I'm trying to understand the proof of Quadratic Reciprocity on Wikipedia (this link: https://en.wikipedia.org/wiki/Proofs_of_quadratic_reciprocity#Proof_using_algebraic_number_theory, then the section "proof using Gauss sums" and then "general case"). I believe it's the proof in Ireland and Rossen.
Let $p$ and $q$ be primes, let $\zeta _p=e^{2\pi i/p}$ and let $g_p$ be the quadratic Gauss sum: \[ g_p=\sum _{k=1}^{p-1}\left (\begin {array}{l}k\\ p\end {array}\right )\zeta _p^k.\]
In the article $g_p^{q-1}$ is worked out in two different ways. First it is shown that
\[ g_p^{q-1}\equiv \left (\begin {array}{l}p^*\\ q\end {array}\right )\text { mod }q\]
where
\[ p^*=\left (\begin {array}{l}\overline -1\\ p\end {array}\right )p.\]
This is just a usual congruence in integers. It is then shown that we also have
\[ g_p^q\equiv \sum _{k=1}^{p-1}\left (\begin {array}{l}k\\ p\end {array}\right )^q\zeta ^{qk}\text { mod }q\]
which is a congruence in the ring $\mathbb Z[\zeta _p]$, meaning literally \[ g_p^q-\left (\begin {array}{l}\overline q\\ p\end {array}\right )g_p=g_p^q-\sum _{k=1}^{p-1}\left (\begin {array}{l}k\\ p\end {array}\right )^q\zeta ^{qk}=qZ\] for some
\[ Z=\sum _{r=0}^{p-1}A_r\zeta ^r\in \mathbb Z[\zeta _p].\]
Consequently
\[ g_p^{q-1}-\left (\begin {array}{l}\overline q\\ p\end {array}\right )=qg_p^{-1}Z\]
and we can compare the two expressions for $g_p^{q-1}$ to deduce
\[ \left (\begin {array}{l}\overline q\\ p\end {array}\right )\equiv \left (\begin {array}{l}p^*\\ q\end {array}\right )\text { mod }q.\]
Where exactly thist last congruence is taken isn't clear to me though - whether it's a congruence in integers or in $\mathbb Z[\zeta _p]$. If it's in $\mathbb Z[\zeta _p]$, then we need to know that $1\not \equiv -1\text { mod }q$ which isn't obvious. But if I look at one of the oldest proofs, due to Cauchy, where no notion of $\mathbb Z[\zeta _p]$ was available, then it seems to say it's actually a congruence in integers? But I can't read French (or maths xD) so maybe I'm just not understanding what's going on. Here's the text:
https://archive.org/details/bub_gb_hzIVAAAAQAAJ/page/n411/mode/2up
where towards the bottom of page 393 I think he shows that the difference I'm looking for really is an integer.
Is $g_p^{-1}Z\in \mathbb Z$? And if not how does the above proof work, if we're being precise about congruences?
It may be something simple I'm missing and that's why it's not written out explicitly (cause it's not written in the Rossen and Ireland text either).
Can anyone clarify this for me?
Here "$\bmod q$" means in the ring $\Bbb{Z}[\zeta]/(q)$.
$\Bbb{Z}[\zeta]/(q)$ is a commutative ring of characteristic $q$ (prime) so that for any of its elements we have $(a+b)^q =a^q+b^q$. Also for any integer $n^q=n$. Whence in that ring $$(\sum _{k=1}^{p-1}(\frac{k}p)\zeta^k)^q=\sum _{k=1}^{p-1}(\frac{k}p)^q\zeta^{qk}=\sum _{k=1}^{p-1}(\frac{k}p)\zeta^{qk}$$
You don't care that $g_p^q-g_p= g_p m$ for some integer $m$ (when $p,q$ are odd), and it is obvious from $g_p = \pm \sqrt{\pm p}$. You can also see it from some Galois theory, namely any automorphism $\sigma $ of $\Bbb{Z}[\zeta]$ is such that $\sigma(g_p)=\pm g_p$ (this follows from $\sigma(\zeta)=\zeta^l$ for some $\gcd(l,q)=1$). Then $q$ must divide this integer $m$ because $g_p\ne 0\bmod q$ (this follows from $g_p^2=\pm p$).