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≤mmN,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