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        riddles >> medium >> prove that
        
(Message started by: tony123 on Mar 5th, 2008, 9:25am)
Title: prove that
Post by tony123 on Mar 5th, 2008, 9:25am
cos(pi/14)cos(3pi/14)cos(5pi/14)=sqrt7/8
Title: Re: prove that
Post by Icarus on Mar 5th, 2008, 7:26pm
A little clarification:

cos(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/14) cos(3http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/14) cos(5http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/14) = 71/2 / 8.

By "an amazing coincidence",

cos(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/14) + cos(3http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/14) - cos(5http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/14) = 71/2 / 2.
Title: Re: prove that
Post by Aryabhatta on Mar 6th, 2008, 1:20pm
I think complex numbers will handle this pretty easily. I have tried to avoid using complex numbers... but the ideas are similar I guess

[hide]

Notice that the product is positive. Now square it

and use cos2x = (cos2x+1)/2

We now are looking for

(1+ cos pi/7) (1 + cos 3pi/7) (1 + cos 5pi/7)/8

For x = pi/7, 3pi/7, 5pi/7 and 7pi/7

notice that they all satisfy

cos 3x + cos 4x = 0

Treating this as a polynomial in y = cosx we get:

P(y) = (-3y + 4y3 + 1 - 8y2 + 8y4 )/8 = 0

(we divide by 8 to get a monic polynomial)

This is a 4th degree polynomial whose only roots are cos x for x = pi/7, 3pi/7, 5pi/7 and 7pi/7

i.e P(y) = (y+1)(y - cos pi/7)(y- cos 3pi/7)(y - 5pi/7)

Let H(y) = (y - cos pi/7)(y- cos 3pi/7)(y - 5pi/7)

i.e. P(y) = (y+1) H(y)
Now differentiate wrt y

P'(y) = H'(y)(y+1) + H(y)   ---- ( 1 )

We are interested in finding -H(-1)/8.

Put y = -1 in ( 1 ) we see that

-H(-1) = -P'(-1)

Now -P'(y)  = (32y3 + 12y2 - 16y - 3)/8

So -P'(-1) = (32 - 12 - 16 + 3)/8 = 7/8

Thus -H(-1)/8 = 7/64.
Thus the required answer is sqrt(7/64) = sqrt(7)/8

[/hide]
Title: Re: prove that
Post by Eigenray on Mar 6th, 2008, 3:30pm
We have more generally that

p = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/prod.gifk=1n cos[http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif(2k-1)/(4n+2)] = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif[2n+1]/2n,

since if we write [hide]cos[(2n+1)z]/cos[z] = F(cos z), where F is a polynomial of degree 2n, then the product of the roots of F is (-1)n p2[/hide].

On the other hand,

http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/prod.gifk=1n sin[http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif(2k-1)/(4n+2)] = 1/2n,

http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/prod.gifk=1n cos[http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif(2k-1)/(4n)] = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/prod.gifk=1n sin[http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif(2k-1)/(4n)] = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif2/2n

are not quite as interesting.
Title: Re: prove that
Post by Eigenray on Mar 6th, 2008, 3:36pm
Let m=2n+1.  The first equality shows that http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gifm) is contained in the totally real subfield http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif(cos[2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/(4m)]) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gif+1/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gif) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gif)http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/cap.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbr.gif of http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gif), where http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gif=http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gif4m is a primitive 4m-th root of unity.  We have Gal(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gif)/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif) ~= (Z/(4m))*, and Gal(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gif)/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gif+1/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gif)) ~= {http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pm.gif1} under the same isomorphism.  So Gal(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gif+1/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gif)/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif) ~= (Z/(4m))*/{http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pm.gif1} ~= (Z/m)*.  So http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gif+1/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gif) contains 2r-1 (real) quadratic subfields, where r is the number of distinct prime factors of m.  For example, we can express http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif3, http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif5, and http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif15 as polynomials in cos(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/30).  [If p is an (odd) prime factor of m, then http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gifp http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif(cos[2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/(4p)]) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gif4p)http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/cap.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbr.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subseteq.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gif4m)http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/cap.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbr.gif = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif(cos[2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/(4m)]).]

Now, Gal(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gif4)/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif) ~ (Z/m)* as well, so http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gif4) has the same number of quadratic subfields (and, moreover, an isomorphic lattice of subfields) as http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gif+1/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gif), but they won't all be real; e.g., we can express http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif{-3}, http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif5, and http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif{-15} as polynomials in http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gif4 = e2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gifi/15, but not http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif3, say.

And finally, Gal(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gif)/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif) ~= (Z/(4m))* ~= (Z/4)* x (Z/m)*, so we can find 2r+1 - 1 quadratic subfields of http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gif); e.g., we can express any product of i=http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif{-1}, http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif3, and http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif5 as polynomials in http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gif=e2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gifi/60.
Title: Re: prove that
Post by Aryabhatta on Mar 8th, 2008, 3:53am
Your posts always have me running to my textbooks Eigenray! (That is a good thing   :) ).
Title: Re: prove that
Post by Hippo on Mar 8th, 2008, 10:49am
Aryabhatta: I have absolutely the same feeling unfortunately it seems to me I will end uneducated ;)
Title: Re: prove that
Post by Eigenray on Mar 8th, 2008, 10:18pm
To go into a bit more detail:

Fix an odd integer n.  A question we might ask is: for which d can we express http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gifd as a polynomial in either http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gifn = e2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gifi/n, an n-th root of unity, or maybe C4n = cos(2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/4n) = (http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gif4n+1/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gif4n)/2.

The current problem generalizes to show that for any d | n, we can express http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gifd as a polynomial in C4n.  Thus, http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gifd) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subseteq.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif(C4n).  Another argument shows that if p | n is prime, then http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif{p*} http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gifn), where p* = (-1)(p-1)/2p http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/equiv.gif 1 mod 4.  This gives at least 2r-1 distinct quadratic subfields, where r is the number of prime factors of n, in either case.

On the other hand, we can show that these are the only such d, i.e., http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif(C4n) and http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gifn) both posses only 2r-1 quadratic subfields.

To do this consider for a moment an arbitrary field extension L/K, and let G=Aut(L/K) be the group of automorphisms of L that fix K.  Given an intermediate field E, with K http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subseteq.gif E http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subseteq.gif L, we can form Aut(L/E), a subgroup of G.  Conversely, given a subgroup H of G, we can form LH, the subfield of L consisting of those elements fixed by every element of H.  Clearly E http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subseteq.gif LAut(L/E), and H http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/le.gif Aut(L/LH).

It turns out that in that case K=http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif, if L is obtained by adjoining to http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif all the roots of a polynomial with rational coefficients, then L/K is what's called a Galois extension: the operations E http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/mapsto.gif Aut(L/E) and H http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/mapsto.gif LH are inverses to each other, and give a bijection between the set of subfields of L containing K, and the set of subgroups of G=Aut(L/K).  Moreover, this is an inclusion reversing lattice isomorphism, and one of its properties is that if H=Aut(L/E), or equivalently, E=LH, then [E:K] = [G:H].  So there is a bijection between subfields of L, quadratic over K, and subgroups of G of index 2.

In our case, let m be arbitrary.  Then L=http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gifm) is Galois over http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif, since it's the splitting field of the polynomial zm-1.  Therefore there is a bijection between the subfields of L, and the subgroups of Aut(L/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif).  Any automorphism over L over http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif is determined by what it does to http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gif; since it must take http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gif to http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gifa for some a relatively prime to m, there are no more than http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif(m) automorphisms.  On the other hand, one has [L:http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif] = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif(m), the degree of the m-th cyclotomic polynomial, so by Galois theory, |Aut(L/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif)| = [L:http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif] = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif(m), and therefore Aut(L/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif) = (http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbz.gif/m)*.
Title: Re: prove that
Post by Eigenray on Mar 8th, 2008, 10:18pm
Another property of the Galois connection is that if N = Aut(L/E) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/trianglelefteq.gif G is a normal subgroup, then E is Galois over K, with Galois group G/N; i.e., Aut(L/K)/Aut(L/E) = Aut(E/K).  In our case, since G is abelian, every subgroup is normal, so every intermediate field is also Galois over http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif.

Take E = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif(Cm) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gif+1/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gif).  Since http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gif satisfies http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gif2 - (http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gif+1/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gif)http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gif+ 1 = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gif2 - 2Chttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gif+ 1 = 0, http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gif has degree no more than 2 over E.  On the other hand, E http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subseteq.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbr.gif, so http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/notin.gif E.  Therefore [L:E]=2, and E = Lhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/cap.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbr.gif.  Note that complex conjugation is an automorphism of L over E; since |Aut(L/E)| = [L:E] = 2, this is the only non-trivial element of Aut(L/E).  Since complex conjugation takes http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gif to http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gif-1, Aut(L/E) = {http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pm.gif1}, when we identify Aut(L/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif) = (http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbz.gif/m)*.

So Aut(E/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif) = Aut(L/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif)/Aut(L/E) = (http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbz.gif/m)*/{http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pm.gif1}.  Now suppose m=4n, where n is odd.  Then (http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbz.gif/m)* ~= (http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbz.gif/4)* x (http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbz.gif/n)* by the Chinese remainder theorem, where the isomorphism is (a mod 4n) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/mapsto.gif (a mod 4, a mod n).  Note that (-1) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/mapsto.gif(-1,-1).  Define http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/phi.gif: (http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbz.gif/4)* x (http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbz.gif/n)* http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif (http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbz.gif/n)* by http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/phi.gif(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pm.gif1,x) = (http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pm.gifx).  This is clearly a surjective homomorphism with kernel <(-1,-1)>.  By the first isomorphism theorem,

Aut(E/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif) = (http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbz.gif/4n)*/{http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pm.gif1} = ((http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbz.gif/4)* x (http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbz.gif/n)*)/<(-1,-1)> ~= (http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbz.gif/n)*.

Now by Galois theory, the quadratic subfields of E correspond to the subgroups of Aut(E/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif) of index 2.  But any finite abelian group is self-dual: the subgroups of index 2 correspond (non-canonically) to the subgroups of order 2.  By the Chinese remainder theorem, if n = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/prod.gif pk, then (http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbz.gif/n)* ~= http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/prod.gif(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbz.gif/pk)*.  So the elements of order 2 correspond to picking an element of order 1 or 2 in each factor, but not all 1.  Since each factor is cyclic of even order pk-1(p-1), the number of elements of order 2 is exactly 2r-1.

The same is true for http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gifn), since Aut(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gifn)/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif) = (http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbz.gif/n)* as well.  In this case the quadratic subfields are given by http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gifd*), where d | n, and d* = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pm.gifd http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/equiv.gif 1 mod 4, whereas the quadratic subfields of http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif(cos(2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/4n)) are given simply by http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gifd), with d|n.
Title: Re: prove that
Post by Eigenray on Mar 8th, 2008, 10:19pm
With a bit more work one can consider the case http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gifm) for m even, too.  The general case http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif(cos(2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/m)) is trickier: what is (http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbz.gif/m)*/<-1>?

But we don't actually need to determine the group.  An element of order 1 or 2 in the quotient corresponds to an element of order 1,2, or 4 in (http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbz.gif/m)*.  Write m = 2an, where n is odd, with r distinct prime factors.  The number of elements of order 1 or 2 is 2r, if a<2; 2r+1 if a=2; and 2r+2 if a>2.  The number of elements of order 4 is 0 if -1 is not a square mod m, and 2r otherwise.  But in the quotient, we only have half as many elements (for m>2).  So the number of quadratic subfields of http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif(cos(2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/m)) is the number of elements of (http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbz.gif/m)*/<-1> of order 2, which is:

a=0,1: 2r-1-1 if (-1|n)=-1,   2r-1 if (-1|n)=1.
a=2: 2r-1
a>2: 2r+1-1.

If n is odd, then http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gifn)=http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gif2n), so their intersections with http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbr.gif are equal as well.  So the result is the same for a=0 as a=1.

Let p1,...,ps,q1,...,qt be the primes dividing n, with pi=1 mod 4, qj=3 mod 4.  Then the quadratic subfields of http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gifn) or http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gif2n) are 'generated' by http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gifpi and http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif{-qj}, for a total of 2s+t-1.  The quadratic subfields of http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif(cos(2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/n)) or http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif(cos(2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/2n)) are exactly those of http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gifn) which are real, and these are 'generated' by http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gifpi, and http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif{q1qj}, 1<jhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/le.gif t, and there are 2s+max(t-1,0)-1 of these.  If t=0, then (-1|n)=1, and this is 2r-1.  If t>0, then (-1|n)=-1, and this is 2r-1-1.

If a=2, i.e., m=4n, then the quadratic subfields of http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gifm) are generated by I, http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gifpi, and http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif{-qj}, for a total of 2r+1-1.  Since http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif{-1}http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/in.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gif4n), the real ones are generated by http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gifpi, http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gifqj, for a total of 2r-1.

Finally, if a>2, i.e., 8|m, then we also get http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif2 in both http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gifm) and http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/bbq.gif(cos(2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/m)), so the number of quadratic subfields are 2r+2-1 and 2r+1-1, respectively.

For example:
m=2: cos(2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/2)=-1, degree 1.  a=1,n=1,r=0, #QS = 20-1=0.
m=3: cos(2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gifi/3)=-1/2, degree 1.  a=0,n=3, r=1, (-1|n)=-1, #QS = 21-1-1=0.
m=4: cos(2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gifi/4)=0, degree 1.  a=2,n=1, r=0, #QS = 20-1.
m=5: cos(2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gifi/5)=(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif5-1)/4, degree 2.  a=0,n=5, r=1, (-1|n)=1, #QS = 21-1=1.
m=6: cos(2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gifi/6)=1/2, degree 1.  m=2*3, same field as m=3.
m=7: cos(2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gifi/7) has degree 3.  a=0,n=7, r=1, (-1|n)=-1, #QS = 21-1-1=0.
m=8: cos(2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gifi/8)=1/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif2. a=3,n=1,r=0.  #QS = 20+1-1 = 1.
m=9: cos(2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gifi/9), degree 3.  a=0,n=9, r=1, (-1|n)=-1, #QS = 21-1-1 = 0.
m=10: cos(2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gifi/10)=(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif5+1)/4, degree 2, same field as m=5.
m=12: cos(2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gifi/12)=http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif3/2, degree 2, a=2,n=3, r=1, #QS=21-1=1.


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