系10. 设 gcd(fi(α), f2(α)) = 1 且 fi(α)lg(α), f2(α)lg(α),则 fi(α)f2(α)lg(αc).证明： 设 g(α) = fi(α)s(α) = f2(c)t(a). 由于 gcd(fi(α), f2(c)) = 1, 存在 u(c), (α) 使fi(α)u(c) + f2(c)v(c) = 1.从而g(r) = g(α) · 1= g(α)[fi(α)u(α) + f2(c)v(c)= g(α)fi(α)u(α) + g(α)f2(α)v(c)= f2(α)t(α)fi(α)u(α) + fi()s(c)f2(α)v(α)= fi(α)f2(α)[t(c)u(α) + s(α)(α)l口
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(α),g(c)) = 1 且 f(c)lg(c)h(c),则 f(c)|h(α).证明：存在u(α),u(α)使1 = f(α)u(α) + g(α)v(c)两边同乘h(α）得h(c) = f(c)u(c)h(c) + g()h(c)v(c).口[练习】自已独立证明书上推论(5.3.3),(5.3.4)
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(α),g(c)EKal.则f(α)在Kcl中整除g(a)当且仅当f(α)在K'[] 中整除 g(α).证明：→:f(α)在K[) 中整除 g(α)意味着存在 h(α) E K[al 使 g() = f(α)h(a). 然而 h(c) EK'[cl.所以f(c)在K'[al中也整除 g(α)。←：在K[al中作带余除法得g(α) = f(α)q(α) +r(α),其中 deg(r)< deg(f). 由于q(α),r(c)也属于 K'[],在Kal中f(α)除g(α)的余式也是r(α).从而r(α)=0.所以f(α)在K[cl中整除g(α)
~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(α),g(α),h(α) EK[al为非零多项式，h(α)被 g(α)除的余式是个非零常数 c,f(α)被 g(α)除的余式是r(c),则f(α)h(α)被 g(α)除的余式是c.r(c)。证明：由f(α) = g()q(α) +r(α)和h(α) = g(α)qi(α) + c得f(α)h() = g(α)[q()qi(α)g(α)+c.q(α)+r()qi(c)+cr()而 deg(c · r(α)) = deg(r(α)) < deg(g(c), 所以 c :r(α)是f(α)h(α)被g(α)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91(α)，..:,9n(c）是一组两两互素的非零多项式。对于任给的多项式ai(c),...,an(c)假如 deg(ai(α)< deg(gi(α)对 1 ≤i≤ n 都成立，则存在一个多项式f(c)，使得对每个i，多项式f(α)被 gi(a)除的余式恰好是ai(α).证明：记h;(α) = g1(c)..·gi-1(c)gi+1(α).·gn(α)则 g(α)与hi(α)互素。于是存在 u(α),U()使gi;(c)ui(c) + hi(c)vi(α) = 1.移项得h;(α)v;(r) = 1 - g;(α)u;(α令f(c) = hi(c)vi(c)ai(c)+. .. + hn(c)un(c)an(c)对每个i，多项式f(α）均可写成f(α) = hi(α)vi(c)ai(c) + gi(α)b(c)= ai(α) + gi;(α)[6(c) - u;(α)]这表明f(α）被gi()除的余式恰好是a()
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). ✷