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# 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.

(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 Answers

- IvanLv 51 decade agoFavorite Answer
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

Source(s): PhD Math- Login to reply the answers

- JacobLv 61 decade ago
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！=]

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- myisland8132Lv 71 decade ago
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

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