系10.设gcd(f(x),f(x)=1且f1(x)|g(x),f2(x)!(x), 则f1(x)2(x)!(x) 证明:设g(x)=f1(x)s(x)=f2(x)t(x).由 于gcd(f(x),f(x)=1,存在u(x),v(x)使 f1(x)u(x)+f2(x)(x)=1 从而 9( 9(x)[f1(x)u(x)+f2(x)v(x) g(a)fi(ru(a)+g()f2(a)u(a) f2(x)t(x)f1(x)u(x)+f1(x)s(x)2(x)(x) fi() f2( )t(a)u(a)+s(a)u(a)
X10. gcd(f1(x), f2(x)) = 1 f1(x)|g(x), f2(x)|g(x), Kf1(x)f2(x)|g(x). y²µ g(x) = f1(x)s(x) = f2(x)t(x). d u gcd(f1(x), f2(x)) = 1, 3 u(x), v(x) ¦ f1(x)u(x) + f2(x)v(x) = 1. l g(x) = g(x) · 1 = g(x)[f1(x)u(x) + f2(x)v(x)] = g(x)f1(x)u(x) + g(x)f2(x)v(x) = f2(x)t(x)f1(x)u(x) + f1(x)s(x)f2(x)v(x) = f1(x)f2(x)[t(x)u(x) + s(x)v(x)]. ✷
系11.设gcd(f(x),g(x)=1且f(x)lg(x)h(x), 则f(x)h(x) 证明:存在u(x),v(x)使 1=f(a)u(a)+g(ru() 两边同乘h(x)得 h(a)=f(c)u(rh(a)+g(ah(a)u(a) 陈习自己独立证明书上推论(5.3.3)⑤534)
X11. gcd(f(x), g(x)) = 1 f(x)|g(x)h(x), K f(x)|h(x). y²µ3 u(x), v(x) ¦ 1 = f(x)u(x) + g(x)v(x). ü>Ó¦ h(x) h(x) = f(x)u(x)h(x) + g(x)h(x)v(x). ✷ [öS] gCÕáy²ÖþíØ(5.3.3),(5.3.4)
例6.设K,K是两个数域且满足KCK.设f(x),9(x)∈ Kx-则f(x)在Kx]中整除g(x)当且仅当f(x) 在Kc]中整除9(x 证明:→:f(x)在K]中整除g(x)意味着存 在h(x)∈K]使g(x)=f(x)h(x).然而h(x)∈ Kc小.所以f(x)在K]中也整除g(x :在Kx]中作带余除法得 (a)=f(aq()+r(a) 其中deg(r)<deg(f)由于q(x),r(x)也属于Kxl, 在K]中f(x)除g(x)的余式也是r(x).从而r(x) 0.所以f(x)在K]中整除g(x)口
~6. K, K0 ´üê ÷v K ⊂ K0 . f(x), g(x) ∈ K[x]. K f(x) 3 K[x] ¥Ø g(x) = f(x) 3K0 [x] ¥Ø g(x). y²µ⇒: f(x) 3 K[x] ¥Ø g(x) ¿X 3 h(x) ∈ K[x] ¦ g(x) = f(x)h(x). , h(x) ∈ K0 [x]. ¤±f(x) 3 K0 [x] ¥Ø g(x)" ⇐: 3 K[x] ¥{Ø{ g(x) = f(x)q(x) + r(x), Ù¥ deg(r) < deg(f). du q(x), r(x) áu K0 [x], 3K0 [x] ¥ f(x) Ø g(x) {ª´ r(x). l r(x) = 0. ¤± f(x) 3 K[x] ¥Ø g(x). ✷
例7.设f(x),g(x,h(x)∈K团d]为非零多项式,h(x) 被g(x)除的余式是个非零常数c,f(x)被g(x 除的余式是r(x),则f(x)h(x)被g(x)除的余式 是c:r(x) 证明:由 f(x)=9(x)q(x)+r(x) 和 h(a)=g(a)q1() 得 f(x)hb(x)=9(x)q(x)q1(x)9(x)+cq(x)+r(x)q(x)+cr(x) 而deg(c:r(x)=deg(r(x)<deg(g(x),所以c r(x)是f(x)h(x)被g(x)除的余式。口
~7. f(x), g(x), h(x) ∈ K[x] "õª§ h(x) g(x) Ø{ª´"~ê c, f(x) g(x) Ø{ª´ r(x), Kf(x)h(x) g(x) Ø{ª ´ c · r(x)" y²µd f(x) = g(x)q(x) + r(x) Ú h(x) = g(x)q1(x) + c f(x)h(x) = g(x)[q(x)q1(x)g(x)+c·q(x)+r(x)q1(x)]+c·r(x). deg(c · r(x)) = deg(r(x)) < deg(g(x)), ¤± c · r(x) ´f(x)h(x) g(x) Ø{ª"✷
定理12(中国剩余定理).设 q1(x),,9n(x)是一组两两互素的非零多项 式。对于任给的多项式 (x),,an(x) 假如deg(a(x)<deg(91(x)对1≤i≤m都成 立,则存在一个多项式f(x),使得对每个i,多项 式f(x)被g(x)除的余式恰好是a;(x) 证明:记h(x)=91(x)…91-1(x)9+1(x)…9n1(x) 则9(x)与h(x)互素。于是存在u(x),v(x)使 91(x)1(x)+h(x)v(x) 移项得 h:(x)v(x)=1-9(x)u1(x 令f(x)=h1(x)1(x)/a1(x)+…+hn(x)n(x)an(x) 对每个i,多项式∫(x)均可写成 f(a)=hi(a)vi()ai(a)+gi()b(a) a:(x)+9(x){b(x)-v1(x) 这表明f(x)被9(x)除的余式恰好是a(x).口
½n12 ( ¥I{½n). g1(x), . . . , gn(x) ´|üüp"õ ª"éu?õª a1(x), . . . , an(x), bX deg(ai(x)) < deg(gi(x)) é 1 ≤ i ≤ n Ѥ á§K3õª f(x), ¦éz i, õ ª f(x) gi(x) Ø{ªTд ai(x). y²µP hi(x) = g1(x)· · · gi−1(x)gi+1(x)· · · gn(x). K gi(x) hi(x) p"u´3 ui(x), vi(x) ¦ gi(x)ui(x) + hi(x)vi(x) = 1. £ hi(x)vi(x) = 1 − gi(x)ui(x) -f(x) = h1(x)v1(x)a1(x) + · · · + hn(x)vn(x)an(x). ézi, õª f(x) þ¤ f(x) = hi(x)vi(x)ai(x) + gi(x)b(x) = ai(x) + gi(x)[b(x) − ui(x)]. ùL² f(x) gi(x) Ø{ªTд ai(x). ✷