(x)g(x)) \ d(x)|g(x), w d(x)|d1(x). N,U￾e d1(x)|(f(x) − l(x)g(x)) \ d1(x)|g(x), # d1(x)|f(x) \ d1(x)|g(x), 4 d1(x)|d(x). ~ d(x), d1(x) p
r~ 1, w d(x) = d1(x). ✷ 9? i f(x), g(x) ∈ K[x], #" f(x)  g(x) !33m d(x), \" u(x), v(x) ∈ K[x], l d(x) = u(x)f(x) + v(x)g(x). NB e f(x) = g(x) = 0, ^ d(x) = u(x) = v(x) = 0 <G
e f(x) = 0, g(x) 6= 0, # (f(x), g(x)) = cg(x), [. c n g(x) !p
r!r{H￾e g(x) = 0,f(x) 6= 0 k￾APJ
1