Example
The situation is simplest when n is a prime number p > 2. In that case G is cyclic of order p − 1, and has one subgroup H of order d for every factor d of p − 1. For example, we can take H of index two. In that case H consists of the quadratic residues modulo p. Corresponding to this H we have the Gaussian period
summed over (p − 1)/2 quadratic residues, and the other period P* summed over the (p − 1)/2 quadratic non-residues. It is easy to see that
since the left-hand side adds all the primitive p-th roots of 1. We also know, from the trace definition, that P lies in a quadratic extension of Q. Therefore, as Gauss knew, P satisfies a quadratic equation with integer coefficients. Evaluating the square of the sum P is connected with the problem of counting how many quadratic residues between 1 and p − 1 are succeeded by quadratic residues. The solution is elementary (as we would now say, it computes a local zeta-function, for a curve that is a conic). One has
- (P − P*)2 = p or −p, for p = 4m + 1 or 4m + 3 respectively.
This therefore gives us the precise information about which quadratic field lies in Q(ζ). (That could be derived also by ramification arguments in algebraic number theory; see quadratic field.)
As Gauss eventually showed, to evaluate P − P*, the correct square root to take is the positive (resp. i times positive real) one, in the two cases. Thus the explicit value of the period P is given by
Read more about this topic: Gaussian Period
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