引理7.设fi(α)，..,fn(α)EK[cl 全不为零。假定n>2. 设1 <r< n. 记di(α) = gcd(fi(c), ..., fr(r)),d2(c) = gcd(fr+1(c), .. ., fn(c),d(α) = gcd(di(α), d2(c)).则d(α)是fi(αc)，..,fn(α)的最大公因式。证明：由 d(a)ldi(c),d(a)ld2(c) 得知 d(α) 整除每个fi(α).假定h(a)整除每个f;(α),则h(α)整除fi(c),..,fr(c中任何一个，因而 h(a)ldi(α).同理h(α)ld2(c).所以h(α)|d()
Ún7. f1(x), . . . , fn(x) ∈ K[x] Ø""b ½ n > 2. 1 ≤ r < n. P d1(x) = gcd(f1(x), . . . , fr(x)), d2(x) = gcd(fr+1(x), . . . , fn(x)), d(x) = gcd(d1(x), d2(x)). K d(x) ´ f1(x), . . . , fn(x) úϪ" y²µd d(x)|d1(x), d(x)|d2(x) d(x) Ø z fi(x). b½ h(x) Øz fi(x), K h(x) Ø f1(x), . . . , fr(x) ¥?Û§Ï h(x)|d1(x). Ón h(x)|d2(x). ¤ ±h(x)|d(x). ✷
练习设fi(c)，..,fn(c）不全为零，则它们的最大公因式存在且在相差一个非零常数的意义下唯
öS f1(x), . . . , fn(x) Ø "§K§ úϪ3 3 "~ê¿Âe "
作业：p.194: 1,2,3
µ p.194: 1,2,3
定义5.若 gcd(f(c),g(α))=1，则称 f(α)和 g(α)互素。引理8.f(α)和g(α)互素的充要条件是存在u(α),u(c) EK[c] 使f(α)u() + g(c)v() = 1.证明：→：由定义立得。←: 令 = (a(α)f(α) +b(α)g(α)la(α),b(α) EK[c]].则1ET.而1在T的非零元素中次数达到最小值，所以gcd(f(a),g(α))=1.[注|对任意一组多项式f(α),g(α),u(α),(α),d(α),即使f(c)u(a)+g(α)u(α) =d(a),也不能说d(c)=gcd(f(α), g(α)). [见习题3,p.190]
½Â5. e gcd(f(x), g(x)) = 1§K¡ f(x) Ú g(x) p" Ún8. f(x) Ú g(x) p¿^´3 u(x), v(x) ∈ K[x] ¦f(x)u(x) + g(x)v(x) = 1. y²µ⇒: d½Âá" ⇐: - Γ = {a(x)f(x) + b(x)g(x)|a(x), b(x) ∈ K[x]}. K 1 ∈ Γ. 1 3 Γ "¥gê §¤± gcd(f(x), g(x)) = 1. ✷ [5] é?¿|õª f(x), g(x), u(x), v(x), d(x), =¦f(x)u(x) +g(x)v(x) = d(x), ØU` d(x) = gcd(f(x), g(x)). [SK3,p.190]
系9.如果g(α）与fi(α),...，fn(c)中每个多项式互素，则 g(α)与fi(α)·fn(α)互素。证明：对每个i都存在i(c),(a)使f;(c)ui(α) +g(α)vi(α) = 1.移项得fi(α)ui(α) = 1 - g(c)v;(c).全部相乘得fi(c) ... fn(c)ui(c) ...un(c) = 1 - g(c)b(c)再次移项得fi(α) . .. fn(r)ui(c) -. un(α) + g(α)b(c) = 1口[注]书上推论(5.3.5)是本推论的特殊情形
X9. XJ g(x) f1(x), . . . , fn(x) ¥zõª p§K g(x) f1(x)· · · fn(x) p" y²µéz i Ñ3 ui(x), vi(x) ¦ fi(x)ui(x) + g(x)vi(x) = 1. £ fi(x)ui(x) = 1 − g(x)vi(x). ܦ f1(x)· · · fn(x)u1(x)· · · un(x) = 1 − g(x)b(x). 2g£ f1(x)· · · fn(x)u1(x)· · · un(x) + g(x)b(x) = 1. ✷ [5] ÖþíØ(5.3.5) ´ íØAÏ/