(3)故(2)即得结命.口 例2设∫(x),9(x)是K上的两意非零序定式,证明存基整数N使得形任个的 m1,m2>N,总有(fm(x),9(x)=(fn2(x),9(x) 某令若(f(x),9(x)=1,取N=1.若(f(x),g(x)=d(x)≠1,设d(x)的数准 分解式是d(x)=n1(x)p2(x)…pm(x),其中p(x)是K上两两互成的本分不可换 序定式,e1>0,1≤i≤m.调应的,设f(x)=m(x)n2(x)…pm(x)f1(x),9(x)= m(x)n2(x)…mhm(a)9(a),其中a1>0,b>0,1≤i≤m.取N=max1a,积形 任个的n1,m2>N,总有(fm(x),9(x)=(fm(x,g(x)=m(x)2(x)…pw(x),口 例3设∫(x),9(x)∈Kr.证明(f(x),g(x)≠1的所要条证是存基K上不可 换序定式p(x),使得p(x)f(x)+g(x),p(x)|f(x)9(x) 某令必要盾.设d(x)=(f(x),9(x).故已知d(x)≠1,可取到d(x)的不可换 因式p(x),容易反证p(x)f(x)+g(x),p(x)f(x)g(x). 所分盾.因p(x)f(x)+g(x),所以p(a)f(x)g(x)+92(x).果因p(x)f(x)9(x), 从而p(x)|g2(x)均个到p(x)是K上不可换序定式,由p(x)g(x)或故p(x)f(x)+ g(x),即有p(x)f(x).假不明p(x)是f(x),g(x)的分意公因式,由(f(x),9(x)≠1 例4设f(x)∈Kl],a∈K.明y=x-a,得g(y)=f(y+a).证明f(x)基K 上可换的所要条证是g(y)基K上可换 某令必要盾.设∫(x)次数为n.因f(x)基K上可换,由存基K上次数都过n 的序定式u(x),v(x),使得∫(x)=u(x)v(x),适x用y+a次约,得9(y)=s(y)(y), 其中s(y)=u(y+a),t(y)=t(y+a)设∫(x)的本定为anxn,简为计乘知g(y)的 本定是any,假不明degg(y)=degf(x)=n.存理可得degs(y)=degu(x)<n, degt(y)=degu(x)<n.假质证明理g(y)基K上妨可换 所分盾类似可得.口 作业使P19s.1(1),2,4,6;P27.3 顺所1求证f(x)=1+x++哥+…+无重因式 顺所2:求x3+px+q有重因式的条证(3) : (2) F M`
✷ ^ 2 ~ f(x), g(x)  K }!\61^*&￾JgDI N (x6! n1, n2 > N, V; (f n1(x), g(x)) = (f n2(x), g(x)). j_ { (f(x), g(x)) = 1, u N = 1. { (f(x), g(x)) = d(x) 6= 1, ~ d(x) ! T 2N d(x) = p e1 1 (x)p e2 2 (x)· · · p em m (x), qQ pi(x)  K }\\A!2 UB *&￾ ei > 0, 1 ≤ i ≤ m. %8!￾~ f(x) = p a1 1 (x)p a2 2 (x)· · · p am m (x)f1(x), g(x) = p b1 1 (x)p b2 2 (x)· · · p bm m (x)g1(x), qQ ai > 0, bi > 0, 1 ≤ i ≤ m. u N = maxm i=1 bi ai , E( x6! n1, n2 > N, V; (f n1(x), g(x)) = (f n2(x), g(x)) = p b1 1 (x)p b2 2 (x)· · · p bm m (x). ✷ ^ 3 ~ f(x), g(x) ∈ K[x]. Jg (f(x), g(x)) 6= 1 !/JD K } U B*& p(x), p(x)|f(x) + g(x), p(x)|f(x)g(x). j_ /)
~ d(x) = (f(x), g(x)). :3K d(x) 6= 1, Uu d(x) ! UB 7 p(x), y5.J p(x)|f(x) + g(x), p(x)|f(x)g(x). 2)
7 p(x)|f(x) + g(x), 4 p(x)|f(x)g(x) + g 2 (x). =7 p(x)|f(x)g(x), + p(x)|g 2 (x). S6 p(x)  K } UB*&￾: p(x)|g(x). C: p(x)|f(x) + g(x), F; p(x)|f(x). H g p(x)  f(x), g(x) !2697￾: (f(x), g(x)) 6= 1. ✷ ^ 4 ~ f(x) ∈ K[x], a ∈ K. _ y = x − a, g(y) = f(y + a). Jg f(x) D K }UB!/J g(y) D K }UB
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~ f(x)   n. 7 f(x) D K }UB￾:D K } '> n !*& u(x), v(x), f(x) = u(x)v(x),  x 9 y + a B￾ g(y) = s(y)t(y), qQ s(y) = u(y + a), t(y) = v(y + a). ~ f(x) !& anx n , IGK g(y) ! & any n , H g degg(y) = degf(x) = n. WU degs(y) = degu(x) < n, degt(y) = degv(x) < n. HPJg℄ g(y) D K }0UB
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