myisland8132 發問於 科學及數學數學 · 1 十年 前

Question about Q(cos(2π /n))

The following question is related to Q(cos(2π /n)) and Q(ζn) where ζn is the n th root of unity.

(a) Using the fact that 2cos(2π /n)= ζn +ζn-1 , or otherwise, show that all the numbers cos(2π /n), cos(4π /n), cos(6π /n)… belong to the field Q(cos(2π /n))

(b) Show that (Q(ζn): Q(cos(2π /n)))=2 for n>2, and hence show that the degree of cos(2π /n) is φ(n)/2. Use this to prove that cos(2π /7) and cos(π /9) has degree 3

(c) For which integer n is cos(2π /n) rational

(d) Use (b) to show that cos(2π /n) is of degree 3 only for n=7,9,14,18. Also find the n for which cos(2π /n) is of degree 4.

(e) Show that any isomorphisms of Q(cos(2π /n) are isomorphisms of Q(ζn).

(f) Show that for each k relatively prime to n there is an automorphism σk of Q(cos(2π /n) defined by σk(cos(2π /n))=cos(2kπ /n), and that every automorphisms of Q(cos(2π /n) is of this form.(compared with (e))

(g) How many of the automorphisms σk of Q(cos(2π /n) are actually distinct? Compare this number with the value of (Q(cos(2π /n) : Q) found in (b)

(h) Use (g) or otherwise, find fields Q(cos(2π /n) with automorphism groups of three and five elements.

更新:

(i) Consider the effect of restricting the automorphism of Q(ζn) to Q(cos(2π /n)) (compare with (f)) What is the kernel of the restriction map ?

4 個解答

評分
  • Ivan
    Lv 5
    1 十年 前
    最佳解答

    好長@@“”

    冇人答我可以後補

    2009-02-11 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?

    2009-02-12 01:52:15 補充:

    (a)

    By induction:

    ζn +ζn-1 is in the field.

    ζnk +ζn-k = (ζn +ζn-1)k - multiplies of (ζnr+ζn-r) with lower r, is in the field.

    hence cos(2kπ /n) = ζnk +ζn-k 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 +ζn-1 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(n-k)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 --> ζn-1

    2009-02-12 01:53:54 補充:

    我相信(e)跟(f)有更好的方法做

    有錯漏請指正

    資料來源: PhD Math
  • Jacob
    Lv 6
    1 十年 前

    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!=]

  • 1 十年 前

    YES﹐這條習題非常有用

    2009-02-11 15:27:25 補充:

    我估yes﹐因為去到那個chapter都未說到automorphism

    2009-02-12 18:59:50 補充:

    good

  • 貓朋
    Lv 5
    1 十年 前

    Cyclotomic field ?????!!

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