又设d=ms-nt,s,t≥0,则d+nt=ms.于是 rms-1=rd+nr-1=(ard-1)xr+xnr-1 若f(x)∈K[]满足f(x)|xm-1,f(x)|xn-1,则(f(x),x)=1,且f(x)|xm-1,f(x)|xm-1,于是 f(x)|(x4-1)xm.由f(x)与x互素可得f(x)|x2-1.因此(xm-1,xm-1)=x4-1,其中d=(m,n 1.证明只要(1,=0(的次数都大于零,就可以适当选择适合等式 u(x)f(x)+v(x)g(x)=(f(x),9(x) 的u(x)与v(x),使 deg u(a)< deg g(a) deg v(a)< deg f(r) (f(x),9(x) (f(x),9(x) 证明:存在多项式s(x),t(x)∈K[x]使 s(x)f(x)+t(x)g(x)=(f(x),9(x) s(a)z (f(x),g(r)+t(。9(x) f(a) (f(x),g(x) 令 s(x)= f(a),9(x)9(x)+(x) 其中()=0或gu()<d f q(x)+t(x),则由(*)知 (0+ 由假可(2,9(与7mm的次数都大于零,所以a(a()都不是零多项式于是 deg u(er)< deg g(r) (f(x),g(x) 由(**)知 deg(u(a) (f(x),g(x) g(()a) f(x),9(x)) 从而 degv(ar)< deg>f(r) f(r),g(a)) 习题10-4 1.设(f(x),m(x)=1,证明:对任何的多项式g(x),都存在多项式h(x),使 h(a)f(a)=g(r)(mod m(r)) 证明:由假设,存在u(x),v(x)∈K],使 u(a)f(a)+v(a)m(a)=1 g(a)u(a)f(r)+g(a)v(e)m(a)=g(a). 于是 g(r)u(a)f(=g(r(mod m())
C d = ms − nt, s, t > 0, d + nt = ms. ! x ms − 1 = x d+nr − 1 = (x d − 1)x nr + x nr − 1. f(x) ∈ K[x] f(x) | x m − 1, f(x) | x n − 1, (f(x), x) = 1, F f(x) | x ms − 1, f(x) | x nt − 1, ! f(x) | (x d − 1)x nr . f(x) x @O f(x) | x d − 1. l (x m − 1, xn − 1) = x d − 1, -* d = (m, n). ∗14. ^_: lE f(x) (f(x), g(x)) , g(x) (f(x), g(x)) y2|{, T@$P./$% u(x)f(x) + v(x)g(x) = (f(x), g(x)) u(x) v(x), N deg u(x) < deg µ g(x) (f(x), g(x))¶ , deg v(x) < deg µ f(x) (f(x), g(x))¶ . '(: DqI~ s(x), t(x) ∈ K[x] N s(x)f(x) + t(x)g(x) = (f(x), g(x)). s(x) f(x) (f(x), g(x)) + t(x) g(x) (f(x), g(x)) = 1. (*) S s(x) = g(x) (f(x), g(x)) q(x) + u(x), -* u(x) = 0 k deg u(x) < deg g(x) (f(x), g(x)) . G v(x) = f(x) (f(x), g(x)) q(x) + t(x), (*) u, u(x) f(x) (f(x), g(x)) + v(x) g(x) (f(x), g(x)) = 1. (**) 0 , f(x) (f(x), g(x)) g(x) (f(x), g(x)) y2|{, u(x), v(x) 2`!{I~. ! deg u(x) < deg g(x) (f(x), g(x)) . (**) u deg µ u(x) f(x) (f(x), g(x)) ¶ = deg µ v(x) g(x) (f(x), g(x)) ¶ , IJ deg v(x) < deg f(x) (f(x), g(x)) . % & 10–4 1. (f(x), m(x)) = 1, ^_: 4LI~ g(x), 2DqI~ h(x), N h(x)f(x) ≡ g(x) (mod m(x)). '(: 0 , Dq u(x), v(x) ∈ K[x], N u(x)f(x) + v(x)m(x) = 1. g(x)u(x)f(x) + g(x)v(x)m(x) = g(x). ! g(x)u(x)f(x) ≡ g(x) (mod m(x)). · 6 ·
h(a)f(a)=g(r)(mod m(r)) 2.设m1(x),……,m2(x)为一组两两互素的多项式,证明:对任何的多项式f1(x),……,f3(x),都存在 多项式F(x),使 F(x)≡f(x)( mod mi(x),i=1,……,s 证明令M()=m(mn()…mn(x),B()=m1(x,则(B(),m(x)=1,m()|B() i≠j.存在h(x)使(习题1) hi()ri(a)=fi(e)(mod mi(a)) 令 F(a)=∑h(x)R(x) F(a)=>hi(r)R;(r)(mod mk(a)) ≡k(x)( mod mk(x) 3.设m(x)为复系数多项式,且m(0)≠0.证明:存在复系数多项式f(x),使 f2(x)≡x(modm(x) 证明:(a)首先证明对任意的a≠0,同余式 f2(x)≡x(mod(x-a)m) 有解设√a是a的任意一个平方根,则 (r-a)m=((va-va(va+vam=(vr-vam(v+va) =((x)V-g(x)(h(x)√G+g(x)=h2(x)x-9(x) 于是 92(a)=h(r)r(mod (a-a)m) 而h(a)√a+9(a)=(va+√am≠0,而h(a)√a-g(a)=(a-√am=0.因此g(a)h(a)≠0,从而 (h(x),(x-a厘m)=1,存在h1(x)∈K[x使h1(x)h(x)≡1(mod(x-a)m).于是 (hi(r)g(a -=r (mod (r-a)m) 取∫(x)=h1(x)9(x),则有 ≠a;对i≠j.则(x-a1) 两两互素.由(a),存在f(x)∈K],使 f2(x)≡x(mod(x-a)m) 由习题2,存在f(x)使 f(a)=fi(r)(mod (a-ai"o) 于是 f2(x)≡x(mod(x-a1)
S h(x) = g(x)u(x), h(x)f(x) ≡ g(x) (mod m(x)). ∗2. m1(x), · · · , ms(x) .f]@@ I~, ^_: 4LI~ f1(x), · · · , fs(x), 2Dq I~ F(x), N F(x) ≡ fi(x) (mod mi(x)), i = 1, · · · , s. '(: S M(x) = m1(x)m2(x)· · · ms(x), Ri(x) = M(x) mi(x) . (Ri(x), mi(x)) = 1, mj (x) | Ri(x), i 6= j. Dq hi(x) N (NO 1) hi(x)Ri(x) ≡ fi(x) (mod mi(x)) S F(x) = Xs i=1 hi(x)Ri(x), F(x) ≡ Xs i=1 hi(x)Ri(x) (mod mk(x)) ≡ hk(x)Rk(x) (mod mk(x)) ≡ fk(x) (mod mk(x)). ∗3. m(x) .syI~, F m(0) 6= 0. ^_: DqsyI~ f(x), N f 2 (x) ≡ x (mod m(x)). '(: (a) !{^_45 a 6= 0, am f 2 (x) ≡ x (mod (x − a) m) $. √ a ! a 45fv/>x, (x − a) m = ((√ x − √ a)(√ x + √ a))m = (√ x − √ a) m( √ x + √ a) m = (h(x) √ x − g(x))(h(x) √ x + g(x)) = h 2 (x)x − g 2 (x). ! g 2 (x) ≡ h 2 (x)x (mod (x − a) m) J h(a) √ a + g(a) = (√ a + √ a) m 6= 0, J h(a) √ a − g(a) = (√ a − √ a) m = 0, l g(a)h(a) 6= 0, IJ (h(x),(x − a) m) = 1, Dq h1(x) ∈ K[x] N h1(x)h(x) ≡ 1 (mod (x − a) m). ! (h1(x)g(x))2 ≡ x (mod (x − a) m) Q f(x) = h1(x)g(x), $ f 2 (x) ≡ x (mod (x − a) m). (b) m(x) = (x − a1) m1 (x − a2) m2 · · ·(x − as) ms , ai 6= aj i 6= j. (x − a1) m1 , · · · ,(x − as) ms @@ . (a), Dq fi(x) ∈ K[x], N f 2 i (x) ≡ x (mod (x − ai) mi ). NO 2, Dq f(x) N f(x) ≡ fi(x) (mod (x − ai) mi ) ! f 2 (x) ≡ x (mod (x − ai) mi ) · 7 ·