Aug 212012
 

If \(p,q\) are distinct odd primes,  then

\[\left( \frac pq\right) \left( \frac qp\right) = (-1)^{\frac{(p-1)(q-1)}4},\]

where \(\left( \frac{}{}\right)\) is the Legendre symbol. 这就是被 Gauss 称为”数论酵母” 的二次互反律.

自 Legendre 的那个没有完成的证明以来, 据 Reciprocity Laws: From Euler to Eisenstein 的作者 Franz Lemmermeyer 统计, 发表的证明是 \(240\) 个. 可以预见的到, 这个数字还会不断增加. 这些证明的作者, 发表年份,使用的方法以及文献的详细列表可在 Lemmermeyer 的个人主页找到. 其中, \(1889\) 这一年就有 \(6\) 个证明公布, \(1893,1951,1961\) 年也各有 \(5\) 个证明发表在不同的期刊上. 仅看从 \(1950\) 年到今天的这 \(60\) 多年, 只有 \(1956,1959,1968,1970,1975,1977,1982,1986,1988,1996,2002\) 这 \(11\) 年没有证明发表.不过, Lemmermeyer 的统计好像有小错误, 例如 Wouter Castryck 在 \(2008\) 年发表了一篇文章, 办法类似于 V. A. Lebesgue 在 \(1838\) 年的论证, 但 Lemmermeyer 的统计认为 Castryck 的证明在 \(2007\) 年给出.

这么多的证明, 想要全部分门别类, 整理好写出来, 肯定是困难的. 文献多, 不容易都找到. 即便都找齐了, 也因为是不同的语言, 也不能都看懂. 这些方法各有繁简. 哪个证明才是最简单的呢? 显然, Proofs from THE BOOK 给出的两个途径, 是比较繁琐的; Jean-Pierre Serre 在他的 A Couse in Arithmetic 第一章给出的都依靠 Gauss 引理的两个证明, 也不算太简单. 那么, 到底哪个才是最简单的呢? 各人的看法可能有不同. 但, 美妙的, 能给人以深刻印象, 令人荡气回肠的证明, 应该有一些共同的特征, 或者说一个证明要能成为好的证明, 美丽的证明, 应该有一定的门槛, 满足一定的条件. 我们试着列出这些条件:

  • 简单是首要条件. 基于简单的想法, 能揭示问题的本质, 加深对事物的理解.
  • 美. 构思巧妙, 论证精妙方能展现出深刻. 当然, 简单其实也是一种美.
  • 自然. 方法能用在更广泛的地方, 解决更多的问题.

同时满足这些要求的证明不容易找到, Castryck 的证明大概可以满足前两条. 下面是他的详细论证:

For any odd \(n \in\Bbb N\), denote by \(N_n\) the number of solutions in \(\left( \Bbb Z / (q)\right)^n\) to the equation

\[x_1^2-x_2^2+x_3^2-\dotsb + x_n^2 = 1.\]

If we substitute \(x_1 \gets x_1+x_2\), then we get

\[x_1^2+x_3^2-\dotsb + x_n^2-1 = -2x_1x_2.\]

For any non-zero \(x_1\)-value and any values of \(x_3,\dotsc,x_n\), there is a unique corresponding \(x_2\)-value. If \(x_1=0\), there are no solutions, except if \(x_3^2-\dotsb + x_n^2 = 1\) (which happens in \(N_{n-2}\) cases): then all possible values of \(x_2\) do the job. We find that

\[N_n = q^{n-2}(q-1)+qN_{n-2}.\]

and hence \(N_n = q^{n-1}+q^{\frac{n-1}{2}}(N_1-1) = q^{n-1}+q^{\frac{n-1}{2}}.\) In particular,

\begin{equation}N_p\equiv1+\left(\frac qp\right) \pmod p.\end{equation}

Next, \(N_p\) can be classically determined as

\[\sum_{t_1+\dotsb + t_p = 1}N(x_1^2 = t_1)N(x_2^2 = -t_2)N(x_3^2=t_3)\dotsm N (x_p^2=t_p),\]

where the \(t_i\) are in \(\Bbb Z/(q)\) and \(N(\cdots)\) denotes the number of solutions to the corresponding univariate equation. This can be rewritten as

\[\sum_{t_1+\dotsb+t_p=1}\left(1+\left(\frac{t_1}q\right)\right)\left( 1+\left( \frac{-t_2}q\right)\right)\left( 1+\left( \frac{t_3}q\right)\right)\dotsm \left( 1+\left( \frac{t_p}q\right)\right),\]

When expanding out the product, only the terms \(1\cdot 1 \cdot 1 \cdots 1\) and \(\left(\frac{t_1}q\right)\cdot\left(\frac{-t_2}q\right)\cdot\left( \frac{t_3}q\right)\cdots\left( \frac{t_p}q\right)\) should be taken into consideration; the other terms disappear because Legendre symbols sum up to zero: \(\sum\limits_{t\in\Bbb Z/(q)}\left(\frac tq\right) = 0.\) Therefore, the above expression simplifies to

\[q^{p-1}+\left( \frac{(-1)^{\frac{p-1}2}}q\right)\sum_{t_1+\dotsb+t_p = 1}\left(\frac{t_1t_2t_3\dotsm t_p}q\right).\]

Modulo \(p\), the latter sum almost completely vanishes, since the tuples \((t_1,\dotsc,t_p)\) satisfying \(t_1+\dotsb + t_p = 1\) with not all \(t_i\) equal to \(p^{-1}\) can be collected in groups of size \(p\) by cyclic permutation. Note that \(p\) is indeed a multiplicative unit in \(\Bbb Z/(q)\). We thus obtain

\begin{equation}N_p\equiv1+\left( \frac{(-1)^{\frac{p-1}2}}q\right) \left( \frac{p^{-p}}q\right) \equiv1+(-1)^{\frac{p-1}2\frac{q-1}2}\left( \frac pq\right) \pmod p.\end{equation}

The last congruence follows from Euler’s criterion \(\left(\frac aq\right) \equiv a^{\frac{q-1}2} \pmod q\) and the observation that \(p^{-p}\) is a square in \(\Bbb Z/(q)\) if and only if \(p\) is a square in \(\Bbb Z/(q)\).

Comparing \((1)\) and \((2)\), the reciprocitylaw follows. \(\Box\)

接下来的论证, 是 Aurelien Bessard (2010) 改编自 V. A. Lebesgue 和 Eisenstein的一个证明:

Let \(p = 2m+1\) and \(q\) be distinct odd primes and let \(N\) denote the number of solutions of the equation

\[ x_1^2 + \ldots + x_p^2 = 1 \]

in the finite field \(\Bbb F_q\).

The group \(\Bbb Z/p\Bbb Z\) acts on the solution space \(X\) by shifting indices: if \((x_1, \ldots, x_p) \in X\), then so is \((x_a,x_{a+1}, \dotsc)\) for each \(a \in {\mathbb Z}/p{\mathbb Z}\), where the indices have to be read modulo \(p\). Each orbit has exactly \(p\) elements except if there is an \(x\) with \((x,x,\ldots,x) \in X\): the orbit of this element has \(1\) element. Now \((x,x,\ldots,x) \in X\) if and only if \(px^2 = 1\) is solvable in \(\Bbb F_q\), hence

\begin{equation} N \equiv \Big( \frac pq \Big) + 1 \pmod p.\end{equation}

We make a change of variables to transform the diagonal equation into an equation where counting the number of solutions is easier. To this end, consider the matrix

\[ A = \left( \begin{matrix} 0 & 1 & & & & & & \\ 1 & 0 & & & & & & \\ & & 0 & 1 & & & & \\ & & 1 & 0 & & & & \\ & & & & \ddots & & & \\ & & & & & 0 & 1 & \\& & & & & 1 & 0 & \\& & & & & & & a\end{matrix} \right) \]

with \(a = (-1)^{\frac{p-1}2}\). Since \(\det A = 1\), this matrix is congruent to the unit matrix, hence \(X\) and the solution spaces \(X^\prime\) of the equation \(x^T A x = 1\),i.e.,recall that \(p=2m+1\), of

\[2(y_1z_1+\dotsb+y_mz_m)+ax_p^2=1\]

are isomorphic.

For counting the number of solutions of \(X^\prime\), observe that if \((y_1, \dotsc, y_m) = 0\), we have \(q^m(1+a^{\frac{q-1}2})\) possibilities for choosing \(z_1, \dotsc, z_m\) and \(x_p\).

If \(y = (y_1, \dotsc, y_m) \ne 0\), on the other hand, then for each choice of \(y\) and \(x_p\) we have to count the number of points on a hyperplane of dimension \(m\); there are \(q^{m-1}\) points on such a hyperplane, and the number of overall possibilities in this case is \((q^m-1) \cdot q \cdot q^{m-1} = q^m(q^m-1)\).

Thus we find

\begin{equation}\begin{split} N & = q^m (1+a^{\frac{q-1}2})+q^m(q^m-1) \\&=q^m(q^m+(-1)^{\frac{p-1}2\frac{q-1}2})\\ & \equiv\Big(\frac qp\Big)\bigg(\Big(\frac qp\Big)+(-1)^{\frac{p-1}2\frac{q-1}2}\bigg)\\&\equiv1+(-1)^{\frac{p-1}2\frac{q-1}2}\Big(\frac qp\Big)\pmod p.\end{split}\end{equation}

Comparing \((3)\) and \((4)\), gives the quadratic reciprocity law. \(\Box\)

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