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Quadratic reciprocity theorems states that for two different odd prime p and q, we have (p/q)(q/p)=(-1)^(p-1)(q-1)/4.

What is the statement of this theorem in L-function?

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You can consider $\zeta(s)L(s,\chi)=\zeta_K(s)$ to be a reformulation of quadratic reciprocity, where $\chi$ is a primitive quadratic character of conductor $N$ and $K=\mathbb{Q}\left[\sqrt{\chi(−1) N} \right]$. This follows from the relation between Legendre symbols and splitting in quadratic extensions. –  Sam Derbyshire Oct 22 '10 at 22:35
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Maybe you find this beautiful article by T. Honda helpful. numdam.org:80/numdam-bin/fitem?id=RSMUP_1973__49__323_0 –  user5831 Oct 22 '10 at 23:02
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2 Answers 2

Say $K=\mathbf{Q}(\sqrt{p^{\*}})$ is the unique quadratic field in which $p$ is the only ramified finite prime. That is, $p^*=(-1)^{(p-1)/2}p$. There are two L-functions one can concoct out of $K$, and quadratic reciprocity manifests in the fact that the L-functions coincide.

First, there is the L-function arising from the Galois character $\chi$ which cuts out $K$. This is a one-dimensional Artin L-function. Its Euler factor at a prime $q$ is $(1\pm p^{-s})^{-1}$, where you place $-$ whenever $q$ splits in $K$, and $+$ whenever $q$ is inert in $K$.

The other L-function is a Dirichlet L-function. Let $\varepsilon$ be the unique character of $(\mathbf{Z}/p\mathbf{Z})^\times$ of order exactly two. This extends to a Dirichlet character $\varepsilon$ on $\mathbf{Z}$, and the Dirichlet L-function is $\sum_{n\geq 1} \varepsilon(n)n^{-s}$.

Let us see that if the two L-functions are equal, then quadratic reciprocity holds. Indeed, look at the coefficient of $q^{-s}$. The coefficient in the first L-function is $\left(\frac{p^*}{q}\right)$. The coefficient in the second L-function is $\left(\frac{q}{p}\right)$. (This requires some explanation: $x\mapsto \left(\frac{x}{p}\right)$ is a character of $(\mathbf{Z}/p\mathbf{Z})^\times$ of order two, so it must equal $\varepsilon$ by uniqueness.) We find

$$\left(\frac{q}{p}\right)=\left(\frac{p^{\*}}{q}\right)=\left(\frac{(-1)^{(p-1)/2}}{q}\right)\left(\frac{p}{q}\right) = (-1)^{\frac{p-1}{2}\frac{q-1}{2}}\left(\frac{p}{q}\right).$$

Of course you didn't need L-functions to state any of this, but they are nonetheless very important. For instance, you can't prove that the Artin L-function has an analytic continuation and functional equation without knowing that it equals a Dirichlet L-function. The generalization of this coincidence to higher-degree Artin L-functions is still quite conjectural!

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What is the version of this for the $l$-tic reciprocity law, where $l$ is a prime $\neq2\;$? –  Chandan Singh Dalawat Oct 23 '10 at 6:06
    
See for example Honda (Taira), Invariant differentials and L-functions. Reciprocity law for quadratic fields and elliptic curves over Q. Rend. Sem. Mat. Univ. Padova 49 (1973), 323–-335, available as numdam.org/item?id=RSMUP_1973__49__323_0 –  Chandan Singh Dalawat May 12 '12 at 7:02
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Let me try to say a few things from my point of view concerning Chandan Singh Dalawat's question about higher reciprocity.

The "Galois character" of the quadratic extension $K$ ramified only at $p$ is given by the Kronecker symbol $\chi(q) = (p^*/q)$; the associated relation $\zeta(s) L(s,\chi) = \zeta_K(s)$ between zeta functions of the rationals, the Dirichlet L-function belonging to $\chi$, and the Dedekind zeta function of the field $K$ encodes the decomposition law of quadratic extensions (see the comment by Sam Derbyshire).

The quadratic reciprocity law in Euler's form states that this Kronecker symbol is a Dirichlet character with conductor $p$, i.e. that $\chi(q) = \chi(r)$ for positive primes $q \equiv r \bmod p$. Thus it must be one of the $p-1$ Dirichlet characters modulo $p$, and since $({\mathbb Z}/p{\mathbb Z})^\times$ is cyclic, is must be the quadratic Dirichlet character modulo $p$.

This has an interpretation in terms of L-functions for Dirichlet characters; the zeta relation $\prod L(s,\chi) = \zeta_L(s)$, where the product on the left hand side is over all Dirichlet characters mod $p$ (suitably interpreted at the bad prime $p$), and where $L$ is the field of $p$-th roots of unity, can be interpreted as the decomposition law for primes in cyclotomic extensions. The reason why there is a connection between these objects and the Kronecker character is, ultimately, the fact that $K \subset L$, something that can be proved directly e.g. with Gauss sums, another tool for proving reciprocity laws. In any case, the ultimate source of this proof is the realization of the Kummer extension $K/{\mathbb Q}$ as a class field (inside the ray class field $L/{\mathbb Q}$).

For getting the cubic reciprocity law, you can work with Kummer extensions over the field of cube roots of unity; comparing the decomposition laws for Kummer extensions and class fields gives you something like $(\pi/q)_3 = (q/\pi)_3$ for primary primes $\pi$ and rational primes $q$, which is strong enough to give you the full cubic reciprocity law with a couple of formal manipulations. Writing down the corresponding Artin and Dirichlet L-series is straightforward.

Something similar works for fourth powers, but then you run into trouble since you only get Eisenstein reciprocity out of this comparison of decomposition laws.

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Many thanks for your answer, Franz. –  Chandan Singh Dalawat Oct 25 '10 at 3:33
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