[功課] 代數習題

看板NCCU08_Math作者jacky7987 (憶)時間14年前 (2011/03/24 23:20)推噓5(5推 0噓 3→)

留言8則, 7人參與討論串1/1

下面說明不是我太過高傲或是什麼只是害怕有心人士會這樣先聲明我們提供的習題解答可能都有錯寫解答是基於大家可以有所參考的答案歡迎告訴我們有哪裡錯我們會盡量更正所以不要背了之後寫考卷錯了來怪我們我們也不知道應該要生氣還是抱歉才是希望大家可以順利的通過各代數大小考謝謝大家 ===== 以下是勘誤XD 3.72 請忽略紙上的答案以下也許才是正確的 inf i Let I be any ideal in K[[x]]={f(x)=\sum a_i*x | a_i\in K} i=0 By Well-Ordering Principle ,let n={ord(f)|f\in I}, and defined ord(f(x))=N (The defn of ord, just see exercise 3.40) Claim: I=(f(x)) n First, by exercise 3.40,f(x)=x *u_1,where u_1 is an unit. Given g(x)\in I Clearly, ord(g(x))≧ord(f(x)) m Hence, g(x)=x *u_2, m≧n m n m-n Hence,g(x)=x *u_2=x *x *u_2 n -1 m-n =(x *u_1)*u_1 *x u_2 -1 m-n =f(x)*(u_1 *x *u_2)\in (f(x)) then we are done. ==== 3.77 Suppose that \pi,\beta are not relative prime i.e. gcd(\pi,\beta)=d≠1 在這後面加上以下這段 The reason why we can just simplify d≠"1" because In general,if u is an unit and u=gcd(x,y) then (u)=R=(1) since u^(-1)*u=1\in (u),and R is PID Hence,we just simplify it 我解釋一下這段因為在一般的domain裡面 gcd不是唯一的! (在Z裡面特地調成正的,在polynimail裡面特地用monic的那個) 但是在PID裡面因為兩個gcd只差一個unit(上面:u跟1只差u^(-1)) 所以By Prop.3.41這兩個生成的ideal會一樣,所以我們把他都當成一樣的再繼續下面那幾行 ==== 3.37的完整證明 (<=) By division algorithm 2 f(x)=(x-a) *q(x) + cx+d, for some q(x)\in R[x] 2 then f'(x)=2(x-a)q(x)+(x-a) *q'(x)+c Because (x-a)|f' =>f'(a)=0 => c=0 2 =>f(x)=(x-a) *q(x)+d By assumption (x-a)|f => f(a)=0 => d=0 2 therefore d=0 => (x-a) 整除f(x) -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 123.194.201.206 ※ 編輯: jacky7987 來自: 123.194.201.206 (03/24 23:29)

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