# Gaussian Periods

Let \(\zeta = e^{{2\pi i / 17}}\) be a primitive 17th root of unity. Define \(x_1, x_2\) by

The exponents of the terms of \(x_1\) are the quadratic residues and those of \(x_2\) are the nonresidues.

These are examples of *Gaussian periods*.
We stated before that \(x_1 x_2\) is a rational that is easy
to compute. Why?

Write \((x_1 - x_2)^2\) as

for integers \(a_i\). Replacing \(\zeta\) with \(\zeta^3\) merely swaps \(x_1\) and \(x_2\) due to their construction, whence

Thus looking at the coefficients of the powers of \(\zeta\) we have \(a_{16} = a_1, a_1 = a_2, ..., a_{15} = a_{16}\), that is, they are all equal to some integer \(a\):

Thus \(x_1 x_2 = ((x_1+x_2)^2 - (x_1-x_2)^2)/4\) is some rational number. (Recall \(x_1 + x_2 = -1\).)

Each of \(x_1\) and \(x_2\) consists of 8 terms so their product \(x_1 x_2\) has 64 terms of the form \(\zeta^k\) for some \(k\). Since \(17 = 1 \pmod{4}\) we know \(-1\) is a quadratic residue, hence if \(r\) is a quadratic residue so is \(-r\). As each \(k\) can be viewed as the sum of a residue and a nonresidue modulo 17, this means \(k\) is never zero.

Hence if we write:

we must have \(b_0 = 0\), and the 64 terms must be evenly distributed among \(\zeta ,..., \zeta^{16}\), that is, \(b = 4\). Therefore \(x_1 x_2 = -4\). (Thanks to Dennis Westra for bringing this argument to my attention.)

Exercise: what happens when \(-1\) is a quadratic nonresidue?

We can generalize these statements.

## A Loose End

Before we argued that given

we can equate coefficients of the powers of \(\zeta\). Let us see why. In the above equation, we can move all terms to one side then divide through by \(\zeta\) to find \(b_i\) such that

(each \(b_i\) is the difference between some \(a_j\) and \(a_k\)). Then consider the polynomial \(g(x) = b_1 + b_2 x + ... + b_{16} x^{15}\). Now \(\zeta\) is a root of \(g\) as well as the polynomial

Hence \(\zeta\) must also be a root of \(d = \gcd(f, g)\). If \(g \ne 0\), then the polynomial \(d\) is a nonzero polynomial dividing \(f\) with degree at most \(g\), which is smaller than the degree of \(f\). This is a contradiction since \(f\) has no factors with rational coefficients (by Eisenstein).

Thus \(g\) must be the zero polynomial, and we may equate the coefficients of the powers of \(\zeta\).

[In abstract algebra, we say all this in one line: \(\mathbb{Q}[\zeta]\) is a degree 16 extension of \(\mathbb{Q}\) thus \(g = 0\).]

*blynn@cs.stanford.edu*💡