Gaussian Periods

As before, 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.

Since \(17 = 1 \pmod{4}\) we know \(-1\) is a quadratic residue, hence if \(r\) is a quadratic residue so is \(-r\). Thus any term of \(x_1\) multiplied by any term of \(x_2\) is never 1, hence \(a_0 = 0\).

There are 8 terms each in \(x_1\) and \(x_2\). Then \(x_1 x_2\) has 64 terms, which must be evenly distributed among \(\zeta ,..., \zeta^{16}\), which implies \(a = 4\). (Thanks to Dennis Westra for pointing this out to me.) Hence \(x_1 x_2 = -4\).

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 17 extension of \(\mathbb{Q}\) thus \(g = 0\).]


Ben Lynn blynn@cs.stanford.edu 💡