# Confusing application of power residue reciprocity in Milne's CFT

Hey I am trying to figure out the details of the proof of Theorem 5.14 (p.246) in Milne's CFT (see here). I hope somebody is familiar with this. But let me sketch the proof and what I don't understand. Please feel free to answer any of the questions below.

Theorem 5.14 Let $x, y, z$ be relative prime positive integers such that $p$ does not divide $xyz$ and $x^p + y^p = z^p$. Then for every prime $q$ dividing $xyz$, the equation $q^{p-1} \equiv 1 \mod{p^2}$ holds.

Proof

Let $\zeta\in \mathbb C$ be a primitive $p$-th root of unity. Now we consider $K = \mathbb Q[\zeta]$ and its ring of integers $\mathcal O_K = \mathbb Z[\zeta]$. Note that $(p) = \mathfrak p^{p-1}$ where $\mathfrak p$ is the prime ideal generated by $\pi = 1-\zeta$.

We know $z^p = \prod\limits_{i=0}^{p-1} (x+\zeta^i y)$. Because $x+\zeta^iy$ are pairwise relatively prime, their principal ideals are $p$-th powers of ideals in $\mathbb Z[\zeta]$.

This implies that the $p$-th power residue symbol $\left(\frac{\alpha}{\beta}\right) = 1$ for all $\beta \in \mathbb Z[\zeta]$ relatively prime to $\alpha = \frac{x+\zeta y}{x+y} = 1 - \frac{y\pi}{x+y}$ with the prime element $\pi=1-\zeta$. (Q1)

Setting $\beta = q^{p-1}$ one can also check $\left(\frac{\beta}{\alpha}\right) = 1$. (Q2)

The power residue reciprocity theorem says $\left(\frac{\beta}{\alpha}\right)\left(\frac{\alpha}{\beta}\right) ^{-1} = (\alpha,\beta)_{\mathfrak p}$ where $(\alpha,\beta)_{\mathfrak p}$ is the Hilbert symbol in $K$ ($=\mathbb Q[\zeta]$).

One can compute $(\alpha,\beta)_{\mathfrak p} = \zeta^{\mathrm{Tr}_{K/\mathbb Q}(\eta)}$ where $\eta = \frac{\beta-1}{p}\frac{\alpha-1}{\pi}$. (Q3)

From the triviality of the power residue symbols above we get $(\alpha,\beta)_{\mathfrak p} = 1$ or equivalently $p$ divides the integer $\mathrm{Tr}_{K/\mathbb Q}(\eta)$.

We calculate $\mathrm{Tr}(\frac{\alpha-1}{\pi}) = \mathrm{Tr}(-\frac{y}{x+y}) = -\frac{y}{x+y}\cdot (p-1)$, which is relatively prime to $p$. Hence $\frac{\beta-1}{p} = \frac{q^{p-1}-1}{p}$ is divisible by $p$.

Questions

(Q1) I understand that $\alpha$ generates a principal (fractional) ideal that is a $p$-th power of another (fractional) ideal. Why does this imply $\left(\frac{\alpha}{\beta}\right) = 1$ for all $\beta \in \mathbb Z[\zeta]$ relatively prime to $\alpha$?

(Q2) Clearly $\left(\frac{\beta}{\alpha}\right) = \left(\frac q\alpha\right)^{p-1} = \left(\frac q\alpha\right)^{-1} = 1 \iff \left(\frac q\alpha\right) = 1$, but how do you continue from there?

(Q3) Just before in Milne's CFT on pages 244f (in particular Proposition 5.12), there is a description how to compute exactly these Hilbert symbols. But I cannot relate that description to $(\alpha,\beta)_{\pi}=(\alpha,\beta)_{\mathfrak p} = \zeta^{\mathrm{Tr}_{K/\mathbb Q}(\eta)}$.

Last remark

Literature that I am familiar with is Milne's course notes (see here), Serre's "Local fields", and Neukirch's "Algebraic Number Theory" (which includes his book on class field theory). Any advice on the questions will be welcomed but I can parse similar language as in those books fastest.

• I have a vague impression that Wiles' paper on explicit reciprocity laws contains a similar formula to your question $3.$ But I am not very sure. Feb 9, 2014 at 8:22