I was told that Weil reciprocity theorem (one has two meromorphic function $f,g$ on a complex curve $C$, so $\prod\limits_{x\in C} g(x)^{ord_xf}=\prod\limits_{x\in C}f(x)^{ord_xg} \ $ where $ord_xf$ is the smallest degree in Taylor expansion of $f$ at $x$, product is taken only by points in divisors of $f,g$, we assume that these divisors are not intersected with each other) was introduced by Weil after thinking about quadratic reciprocity. Could you explain me the connection between them?

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    $\begingroup$ The hypothesis that the divisors of $f$ and $g$ are disjoint is not necessary : you can make sense of $f^{\mathrm{ord}_x(g)}/g^{\mathrm{ord}_x(f)}$ evaluated at $x$. However, I think you then need to introduce a sign $(-1)^{\mathrm{ord}_x(f) \mathrm{ord}_x(g)}$ so that the product formula holds. $\endgroup$ Jun 7 '12 at 14:47
  • $\begingroup$ @ François thanks, I know) $\endgroup$ Jun 8 '12 at 10:07

There are already good answers by quid and by Dustin Clausen here. I thought, though, that I'd take the time to write out something more leisurely and expository. To get from Weil Reciprocity to Quadratic Reciprocity, one must make some things more general and some things less general, and there is a choice of which order to do these things in. I will first describe the less general route and will then make some comments about what happens when you make everything as general as possible.

First, we specialize from a general curve $C$ to $\mathbb{CP}^1$. Let $f$ and $g$ be polynomials in $\mathbb{C}[x]$, with roots at the disjoint sets $\alpha_1$, ..., $\alpha_a$ and $\beta_1$, ..., $\beta_b$ and with leading terms $f(x) = f_{\infty} x^{a} + \cdots$ and $g(x) = g_{\infty} x^b + \cdots$. Then Weil reciprocity says $$\prod_{j=1}^b f(\beta_j) = \prod_{i=1}^a g(\alpha_i) \cdot (-1)^{ab}\ f_{\infty}^{b} \ g_{\infty}^{-a}. \quad (1)$$ (The extra terms on the right come because $f$ and $g$ both have poles at $\infty$; Francois discusses this in a comment above.)

Of course, $(1)$ is easy to prove directly by writing $f(x) = f_{\infty} \prod (x-\alpha_i)$ and $g(x) = g_{\infty} \prod (x-\beta_j)$. Also, this makes it clear that the above identity holds over any algebraically closed field.

We now generalize to a non-algebraically closed field $k$. Let $f$ and $g$ be relatively prime polynomials in $k[x]$. Let $f(x) = f_{\infty} r_1(x) r_2(x) \cdots r_c(x)$ be the factorization of $f$ into monic irreducibles. Similarly, let $g(x) = g_{\infty} s_1(x) \cdots s_d(x)$. Let $K_i$ be the field $k[x]/r_i(x)$ and let $L_j = k[x]/s_j(x)$. For $u$ and $v \in k[x]$, I'll write $(u \bmod v)$ for the image of $u$ in $k[x]/v$. The generalization of $(1)$ is $$\prod_{j=1}^d N_{L_j/k}(f \bmod s_j) = \prod_{i=1}^c N_{K_i/k}(f \bmod r_i) \cdot (-1)^{ab}\ f_{\infty}^{b} \ g_{\infty}^{-a}. \quad (2)$$

Exercise 1: See that, when $k$ is algebraically closed, $(2)$ specializes to $(1)$. Exercise 2: Deduce (2) from (1), by grouping together terms from $(1)$ over $k^{\mathrm{alg}}$.

We now specialize to the case $k = \mathbb{F}_p$. The norm map from $\mathbb{F}_{p^n}$ to $\mathbb{F}_p$ is raising to the $(p^n-1)/(p-1)$ power. So $(2)$ becomes

$$\prod_{j=1}^d (f \bmod s_j)^{\frac{p^{\deg s_j}-1}{p-1}} = \prod_{i=1}^c (f \bmod r_i)^{\frac{p^{\deg r_i}-1}{p-1}} \cdot (-1)^{ab}\ f_{\infty}^{b} \ g_{\infty}^{-a}. \quad (3)$$

We specialize further to the case that $f$ and $g$ are irreducible to get $$(f \bmod g)^{\frac{p^{b}-1}{p-1}} = (g \bmod f)^{\frac{p^{a}-1}{p-1}} \cdot (-1)^{ab}\ f_{\infty}^{b} \ g_{\infty}^{-a}. \quad (4)$$

Let $p$ be odd, and raise both sides to the $(p-1)/2$ to get $$(f \bmod g)^{(p^b-1)/2} = (g \bmod f)^{(p^a-1)/2} \cdot (-1)^{ab(p-1)/2} \ f_{\infty}^{b (p-1)/2} g_{\infty}^{-a(p-1)/2}. \quad (5)$$

Recall Euler's criterion: For $u \in \mathbb{F}_p$, we have $u^{(p-1)/2} = \left( \frac{u}{p} \right)$. It has a generalization: For $u \in \mathbb{F}_{p^n}$, the power $u^{(p^n-1)/2}$ is $\pm 1$ according to whether or not $u$ is square. So, defining the quadratic residue symbol in $\mathbb{F}_p[x]$ in the obvious way, equation $(5)$ says $$\left( \frac{f}{g} \right) = \left( \frac{g}{f} \right) \cdot (-1)^{ab(p-1)/2} \ \left( \frac{f_{\infty}}{p} \right)^b \left( \frac{g_{\infty}}{p} \right)^{-a}. \quad (6)$$

In short, $\left( \frac{f}{g} \right)$ is equal to $\left( \frac{g}{f} \right)$ up to some elementary terms, just like in quadratic reciprocity. One can rewrite the elementary correction terms to make them look more like the terms that show up in standard QR, but I'll leave this as is.

We can get more general statements by (a) not restricting ourselves to the case that $f$ and $g$ are irreducible (b) working with curves $C$ other than $\mathbb{A}^1$ (c) raising both sides of $(4)$ to the $(p-1)/g$ power for some other $g$ dividing $p-1$. I was going to write more about this, but I think it is long enough as it is.

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    $\begingroup$ Write $f = f_\infty{F}$ and $g = g_\infty{G}$, so $F$ and $G$ are monic. For $c \in {\mathbf F}_p^\times$, $(c|f) = (c|p)^a$ and $(c|g) = (c|p)^b$. Then (6) is equivalent to $(F|G) = (G|F)(-1)^{ab(p-1)/2}$ for monic relatively prime $F$ and $G$, which is the usual way on sees Jacobi reciprocity written out (e.g., just like ordinary Jacobi reciprocity is usually written out for a pair of rel. prime positive integers, as opposed to a pair of nonzero rel. prime integers). $\endgroup$
    – KConrad
    Jun 7 '12 at 22:44

John Milnor's Introduction to Algebraic K-Theory, has (on p. 101) some useful background on the connection between $K$-theory and quadratic reciprocity:

Theorem 11.6 (Tate) The group $K_2\mathbb Q$ is canonically isomorphic to the direct sum $$ A_2\oplus A_3\oplus A_5\oplus\ldots $$ where $A_2=\{\pm1\}$ and for $p$ odd, $A_p=(\mathbb Z/p\mathbb Z)^\times$.
Milnor explains:

In fact the isomorphism will be given by the correspondence $$ > \{x,y\}\to(x,y)_2\oplus(x,y)_3\oplus(x,y)_5\oplus\ldots > $$ Tate remarks that his proof of the theorem is lifted directly from the argument which was used by Gauss in his first proof of the quadratic reciprocity law.

(The whole section beginning on p.99 is titled Gauss and Quadratic Reciprocity)

Forgive me if this is already familiar to you.

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    $\begingroup$ Perhaps it is worth adding that quadratic reciprocity follows from Tate's computation of $K_2(\mathbf{Q})$ --- the universal symbol on $\mathbf{Q}$. By universality, the symbol $(x,y)_\infty$ --- defined as $+1$ (resp. $−1$) if and only if the conic $xT^2+yS^2=1$ has (resp. does not have) a solution in $\mathbf{R}$ --- must be expressible in terms of the symbols $(x,y)_2,(x,y)_3,(x,y)_5,\ldots$ Doing so explicitly leads to quadratic reciprocity. $\endgroup$ Jun 9 '12 at 5:23

I am not sure this is the type of answer you are looking for, but perhaps it is, so:

On the wikipedia page linked to in the question it is explained that Weil's law can be restated as the fact that the product over the local symbols $(f,g)_p$ (see the same page for a definition) is $1$.

Hilbert's reciprocty law (scroll down to the relevant section) for an algebraic number field can be stated as $\prod_v (a,b)_v = 1$ where the product is over all places (finite and infinite) and $(a,b)_v$ is the Hilbert symbol (for the respective completion).

The Hilbert symbol $(a,b)$, for a local field $K$, is defined to be $1$ if $z^2 = a x^2 + by^2$ has a solution in $K$ and $-1$ if not.

Now, for the rationals Hilbert's reciprocty law gives quadratic reciprocity (see the two pages I linked to for some details).

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    $\begingroup$ I'm not historian, but I'm pretty sure this is indeed what motivated Weil to state and prove his reciprocity law on curves. $\endgroup$ Jun 7 '12 at 15:49

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