Yahoo Answers: Answers and Comments for Question about Q(cos(2π /n)) [數學]
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From myisland8132
enUS
Fri, 06 Feb 2009 19:18:29 +0000
3
Yahoo Answers: Answers and Comments for Question about Q(cos(2π /n)) [數學]
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From Ivan: 好長＠＠“”
冇人答我可以後補
20090211 13:50...
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https://answers.yahoo.com/question/index?qid=20090206000051KK01471
Thu, 12 Feb 2009 01:52:15 +0000
好長＠＠“”
冇人答我可以後補
20090211 13:50:28 補充：
Part e)
是想問: 所有 *automorphism* of Q(cos(2π /n) 都是某個 *automorphism* of Q(ζn) 的 restriction 嗎?
20090212 01:52:15 補充：
(a)
By induction:
ζn +ζn1 is in the field.
ζnk +ζnk = (ζn +ζn1)k  multiplies of (ζnr+ζnr) with lower r, is in the field.
hence cos(2kπ /n) = ζnk +ζnk is in the field.
(b)
(Q(ζn): Q(cos(2π /n)) > 1 because ζn is complex but cos(2π /n) is real.
Since 2cos(2π /n)ζn= ζn2 +1, ζn satisfies an irreducible (when n>2) quadratic equation of degree 2 with coefficients in Q(cos(2π /n))
Hence (Q(ζn): Q(cos(2π /n)))=2.
Since (Q(ζn): Q) = φ(n), degree of cos(2π /n) is φ(n)/2 by subfield properties.
φ(7) = φ(18) = 6. easily checked. (φ(n) = #k<n relatively prime to n)
(c)
rational means Q(cos(2π /n)) = Q, i.e. φ(n)=2 when n>2.
Only n = 3, 4, 6. n=2 also.
(d)
Using formula for Euler phi function:
φ(n)=6:
n can only have 2,3,7 as prime factors,2,7 appears at most once.
Check: only when n=7,9,14,18
φ(n)=8:
n can only have 2,3,5 as prime factors,3,5 appears at most once. Check: only when n=16,20,24
(e)
Let K=Q(cos(2π /n), then Q(ζn) = K[ζ] where ζ is degree 2.
An isomorphism s of K induces an isomorphism of K[ζ] by sending k to s(k) and ζ to ζ' where ζ' satisfy the same minimal polynomial of ζ. However, since ζ is root of unity, so is ζ', hence K[ζ] = K[ζ'] and we get isomorphism up there.
(f)
By part (e), any isomorphism is defined by isomorphism from Q(ζn). However, isomorphism in Q(ζn) is only defined by ζn > ζnk where k is relative prime to n. Restricting to Q(cos(2π /n)) , we see that cos(2π /n) = ζn +ζn1 maps to cos(2kπ /n). Since cos(2kπ /n) belongs to Q(2π /n) by part (a), this is an isomorphism of Q(2π /n).
(g)
There are φ(n) distinct number < n that is relative prime to n. But cos(2kπ /n) = cos(2(nk)π /n), hence half of them are repated.
So there are φ(n) /2 distinct isomorphisms, same as part (b).
(h)
Same as finding φ(n) = 6 and 10. Using technique in part d):
φ(n) = 6: n=7,9,14,18
φ(n) = 10: n=11
(i)
The kernel are the automorphism that fixes cos(2π /n).
Hence they correspond to ζn > ζn and ζn > ζn1
20090212 01:53:54 補充：
我相信(e)跟(f)有更好的方法做
有錯漏請指正

From Jacob: Abstract Algebra我都好鍾意，可惜未學到呢樣野……...
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https://answers.yahoo.com/question/index?qid=20090206000051KK01471
Sat, 07 Feb 2009 08:40:48 +0000
Abstract Algebra我都好鍾意，可惜未學到呢樣野……=3=
呢兩日睇下D書自修下，如果幾日都冇人答我就試下啦！=]

From myisland8132: YES﹐這條習題非常有用
20090211 15:27:25 補充：
我估yes﹐因...
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https://answers.yahoo.com/question/index?qid=20090206000051KK01471
Fri, 06 Feb 2009 22:58:48 +0000
YES﹐這條習題非常有用
20090211 15:27:25 補充：
我估yes﹐因為去到那個chapter都未說到automorphism
20090212 18:59:50 補充：
good

From 貓朋: Cyclotomic field ?????!!
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https://answers.yahoo.com/question/index?qid=20090206000051KK01471
Fri, 06 Feb 2009 21:18:51 +0000
Cyclotomic field ?????!!