Best Answer:
厄~~

如果他真的寫的這麼少，就真的有點難懂了

要證明A/A1=A2+...+As是direct sum

需要一個輔助定理:

Given any x belong to A/A1, and x has period p^r. There exist a x' belong to A, and x' also has period p^r.

pf: Given any x belong to A/A1, and has period p^r. There exist a x1 belong to A, such that (p^r)x1=na1

if n=0, trivial

if not, write n as (p^k)t, s.t. (p,t)=1 Since t is relative prime to p, ta1 also has period p^r1.

we have (p^r)x1=(p^k)ta1 imply p^(r+r1-k)x1=0

notice that (r+r1-k)<=r1, so r<k.

we have (p^r)x1=(p^r)[p^(k-r)]ta1, notice that [p^(k-r)]ta1,said c, belong to A1.

If x1 belong to A1, trivial

If not, there exist (p^r)(x1-c)=0 and (period of x1-c)<p^r

(p.s. remember x has period p^r)

choose x'=x1-c, and we have done.

證完以後就可以用induction了

最後再補充一下

如果(A2/A1)和(A3/A1)有非0 交集(這裡A2/A1就是你所說的A2,不要搞混囉)

讓order of (A2/A1)>= (A3/A1)好了, let (A3/A1) genetated by a3=(a3)'A1

再用一次上面的lemma, let A/A2 be the factor group, (a3)'A2 有period p^r

則存在 (a3)''屬於A 有period p^r

那我們就選一個新的A3/A1, generated by (a3)''A1

不就是direct sum了嗎?

by induction 可以一直不斷的做下去, 因為是finite group,總會做完

它的意思就是降

2010-01-25 00:05:37 補充：

他媽的,預覽還是好的,只要有prime就變亂碼 氣死我了

2010-01-25 00:08:53 補充：

厄~~

如果他真的寫的這麼少，就真的有點難懂了

要證明A/A1=A2+...+As是direct sum

需要一個輔助定理:

Given any x belong to A/A1, and x has period p^r. There exist a x' belong to A, and x' also has period p^r.

pf: Given any x belong to A/A1, and has period p^r. There exist a x1 belong to A, such that (p^r)x1=na1

if n=0, trivial

2010-01-25 00:09:49 補充：

if not, write n as (p^k)t, s.t. (p,t)=1 Since t is relative prime to p, ta1 also has period p^r1.

we have (p^r)x1=(p^k)ta1 imply p^(r+r1-k)x1=0

notice that (r+r1-k)<=r1, so r

2010-01-25 00:10:42 補充：

notice that (r+r1-k)<=r1, so r

2010-01-25 00:12:14 補充：

notice that (r+r1-k)<=r1, so r<=k.

we have (p^r)x1=(p^r)[p^(k-r)]ta1, notice that [p^(k-r)]ta1,said c, belong to A1.

If x1 belong to A1, trivial

If not, there exist (p^r)(x1-c)=0 and (period of x1-c)

2010-01-25 00:17:51 補充：

(p.s. remember x has period p^r)

choose x``;=x1-c, and we have done.

證完以後就可以用induction了

最後再補充一下

如果(A2/A1)和(A3/A1)有非0 交集(這裡A2/A1就是你所說的A2,不要搞混囉)

讓order of (A2/A1)>= (A3/A1)好了, let (A3/A1) genetated by a3=(a3)``A1

再用一次上面的lemma, let A/A2 be the factor group, (a3)``A2 有period p^r

則存在 (a3)``屬於A 有period p^r

2010-01-25 00:18:26 補充：

那我們就選一個新的A3/A1, generated by (a3)``A1

不就是direct sum了嗎?

by induction 可以一直不斷的做下去, 因為是finite group,總會做完

它的意思就是降

2010-01-25 00:19:44 補充：

不好意思我打了一個多小時

預覽還是好的

貼上來就全部都變亂碼

用補充的變得很亂

我覺得這是yahoo的問題

預覽看到的和po出來的應該要一樣

Source(s):
Undergraduate Algebra