因此P2(x)…p、(x)=c-g2(x)…(x).左式为8-1个不可约因式之积,由归纳假 设s-1=t-1,即s-t,且p(x 根据定理,我们知道对于任意次数大于1的多项式有如下的标准分解式: ∫(x)=qp(x)p2(x)…pm(x) 其中p(x)是首项系数为1的两两互素的不可约多项式,e1≥1,1≤i≤m. 例1f(x)=cp(x)p2(x)…pm(x),g(x)=m(x)2(x)…pm(x) 2(x)为首一两两互素的不可约多项式,a1≥0,b1≥0且a1+b>0,1≤i≤m,则 (1)f(x)g(x)的充分必要条件是a1≤b;,1≤i≤m; (2)(f(x),9(x)=n(x)p2(x)…pm(x),G=min{a,b},1≤i≤m; (3)[f(x),9(x)]=n(x)(x)…m(x),d2=max{a1,b},1≤i≤m; (4)(f(x),g(x)[f(x),9(x)=c-1d-lf(x)9(x) 三.重因式 定义设f(x),p(x)∈K[x],p(x)是不可约多项式,若存在k>1使得p(x)f(x) p+1(x)f(x),则称p(x)是f(x)的k重因式 定理(1)不可约多项式p(x)是f(x)的k重因式,则p(x)是f(x)的k-1重 因式; (2)aa没有重因式,且它的不可约因式与∫(a)的不可约因式相同; (3)f(x)没有重因式的充分必要条件是(f(x),f(x)=1 证明(1)由设f(x)=p(x)h(x),其中p(x)h(x).直接计算得,f'(x)= kp-1(x)p(x)h(x)+p(x)h(x).显然,p-1(x)f(x),但p(x)十kp(x)h(x)+p(x)h(x) 否则有p(x)|p(x)h(x),而p(x)不可约,p(x)十h(x),从而p(x)|p(x),这是不可能 的,因p(x)非零且次数小于p(x)的次数 (2)设∫(x)的标准分解式为f(x)=e1(x)2(x)…F(x),只要证 cP(xr)p2(x)…p(x)命题即得证事实上,f"(x)=ce1p2-1(x)2(x)…p(x)1(x)+ e2n(x)2-1(x)…p(x)2(x)+…+cen(x)p2(x)…p-l(x),(x).所以p-(x) n2-1(x)…p-1(x)是f(x)和f(x)的公因式.注意到p(x)可整除右项中除第 项外的所有各项,但不能整除第一项,故n(x)十∫(x).同理p(x)十f"(x) d(a)=(f(x),f()=n1-1(x)n2-(x)…(),m=ep()p()…p()7 p2(x)· · · ps(x) = c −1 q2(x)· · · qt(x). W s − 1 6 UB7LE￾:<lH ~ s − 1 = t − 1, F s − t, r pi(x) ∼ qi(x). ✷ 8Q&W￾eK(>x6 > 1 !*&;z!! T2N f(x) = cp e1 1 (x)p e2 2 (x)· · · p em m (x), qQ pi(x) &  1 !\\A! UB*&￾ ei ≥ 1, 1 ≤ i ≤ m. ^ 1 f(x) = cp a1 1 (x)p a2 2 (x)· · · p am m (x), g(x) = dpb1 1 (x)p b2 2 (x)· · · p bm m (x), pi(x) 2\\A! UB*&￾ ai ≥ 0, bi ≥ 0 r ai + bi > 0, 1 ≤ i ≤ m, E (1) f(x)|g(x) !2 /J ai ≤ bi , 1 ≤ i ≤ m; (2) (f(x), g(x)) = p c1 1 (x)p c2 2 (x)· · · p cm m (x), ci = min{ai , bi}, 1 ≤ i ≤ m; (3) [f(x), g(x)] = p d1 1 (x)p d2 2 (x)· · · p dm m (x), di = max{ai , bi}, 1 ≤ i ≤ m; (4) (f(x), g(x))[f(x), g(x)] = c −1d −1f(x)g(x). |
R7 [h ~ f(x), p(x) ∈ K[x], p(x)  UB*&￾{D k > 1 p k (x)|f(x), p k+1(x) ∤ f(x), E p(x)  f(x) ! k- R7
[℄ (1) UB*& p(x)  f(x) ! k- R7￾E p(x)  f ′ (x) ! k − 1 R 7 (2) f(x) (f(x),f′(x)) ;R7￾r! UB7? f(x) ! UB7% (3) f(x) ;R7!2 /J (f(x), f′ (x)) = 1. j_ (1) :~ f(x) = p k (x)h(x), qQ p(x) ∤ h(x). MLG ￾ f ′ (x) = kpk−1 (x)p ′ (x)h(x)+p k (x)h ′ (x). #w￾p k−1 (x)|f ′ (x),  p(x) ∤ kp′ (x)h(x)+p(x)h ′ (x), 3E; p(x) | p ′ (x)h(x), + p(x) UB￾ p(x) ∤ h(x), + p(x) | p ′ (x), H Um !￾7 p ′ (x) 1^r '> p(x) !
(2) ~ f(x) ! T2N f(x) = cp e1 1 (x)p e2 2 (x)· · · p es s (x), O/J f(x) d(x) = cp1(x)p2(x)· · · ps(x) iF J

}￾f ′ (x) = ce1p e1−1 1 (x)p e2 2 (x)· · · p es s (x)p ′ 1 (x)+ ce2p e1 1 (x)p e2−1 2 (x)· · · p es s (x)p ′ 2 (x)+· · ·+cesp e1 1 (x)p e2 2 (x)· · · p es−1 s (x)p ′ s (x). 4 p e1−1 1 (x) p e2−1 2 (x)· · · p es−1 s (x)  f(x) ? f ′ (x) !97
S6 p e1 1 (x) UI<&Q$ 2&!;7&￾ mI$2&￾: p e1 1 (x) ∤ f ′ (x). W p ei i (x) ∤ f ′ (x). d(x) = (f(x), f′ (x)) = p e1−1 1 (x)p e2−1 2 (x)· · · p es−1 s (x), f(x) d(x) = cp1(x)p2(x)· · · ps(x). 3