Quadratic Gauss sums, part 1 of whatever

by math_explorer, Apr 19, 2016, 10:52 PM

Let $p$ be a prime and let $\zeta$ be a primitive $p$ th root of unity. Then automorphisms of $\mathbb{Q}[\zeta]$ are determined by what $\zeta$ maps to, some other power of $\zeta$ , so the automorphism group is $\mathbb{Z}/p\mathbb{Z}^*$ . If $g$ is a primitive root mod $p$ , the automorphism group is generated by $\zeta \mapsto \zeta^g$ . Composing this with itself yields functions like $\zeta \mapsto \zeta^{g^k}$ .

Also, the automorphism group has a subgroup of index 2, which corresponds to an extension field of $\mathbb{Q}$ of degree 2. What elements are in the subgroup? They are things of the form $\zeta \mapsto \zeta^{g^{2k}}$ for some $k$ , so things where the exponent is a square.

Hey, cool! Quadratic residues!

I don't actually understand/remember Galois theory as well as I probably sound like I do, but whatever. Anyway, this motivates us to consider the sum $S = \sum_{k = 1}^{p-1} \left(\frac{k}{p}\right) \zeta^k$ where the fraction is the Legendre symbol. It's fixed by a lot of automorphisms.

A final observation, much simpler than all the complicated stuff up there, is that if $\gcd(x, p) = 1$ and $y$ ranges over $\{1, \ldots, p-1\}$ , then $xy$ also ranges over $\{1, \ldots, p-1\} \bmod{p}$ . You may have seen this technique in a proof of Fermat's little theorem. $(1)(2)\cdots(p-1) \equiv (a)(2a)\cdots((p-1)a) \bmod{p}$ because the list of multiplicands on both sides are permutations of each other mod $p$ ; then divide out $(p-1)!$ .
Anyway, the $\zeta$ -to-the-power-of does the modding by $p$ for us, so we let $\ell = k\ell'$ and get
$\begin{align*} S^2 &= \sum_{k = 1}^{p-1} \sum_{\ell = 1}^{p-1} \left(\frac{k}{p}\right) \left(\frac{\ell}{p}\right) \zeta^{k+\ell} \\ &= \sum_{k = 1}^{p-1} \sum_{\ell = 1}^{p-1} \left(\frac{k\ell}{p}\right) \zeta^{k+\ell} \\ &= \sum_{k = 1}^{p-1} \sum_{\ell' = 1}^{p-1} \left(\frac{k^2\ell'}{p}\right) \zeta^{k(1+\ell')} \\ &= \sum_{k = 1}^{p-1} \sum_{\ell' = 1}^{p-1} \left(\frac{\ell'}{p}\right) \zeta^{k(1+\ell')} \\ &= \sum_{\ell' = 1}^{p-1} \left(\frac{\ell'}{p}\right) \left( \sum_{k = 1}^{p-1} \zeta^{k(1+\ell')} \right) \end{align*}$ For fixed $\ell'$ such that $1 + \ell' \not\equiv 0 \bmod{p}$ , the innermost exponent ranges over $1, \ldots, p-1 \bmod{p}$ . Then the inner sum is $-1$ , since it sums those powers of $\zeta$ .
On the other hand, when $1 + \ell' \equiv 0 \bmod{p}$ , the innermost $\zeta^{k(1+\ell')}$ is just 1, so the sum is just $p - 1$ . So if we separate the $p$ from the $-1$ to take neat advantage of the fact that there are equally many quadratic residues as quadratic nonresidues, we get:
$\begin{align*} S^2 &= \sum_{\ell' = 1}^{p-1}\left( \left(\frac{\ell'}{p}\right)(-1)\right) + \left(\frac{-1}{p}\right)\cdot p = \left(\frac{-1}{p}\right)\cdot p \\ S &= \pm\sqrt{\left(\frac{-1}{p}\right)\cdot p}. \end{align*}$
It's a Hard Problem to figure out the sign of $S$ . More simply, there's something cool you can do to this to get quadratic reciprocity or something, but even this is already too hard. See you next time.

number theory

Quadratic Residues

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