I try to use Galois theory to prove quadratic reciprocity, i.e. for primes $l>2$ and $p>2$, $l\ne p$, $$\left(\frac pl\right)\left(\frac lp\right)=(-1)^{\frac{(p-1)(l-1)}{4}}$$
Consider gaussian sum $\displaystyle S=\sum_{x\in \mathbb{F}_l^\times} \left(\frac x l\right)\xi^x $, where $\xi$ is $l-$ primitive root in $\overline{\mathbb F_p}$.
I've work out that $S^2=(-1)^\frac{l-1}2 l$.
If I can prove that $S\in \mathbb F_p$ if and only if $(\frac pl)=1$. Then do a direct calculation of the Legendre notation of $S^2$ can solve the conclusion. But I have trouble proving this proposition.
So the question is:
How to prove: $S\in \mathbb F_p$ if and only if $(\frac pl)=1$ ?
Here's a direct way to see this.
As leoli1 remarked in their comment, you mean $ S = \sum_{x \in \mathbb{F}_l}(\frac{x}{l})\xi^x. $ You've correctly realised that the square of $S$ is always $\pm l,$ with the sign depending on $(\frac{-1}{l}).$ The trick is to compute $S^p$ within $\overline{\mathbb{F}_p},$ which gives \begin{align} (\frac{p}{l}) \cdot S &= \sum_{x \in \mathbb{F}_l}(\frac{xp^{-1}}{l})\xi = \sum_{x \in \mathbb{F}_l} (\frac{x}{l})\xi^p = S^p = S \cdot (S^2)^{\frac{p-1}{2}} = \\ &= S \cdot [(-1)^{\frac{l-1}{2}}l]^{\frac{p-1}{2}} = S \cdot (-1)^{\frac{l-1}{2} \cdot \frac{p-1}{2}} \cdot (\frac{l}{p}), \end{align} which is exactly what you need because $S \neq 0.$