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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.

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  • $\begingroup$ 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. $\endgroup$ – awllower Feb 9 '14 at 8:22
  • $\begingroup$ In the article by G. Gras & R. Quême, "Vandiver's papers on cyclotomy revisited and FLT", Publ. Math. Besançon, 2012, 2, 47-111, you will find complete calculations on the so called Fermat quotients $q^{p-1} - 1 / p, using power residue symbols, not only in the first case (p not dividing xyz) but also in the second, and in a simpler way than Milne's, if I remember well. Actually the authors study the so called Strong FLT, which is an equation in Z[zeta]. $\endgroup$ – nguyen quang do Dec 26 '16 at 14:34
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Q1: There is a typo in the manuscript: instead of $(\frac{\alpha}{\beta}) = 1$ it should read $(\frac{\beta}{\alpha}) = 1$, and this holds for trivial reasons: $(\frac{\beta}{\alpha}) = (\frac{\beta}{{\mathfrak a}^p}) = (\frac{\beta}{{\mathfrak a}})^p = 1$.

Q2: This problem dissolves since now you need to prove $(\frac{\alpha}{\beta}) = 1$, which follows directly from $(\frac{\alpha}{q}) = 1$.

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