[功課] 3.96(ii)

[jacky7987 (憶)]

如果有閒人會看到這篇的話:

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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※ 編輯: jacky7987       來自: 123.193.89.201       (06/01 01:03)
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