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Weil reciprocity actually holds for arbitrary fields k, $k$, not necessarily algebraically closed: just let x $x$ run over all closed points of X, $X$, and replace your expression for (f,g)_x $(f,g)_x$ by its norm from k(x)^* $k(x)^*$ down to k^*.$k^*$.

Then the connection with Artin reciprocity occurs when k $k$ is a finite field , (of size say $q$), as Chandan suggests. More precisely, if K $K$ denotes the function field of X, $X$, then Weil reciprocity for X $X$ is equivalent to Artin reciprocity for the Galois K-algebra $K$-algebra $L_f = K[t]/(t^{p-1}-f)K[t]/(t^{q-1}-f)$, with Galois group k^* $k^*$ acting by multiplication on t.$t$.

Even more precisely, we mean that for all x $x$ in X $X$ we have (f,g)_x $(f,g)_x = Art(L_f/K)_x(g)Art(L_f/K)_x(g)$. 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 $x$ is not in the support of f. $f$. Then the left-hand side is the norm of f(x)^{v_x(g)} $f(x)^{v_x(g)}$ and the right hand side is Frob_x^{v_x(g)}, $Frob_x^{v_x(g)}$, so it suffices to show that Frob_x $Frob_x$ is the norm of f(x). $f(x)$. But the residue field extension is k(x)[t]/(t^{p-1}-f(x)), $k(x)[t]/(t^{q-1}-f(x))$, so if q = #k and d = [k(x):k] letting $d$ denote the degree of $k(x)$ over $k$ we can calculate the Frobenius as sending t $t$ to t^(q^d) $t^{q^d} = t^(q^d-1) * t^{q^d-1} \cdot t = (t^(q-1))^(1+q+...+q^(d-1)) * t^{q-1})^{1+q+...+q^{d-1}} \cdot t = f(x)^(1+q+...+q^(d-1)) * f(x)^{1+q+...+q^{d-1}} \cdot t = Norm(f(x)) * t\cdot t$, as desired.

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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, 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^{p-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^{p-1}-f(x)), so if q = #k and d = [k(x):k] we can calculate the Frobenius as sending t to t^(q^d) = t^(q^d-1) * t = (t^(q-1))^(1+q+...+q^(d-1)) * t = f(x)^(1+q+...+q^(d-1)) * t = Norm(f(x)) * t, as desired.