# Gaussian Periods

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

$x_1 = \zeta^{3^0} + \zeta^{3^2} +...+ \zeta^{3^{14}}$
$x_2 = \zeta^{3^1} + \zeta^{3^3} +...+ \zeta^{3^{15}}$

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

$(x_1 - x_2)^2 = a_0 + a_1 \zeta^{3^1} + a_2 \zeta^{3^2} + ... + a_{16} \zeta^{3^{16}} .$

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

$(x_2 - x_1)^2 = a_0 + a_1 \zeta^{3^2} + a_2 \zeta^{3^3} + ... + a_{16} \zeta^{3^{17}} .$

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 $$a$$:

$(x_2 - x_1)^2 = a_0 + a(\zeta + ... +\zeta^{16}) = a_0 - a$

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:

$x_1 x_2 = b_0 + b(\zeta + ... + \zeta^{16})$

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?

## A Loose End

Before we argued that given

$a_0 + a_1 \zeta^{3^1} + ... + a_{16} \zeta^{3^{16}} = a_0 + a_1 \zeta^{3^2} + ... + a_{16} \zeta^{3^{17}}$

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

$b_1 + b_2 \zeta + ... + b_{16} \zeta^{15} = 0$

(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

$f(x) = 1 + x + ... + x^{16} .$

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 $$g = 0$$.]

Ben Lynn blynn@cs.stanford.edu 💡