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) !33mi d(x), d1(x) n f(x) g(x) !33m#" c ∈ K, l d(x) = cd1(x). pr~ 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,Ue 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) pr~ 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 <Ge f(x) = 0, g(x) 6= 0, # (f(x), g(x)) = cg(x), [. c n g(x) !pr! r{He g(x) = 0,f(x) 6= 0 kAPJ 1
设f(x)≠0≠g(x),由带余除法有 f(x)=9(x)qn(x)+r1(x),其中 degre(x)<deg(x); g(a)=ri(a)q2(r)+r2(a), It'fdegr2()< degr1(r) rI(a)=r2(a)a3(c)+r3(x),It fdegrs(a)< degre(r); rs-2=rs-19s(a)+rs(r), It fdegrs( ) degrs-1 (a rs-1(x)=7s(x)qs+1(x)+0. 由引理知(f(x),9(x)=(f(x)-g(x)q1(x),9(x)=(r1(x),g(x)=(r1(x),r2(x) =(rs-1(x),rs(x)=cTs(x),其中c为rs(x)的首项系数的倒数.另一方面, rs(a)=Ts-2(a)-rs-1()qs(a) rs-1(x)=r-3(x)-7-2(x)qs-1(x), 所以r(x)=(1-qs-1(x)r3-2(x)-qs(x)rs-3(x),逐步代入,得 rs(r)=u(a)f()+u(a)g(a). D 下面介绍计算最大公因式的辗转相除法. 例1f(x)=x4-x3-x2+2x-1,g(x)=x3-2x+1求(f(x),g(x)和 u(a), v(a),tu( f(r)+v(r)g=(f(ar),g(a)) x)=x+1 93(a)=- r1(ar)
i f(x) 6= 0 6= g(x), + f(x) = g(x)q1(x) + r1(x), [.degr1(x) < degg(x); g(x) = r1(x)q2(x) + r2(x), [.degr2(x) < degr1(x); r1(x) = r2(x)q3(x) + r3(x), [.degr3(x) < degr2(x); · · · · · · · · · · · · · · · · · · · · · rs−2 = rs−1qs(x) + rs(x), [.degrs(x) < degrs−1(x); rs−1(x) = rs(x)qs+1(x) + 0. H* (f(x), g(x)) = (f(x) − g(x)q1(x), g(x)) = (r1(x), g(x)) = (r1(x), r2(x)) = · · · = (rs−1(x), rs(x) = crs(x)), [. c ~ rs(x) !pr! rN,U rs(x) = rs−2(x) − rs−1(x)qs(x); rs−1(x) = rs−3(x) − rs−2(x)qs−1(x), w rs(x) = (1 − qs−1(x))rs−2(x) − qs(x)rs−3(x), / d rs(x) = u(x)f(x) + v(x)g(x). ✷ UCh=v33m!$1+ 1 f(x) = x 4 − x 3 − x 2 + 2x − 1, g(x) = x 3 − 2x + 1 ℄ (f(x), g(x)) 8 u(x), v(x), l u(x)f(x) + v(x)g(x) = (f(x), g(x)). x 4 −x 3 −x 2 +2x −1 x 3 −2x +1 q1(x) = x − 1 x 4 −2x 2 +x x 3 −x 2 q(x) = x + 1 −x 3 +x 2 +x −1 x 2 −2x +1 −x 3 +2x −1 x 2 −x q3(x) = −x r1(x) = x 2 −x r2 = −x +1 x 2 −x 0 2
因为(f(x),9(x)=x-1=-n2(x),所以r2(x)=9(x)-r1(x)g(x)=9(x) (f(x)-9(x)q1(x)p2(x)=f(x)(-q2(x)+g(x)[1+q(x)q2(x),故u(x)=g2(x) x+1,v(x)=-1-q1(x)g2(x)=-x2.口 下面介绍多个多项式的最大公因式 定义设f(x)∈K,1≤i≤m,d(x)称为f(x),1≤i≤m的最大公因式, 如果 (1)d(x)Jf(x),1≤i≤m; (2)设h(x)f(x),1≤i≤m,则h(x)d(x). 记f(x),1≤i≤m的首项系数为1的最大公因式为(f1(x),f2(x),…,fmn(x) 引理(f(x),9(x),h(x)=(f(x),g(x),h(x)=(f(x),(g(x),h(x) 证明设d1(x)=(∫(x),g(x),d(x)=(f(x),g(x),h(x),要证(d1(x),h(x) (1)若d(x)h(x),d(x)lg(x),d(r)f(x),所以d(x)h(x),d(x)d1(x) (2)若l(x)d4(x),l(x)h(x),则l(x)f(x),l(x)|g(x),l(x)h(x),所以l(x)d(x) 同理可得另一等式 注设f(x)∈K[],1≤i≤m,c(x)称为f(x),1≤i≤m的最小公倍式,如 果 (1)f(x)(x),1≤i≤m; (2)设f(x)h(x),1≤i≤m,则c(x)h(x) 记∫(x),1≤i≤m的首项系数为1的最小公倍式为[f1(x),f2(x),…,fm(x) 互素 定义设∫(x),9(x)∈K],若(f(x),g(x)=1,则称f(x)与9(x)互素,或称 互质 定理设f(x),9(x)∈K[x],则f(x)与g(x)互素的充分必要条件是存在u(x),v(x)∈ Kr,使得∫(x)u(x)+g(x)v(x)=1 证明必要性已证.现证充分性.设d(x)=(f(x),g(x),则d(x)f(x),d(x)g(x 所以d(x川∫(x)u(x)+g(x)(x),即d(x)1.又d(x)首项系数为1,所以d(x)=1.口
~ (f(x), g(x)) = x − 1 = −r2(x), w r2(x) = g(x) − r1(x)q2(x)= g(x) − (f(x) − g(x)q1(x))q2(x) = f(x)(−q2(x)) + g(x)[1 + q1(x)q2(x)], 4 u(x) = q2(x) = x + 1, v(x) = −1 − q1(x)q2(x) = −x 2 . ✷ UCh(1(m!33m 9J i fi(x) ∈ K[x], 1 ≤ i ≤ m, d(x) Æ~ fi(x), 1 ≤ i ≤ m !33m 7 (1) d(x)|fi(x), 1 ≤ i ≤ m; (2) i h(x)|fi(x), 1 ≤ i ≤ m, # h(x)|d(x). > fi(x), 1 ≤ i ≤ m !pr~ 1 !33m~ (f1(x), f2(x), · · · , fm(x)). L? ((f(x), g(x)), h(x)) = (f(x), g(x), h(x)) = (f(x),(g(x), h(x)). NB i d1(x) = (f(x), g(x)), d(x) = (f(x), g(x), h(x)), Æ) (d1(x), h(x)) = d(x). (1) e d(x)|h(x), d(x)|g(x), d(x)|f(x), w d(x)|h(x), d(x)|d1(x); (2) e l(x)|d1(x), l(x)|h(x), # l(x)|f(x), l(x)|g(x), l(x)|h(x), w l(x)|d(x). {HG N"m ✷ Q i fi(x) ∈ K[x], 1 ≤ i ≤ m, c(x) Æ~ fi(x), 1 ≤ i ≤ m !3 3m 7 (1) fi(x)|c(x), 1 ≤ i ≤ m; (2) i fi(x)|h(x), 1 ≤ i ≤ m, # c(x)|h(x). > fi(x), 1 ≤ i ≤ m !pr~ 1 !3 3m~ [f1(x), f2(x), · · · , fm(x)]. *:u 9J i f(x), g(x) ∈ K[x], e (f(x), g(x)) = 1, #Æ f(x) g(x) :u;Æ :- 9? i f(x), g(x) ∈ K[x], # f(x) g(x) :u!- Æz?n" u(x), v(x) ∈ K[x], l f(x)u(x) + g(x)v(x) = 1. NB Æ ))- i d(x) = (f(x), g(x)), # d(x)|f(x), d(x)|g(x), w d(x)|f(x)u(x) +g(x)v(x) , < d(x)|1. d(x) pr~ 1, w d(x) = 1. ✷ 3
推论1设f1(x)lg(x),f2(x)lg(x),且(f(x),f2(x)=1,则f(x)/2(x)9(x) 证明因为(f1(x),f2(x)=1,所以存在u(x),t(x)∈Kr,使得f(x)u(x)+ f2(x)(x)=1.由f1(x)lg(x),f2(x)lg(x),知道存在s(x),t(x)∈K[x],使得g(x)= f1(x)s(x)=f2(x)t(x).所以g(x)=9(x)(1(x)u(x)+f2(x)(x)=(/2(x)t(x)f(x)ux +f1(x)s(x)f2(x)v(x)=f1(x)f2(x)(t(x)u(x)+s(x)v(x).因此f(x)f2(x)9(x).口 推论2设f(x)g(x)h(x),(f(x),g(x)=1,则f(x)h(x) 证明因为(f(x),9(x)=1,所以存在u(x),v(x)∈K[,使得∫(x)u(x)+ g(x)v(x)=1.因而f(x)u(x)h(x)+g(x)v(x)h(x)=h(x).又因为f(x)g(x)h(x), 故∫(x)h(x).口 推论3设(f(x),g(x))=d(x),f(x)=f1(x)d(x),g(x)=91(x)d(x),则(f(x) g1(x)=1 证明因为(f(x),g(x)=d(x),所以存在u(x),v(x)∈K[a,使得∫(x)u(x)+ 9(x)v(x)=d(x).进一步有f1(x)u(x)+91(x)(x)=1.因而(f1(x),91(x)=1.口 推论4设(f(x),9(x)=d(x),t(x)为首项系数为1的多项式,则(f(x)t(x) g(at(a))=d(a)t(a) 证明因为(f(x),g(x))=d(x),显然有d(r)t(r)f(x)t(x),d(x)(x)lg(x)t(x).同 时存在u(x),(x)∈K],使得∫(x)u(x)+g(x)v(x)=d(x).所以f(x)t(x)u(x)+ g(r)t(x)u(x)=d(x)(x).若h(r川∫(x)t(x),h(x)g(x)t(x),则h(x)ld(x)t(x).这样 (f(a)t(a), g(a)t(a))=d(r)t(). D 推论5设(f1(x),g(x)=1,(f2(x),9(x)=1,则(f1(x)f2(x),9(x)=1 证明存在u(x),(x,u1(x),n(x),使得f1(x)(x)+g(x)v(x)=1,f2(x)u1(x)+ g(x)t1(x)=1.所以f1(x)2(x)(u(x)u1(x)+g(x)(f1(x)u(x)1(x)+f2(x)u1(x)u(x)+ g(x)v(x)1(x)=1.故有(f1(x)f2(x),9(x)=1.口 推论6设∫(x)9(x)≠0,则∫(x)9(x)~(f(x),9(x)[f(x),(x) 证明设(f(x),9(x)=d(x),则f(x)=d(x)f1(x),9(x)=d(x)91(x),f(x)9(x) d(x)(d(x)f1(x)g1(x).记m(x)=d(x)f1(x)91(x),可证m(x)是f(x),9(x)的最小 公倍式,即与[f(x),g(x)相伴.一方面由定义,显然m(x)是f(x),9(x)的公倍
GA 1 i f1(x)|g(x), f2(x)|g(x), \ (f1(x), f2(x)) = 1, # f1(x)f2(x)|g(x). NB ~ (f1(x), f2(x)) = 1, w" u(x), v(x) ∈ K[x], l f1(x)u(x) + f2(x)v(x) = 1. f1(x)|g(x), f2(x)|g(x), * " s(x), t(x) ∈ K[x], l g(x) = f1(x)s(x) = f2(x)t(x). w g(x) = g(x)(f1(x)u(x)+f2(x)v(x)) = (f2(x)t(x))f1(x)ux +f1(x)s(x)f2(x)v(x) = f1(x)f2(x)(t(x)u(x) + s(x)v(x)). f1(x)f2(x)|g(x). ✷ GA 2 i f(x)|g(x)h(x),(f(x), g(x)) = 1, # f(x)|h(x). NB ~ (f(x), g(x)) = 1, w" u(x), v(x) ∈ K[x], l f(x)u(x) + g(x)v(x) = 1. ) f(x)u(x)h(x) + g(x)v(x)h(x) = h(x). ~ f(x)|g(x)h(x), 4 f(x)|h(x). ✷ GA 3 i (f(x), g(x)) = d(x), f(x) = f1(x)d(x), g(x) = g1(x)d(x), # (f1(x), g1(x)) = 1. NB ~ (f(x), g(x)) = d(x), w" u(x), v(x) ∈ K[x], l f(x)u(x) + g(x)v(x) = d(x). D f1(x)u(x) + g1(x)v(x) = 1. ) (f1(x), g1(x)) = 1. ✷ GA 4 i (f(x), g(x)) = d(x), t(x) ~pr~ 1 !(m# (f(x)t(x), g(x)t(x)) = d(x)t(x). NB ~ (f(x), g(x)) = d(x), ` d(x)t(x)|f(x)t(x), d(x)t(x)|g(x)t(x). { k" u(x), v(x) ∈ K[x], l f(x)u(x) + g(x)v(x) = d(x). w f(x)t(x)u(x) + g(x)t(x)v(x) = d(x)t(x). e h(x)|f(x)t(x), h(x)|g(x)t(x), # h(x)|d(x)t(x). ' (f(x)t(x), g(x)t(x))= d(x)t(x). ✷ GA 5 i (f1(x), g(x)) = 1,(f2(x), g(x)) = 1, # (f1(x)f2(x), g(x)) = 1. NB " u(x), v(x), u1(x), v1(x), l f1(x)u(x)+g(x)v(x) = 1, f2(x)u1(x)+ g(x)v1(x) = 1. w f1(x)f2(x)(u(x)u1(x)+g(x)(f1(x)u(x)v1(x)+f2(x)u1(x)v(x)+ g(x)v(x)v1(x)) = 1. 4 (f1(x)f2(x), g(x)) = 1. ✷ GA 6 i f(x)g(x) 6= 0, # f(x)g(x) ∼ (f(x), g(x))[f(x), g(x)]. NB i (f(x), g(x)) = d(x), # f(x) = d(x)f1(x), g(x) = d(x)g1(x), f(x)g(x) = d(x)(d(x)f1(x)g1(x)). > m(x) = d(x)f1(x)g1(x), G) m(x) n f(x), g(x) !3 3m< [f(x), g(x)] ,U%` m(x) n f(x), g(x) !3 4
式;另一方面若h(x)是∫(x),9(x)的公倍式,h(x)=f(x)f2(x)=9(x)g2(x),即 h(x)=d(x)f1(x)2(x)=d(x)gn(x)g(x).因f(x)g(x)≠0,因此d(x)≠0,故 f1(x)f2(x)=91(x)92(x),进而f1(x)9n(x)92(x).注意到(1(x),91(x)=1,所以 f1(x)lg(x),g2(x)=f1(x)t(x),从而h(x)=d(x)g1(x)f1(x)t(x)=m(x)t(x),即 m(x)h(x).命题得证.口 例2(f(x),9(x)=1,则(f(xn),g(xm)=1 证明因为(f(x),g(x)=1,所以存在u(x),v(x)∈K],使得f(x)u(x)+ g(x)v(x)=1.所以∫(xm)u(xm)+9(xmn)(xmn)=1.故(f(xm),g(xm)= 中国剩余定理 弓理设p1(x),p2(x)…,pm(x)∈K[X]两两互素,则存在f(x)∈K[X],1≤i≤ m,使得f(x)=l(x)2(x)+1,f1(x)=ha(x)p1(x),1≤i≠j≤m 证明:当i≠j时,(P(x)(x)=1.所以(p1(x),I=1(x)=1.因此 存在t(x),tl(x)∈K[x使得 pi(a)u (c)+(Ip(a)ui(a)=1 j≠ 令f(x)=1(x)(I1≠P(x),(x)=-(x),hn(x)=(∏1P(x)(x)即得证 注引理中f等同于满足如下条件:f被p(x)除后余1,同时可被所有P(x), ≠i,1≤j≤m整除 中国剩余定理设p1(x),p2(x),…,pm(x)∈K]是两两互素的多项式,a1,a2 ,am∈K,则存在唯一g(x),q(x)∈K1≤i≤m,使得 deg()<∑degp(a) 且 9(x)=p(x)q(x)+a,1≤i≤m 注定理意义在于存在g(x),它被P(x)除后余a,1≤i≤m
mN,Ue h(x) n f(x), g(x) !3m h(x) = f(x)f2(x) = g(x)g2(x), < h(x) = d(x)f1(x)f2(x) = d(x)g1(x)g2(x). f(x)g(x) 6= 0, d(x) 6= 0, 4 f1(x)f2(x) = g1(x)g2(x), D) f1(x)|g1(x)g2(x). 0 (f1(x), g1(x)) = 1, w f1(x)|g2(x), g2(x) = f1(x)t(x), ) h(x) = d(x)g1(x)f1(x)t(x) = m(x)t(x), < m(x)|h(x). Xy ) ✷ 2 (f(x), g(x)) = 1 , # (f(x m), g(x m)) = 1. NB ~ (f(x), g(x)) = 1, w" u(x), v(x) ∈ K[x], l f(x)u(x) + g(x)v(x) = 1. w f(x m)u(x m) + g(x m)v(x m) = 1. 4 (f(x m), g(x m)) = 1. f.6j %H L? i p1(x), p2(x), ...pm(x) ∈ K[X] LL:u#" fi(x) ∈ K[X], 1 ≤ i ≤ m, l fi(x) = li(x)pi(x) + 1, fi(x) = hij (x)pj (x), 1 ≤ i 6= j ≤ m. )V i 6= j k (pi(x), pj (x)) = 1. w (pi(x), Qm j6=i,j=1 pj (x)) = 1. " ui(x), vi(x) ∈ K[x], l pi(x)vi(x) + (Y j6=i pj (x))ui(x) = 1 O fi(x) = ui(x)(Q j6=i pj (x)), l(x) = −v(x), hij (x) = (Q r6=i,j pr(x))ui(x) < ) ✷ Q H. fi "{ Q2 z?fi pi(x) 9 1, {kGw pj (x), j 6= i, 1 ≤ j ≤ m ( P<EM9? i p1(x), p2(x), ..., pm(x) ∈ K[x] nLL:u!(m a1, a2, · · ·, am ∈ K, #"} g(x), qi(x) ∈ K[x], 1 ≤ i ≤ m, l degg(x) < Xm i=1 degpi(x), \ g(x) = pi(x)qi(x) + ai , 1 ≤ i ≤ m. Q %H" " g(x), x pi(x) 9 ai , 1 ≤ i ≤ m. 5
另明由引理,存等∫(x),1≤i≤m,使f(x)=l(x)n(x)+1,f(x) hy(x)p/(x),i≠j令∫(x)=∑m1Jf(x)a.根据带余除法,有f(x)=t(x)II1P1(x)+ g(x),这里degg(x)<degI1p(x)=∑1degp(x).我们指出这里g(x)≠0.否 则,对于固定的j,大为p(x)∑1f(x)a2和p|f(x),r≠j,所以plf(x)a,与 f(x)=l(x)n(x)+1矛盾 9(x)=f(x)-t(x)I=1n2(x) ∑=1f(x)1(x)+f(x)a)-t(x)I=12(x) =()(∑出1()1+()-(x)(2)+2(a) h(x)p(x)a+(l(x)p2(x)+1)a1-t(x) 的证明唯一性:若9(x)≠91(x),g(x)=p(x)(x)+a1,1≤i≤m,g1(x)= n(x)t(x)+a1,1≤i≤m,gi(x)≠t(x),1≤i≤m.则0≠g(x)-91(x) n(x)(q(x)-t(x).故p(x)g(x)-g1(x),1≤i≤m,而p(x),i≠j两两互素, 所以p1(x)p2(x)-pm(x)9(x)-91(x),∑1degp(x)≤max{deg(x), peggy()}< degp(x),此为矛盾.所以g(x)=91(x),即g(x)是唯一的.进一步,根据带 余除法的商的唯一性,二a(x)是唯一的.口 例3( Lagrange插值或式)设a1,a2,…,am∈K为m个不同数,则对任意 b1,b2,…,bmn∈K,存等唯一次数必于m的多项式 L(x)=∑h (x-a3) j≠i 满足L(a)=b,1≤i≤m 另明设p(x)=x-a,1≤i≤m,则p2(x)两两互素,degp(x)=1,显然 L(a1)=b,degL(x)=m-1<∑m1degf(x),L(x)=(x-a1)q(x)+b,根据中国 剩余定理,满足条件的L(x)唯一确定的.口 作充:P1942(1),3,4;P271,2. 补充作充1:设p1(x),P2(x),…,pm(x)∈K]是两两互素的多项式,s1(x),s2(x), sm(x)∈K回]则存等唯一g(x),qa(x),1≤i≤m,使得deg(x)<∑degp(x) 且g(x)=9(x)p(x)+s(x),1≤i≤m
NB H" fi(x), 1 ≤ i ≤ m, l fi(x) = li(x)pi(x) + 1, fi(x) = hij (x)pj (x), i 6= j. O f(x) = Pm i=1 fi(x)ai . 2E + f(x) = t(x) Qm i=1 pi(x)+ g(x), 'I degg(x) < deg Qm i=1 pi(x) = Pm i=1 degpi(x). T,'I g(x) 6= 0. . #& 5%! j, ~ pj (x)| Pm i=1 fi(x)ai 8 pj |fr(x), r 6= j, w pj |fj (x)aj , fi(x) = li(x)pi(x) + 1 R' g(x) = f(x) − t(x) Qm i=1 pi(x) = (Pm j6=i,j=1 fj (x)aj (x) + fi(x)ai) − t(x) Qm i=1 pi(x) = P j6=i hj (x)pi(x)aj + (li(x)pi(x) + 1)ai − t(x) Qm i=1 pi(x) = pi(x)(P j6=i hj (x)aj + li(x) − t(x) Q j6=i pj (x)) + ai . !)V} e g(x) 6= g1(x), g(x) = pi(x)qi(x) + ai , 1 ≤ i ≤ m, g1(x) = pi(x)ti(x) + ai , 1 ≤ i ≤ m, qi(x) 6= ti(x), 1 ≤ i ≤ m. # 0 6= g(x) − g1(x) = pi(x)(qi(x) − ti(x)). 4 pi(x)|g(x) − g1(x), 1 ≤ i ≤ m, ) pi(x), i 6= j LL:u w p1(x)p2(x)...pm(x)|g(x) − g1(x), Pm i=1 degpi(x) ≤ max{degg(x), degg1(x)} < Pm i=1 degpi(x), ~R'w g(x) = g1(x), < g(x) n}!D 2E +!g!} * qi(x) n}! ✷ 3 (Lagrange 6O;F) i a1, a2, ..., am ∈ K ~ m 1 {r#&a b1, b2, ..., bm ∈ K, "}r m !(m L(x) = Xm i=1 bi( Y j6=i (x − aj ) (ai − aj ) ) Q2 L(ai) = bi , 1 ≤ i ≤ m. NB i pi(x) = x − ai , 1 ≤ i ≤ m, # pi(x) LL:u degpi(x) = 1, ` L(ai) = bi , degL(x) = m − 1 < Pm i=1 degfi(x), L(x) = (x − ai)qi(x) + bi . 2E.6 j %HQ2z?! L(x) }_%! ✷ 5 P194 2(1), 3, 4; P227 1, 2. 5 1: i p1(x), p2(x), ...pm(x) ∈ K[x] nLL:u!(ms1 (x), s2(x), · · · , sm(x) ∈ K[x] #"} g(x), qi(x), 1 ≤ i ≤ m, l degg(x) < Pdegpi(x) \ g(x) = qi(x)pi(x) + si(x), 1 ≤ i ≤ m. 6
补充作业2:设f(x),(x)∈K],令 92={u(x)f(x)+u(x)g(x)u(x),v(x)∈K[]} 求证:(1)若a(x),b(x)∈92,则a(x)±b(x)∈92; (2)若a(x)∈9,则对任意h(x)∈K[X],有a(x)h(x)∈9 (3)存在首项系数为1的d(x)∈9,使得对va(x)∈9,有d(x)a(x); 4)d(x)=(f(x),g(x) 思考题:P1945,6,7 选做:设a,b,c两两互异,用x-a,x-b,x-c除∫(x)的余式分别为r,s,t 试求用(x-a)(x-b)(x-c)除∫(x)的余式
5 2: i f(x), g(x) ∈ K[x], O Ω = {u(x)f(x) + v(x)g(x)|u(x), v(x) ∈ K[x]}. ℄) (1) e a(x), b(x) ∈ Ω, # a(x) ± b(x) ∈ Ω; (2) e a(x) ∈ Ω, #&a h(x) ∈ K[X], a(x)h(x) ∈ Ω; (3) "pr~ 1 ! d(x) ∈ Ω, l & ∀a(x) ∈ Ω, d(x)|a(x); (4) d(x) = (f(x), g(x)). tFy P194 5, 6, 7. 4i a, b, c LL: x − a, x − b, x − c f(x) ! m- ~ r, s, t. o℄ (x − a)(x − b)(x − c) f(x) ! m 7