# Weil reciprocity vs Artin reciprocity

This is probably an easy question for the experts:

Given two rational functions $f$, $g$ on a non-singular projective algebraic curve X (over an algebraically closed field $k$) and $p \in X$, one defines the Weil symbol $(f, g)_p$ as the value of $(-1)^{ab} f^a g^{-b}$ at $p$ where $a = v_p(g)$ and $b = v_p(f)$. (Here $v_p$ means order of zero/pole at $p$.)

Weil reciprocity claims that product of $(f, g)_p$ for all $p \in X$ is equal to $1$.

My question is whether the Weil symbol can be realized as a special case of the Artin symbol (for an extension of fields of rational functions)?

(Note that the ground field $k$ is not assumed to be of positive characteristic.)

-
Perhaps you need to assume that $\mathrm{div}(f)$ and $\mathrm{div}(g)$ are disjoint. –  Chandan Singh Dalawat May 8 '12 at 4:34
Strictly speaking, the Artin symbol makes sense only for (number fields and) function fields over finite fields. But there is version of Weil reciprocity for the latter, and your question would still make sense. –  Chandan Singh Dalawat May 8 '12 at 13:55
The Weil symbol can be defined when $\mathrm{div}(f)$ and $\mathrm{div}(g)$ overlap. Let $\pi$ be a uniformizer at $p$ and write $f = f_0 \pi^b$ and $g = g_0 \pi^a$. Then $(f,g)_p = (-1)^{ab} f_0(p)^a g_0(p)^{-b}$. Exercise: This is independent of the choice of $\pi$. –  David Speyer May 8 '12 at 16:12
David - doesn't this agree with what Kiumars wrote? Since $f^ag^{-b}$ has neither a zero nor a pole at p, it can be evaluated at $p$. –  Dustin Clausen May 8 '12 at 17:07
Oh I see. I was thinking we evaluate $f$ and $g$ first and then plug in, but, of course, doing it in the other order makes more sense. So Kiumars' definition is right even when $\mathrm{div}(f)$ and $\mathrm{div}(g)$ do overlap, without modification. My bad. –  David Speyer May 8 '12 at 19:42

Weil reciprocity actually holds for arbitrary fields $k$, not necessarily algebraically closed: just let $x$ run over all closed points of $X$, and replace your expression for $(f,g)_x$ by its norm from $k(x)^*$ down to $k^*$.
Then the connection with Artin reciprocity occurs when $k$ is a finite field (of size say $q$), as Chandan suggests. More precisely, if $K$ denotes the function field of $X$, then Weil reciprocity for $X$ is equivalent to Artin reciprocity for the Galois $K$-algebra $L_f = K[t]/(t^{q-1}-f)$, with Galois group $k^*$ acting by multiplication on $t$.
Even more precisely, we mean that for all $x$ in $X$ we have $(f,g)_x = Art(L_f/K)_x(g)$. To verify this, note that since both sides have a product formula and finite modulus, by weak approximation it suffices to consider the case where $x$ is not in the support of $f$. Then the left-hand side is the norm of $f(x)^{v_x(g)}$ and the right hand side is $Frob_x^{v_x(g)}$, so it suffices to show that $Frob_x$ is the norm of $f(x)$. But the residue field extension is $k(x)[t]/(t^{q-1}-f(x))$, so letting $d$ denote the degree of $k(x)$ over $k$ we can calculate the Frobenius as sending $t$ to $t^{q^d} = t^{q^d-1} \cdot t = (t^{q-1})^{1+q+...+q^{d-1}} \cdot t = f(x)^{1+q+...+q^{d-1}} \cdot t = Norm(f(x)) \cdot t$, as desired.
Could you insert a few \$s (by which I don't mean American money but the symbol \$) to make your answer more easily readable ? –  Chandan Singh Dalawat May 8 '12 at 15:31
It is also worth noting that, if $b | p-1$, then Artin reciprocity for $K[t]/(t^b-f)$ can be expressed in terms of the Weil symbol as well: the relation is $(f,g)_x^{(p-1)/b} = \mathrm{Art}(L_f/K)_x(g)$. In particular, if $p$ is odd, $b$ is $2$, $g$ is irreducible and $x_1$, ..., $x_n \in X$ are the zeros of $g$, then the product $\prod_{x_i} (f,g)_{x_i}^{(p-1)/b}$ is "take $f$ modulo $g$ and raise it to the $(p^{\deg g}-1)/2$ power", an expression which should be familiar from proofs of quadratic reciprocity. –  David Speyer May 8 '12 at 16:05