作者jacky7987 (忆)
看板NCCU08_Math
标题[功课] 3.96(ii)
时间Wed Jun 1 00:55:25 2011
如果有闲人会看到这篇的话:
Claim: F_p has no non-trivial proper subfield
If not, there exists a non-trivial subfield k contained in F_p.
Hence, k is a additive subgroup of F_p.
By Lagrange's theorem |k| divides |F_p|.
However,|k|=1 or p, since k is not trivial;hence, k=F_p.
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Back to 3.96 (ii):
If k is the prime field of K;therefore, k is isomorphic to a subfield of F_p
and F_q, where p≠q.
By the claim, k is isom. to F_p and F_q, which implies F_p is isom to F_q.
But |F_p|=p≠q=|F_q| →←
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And the proof of 3.96 (i) can be as follow:
k is isom to subfield of F_p and by the claim, k is isom to F_p.
Hence, K is char p, but k' is isom to Q and char(Q)=0 →←
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※ 编辑: jacky7987 来自: 123.193.89.201 (06/01 01:03)
1F：推 kevin89126:感谢~ 06/01 03:26