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) ~p
r~ 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