Weil reciprocity vs Artin reciprocity - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T10:12:46Z http://mathoverflow.net/feeds/question/96290 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/96290/weil-reciprocity-vs-artin-reciprocity Weil reciprocity vs Artin reciprocity Kiumars 2012-05-08T04:00:14Z 2012-05-08T17:03:11Z <p>This is probably an easy question for the experts:</p> <p>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 $p$.)</p> <p>Weil reciprocity claims that product of $(f, g)_p$ for all $p \in X$ is equal to $1$.</p> <p>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)? </p> <p>(Note that the ground field $k$ is not assumed to be of positive characteristic.)</p> http://mathoverflow.net/questions/96290/weil-reciprocity-vs-artin-reciprocity/96349#96349 Answer by Dustin Clausen for Weil reciprocity vs Artin reciprocity Dustin Clausen 2012-05-08T15:13:25Z 2012-05-08T17:03:11Z <p>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 $k^*$.</p> <p>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 $t$.</p> <p>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.</p>