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The reciprocity of the title is the following not completely well-posed problem:

Fix $P(X)$ a monic irreducible polynomial of degree $n$, with coefficients in $\mathbb Z$. "Describe" (in some sense) the set of prime $p$ such that $P \mod p$ has a given reduction (for example is irreducible, or on the opposite is a product of $n$ linear factors, etc.).

As is well-known, when one root (hence all roots) of $P$ happens to be a cyclotomic integer (in particular in the very special case where $P$ is quadratic), then the problem has a precise solution, called Artin's reciprocity law, which is the core of class field theory. The sets of primes $p$ such that $P$ has a given reduction are then given by congruences modulo fixed integers satisfied by $p$, that is those sets are union of "set primes in artithemtic sequence". The Galois group of $P$ is abelian in this case.

As the opposite of the complexity spectrum, there are polynomial $P(X)$ of degree $n \geq 5$, whose Galois group is big, say $S_n$, or $A_n$ in particular not solvable. In this case, it is my understanding that our only hope to find a reciprocity law in general is by making huge progress in the Langlands program.

But is it possible, in some special case (for polynomials $P$ of Galois group $S_n$ say satisfying somme assumtions, or for an explicit family of polynomials of variable degree $n \geq 5$, or even for just one explicit polynomials) where some kind of reciprocity law has been worked out (even only partially) by some method that does not involve the Langlands program (i.e. modular forms, automorphic forms, etc.) ?

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I think that Euler already knew the primes $p$ for which $x^3-2$ splits completely mod $p$; one can dress the result up as some weight 1 modular form defined by theta series, and of course the Galois group is soluble, but I am pretty sure that Euler didn't know about the Langlands program. On the other hand of course this doesn't answer the question, because the weight 1 forms that you can access using theta series are precisely those ones giving dihedral Galois reps. –  user30035 Mar 21 '13 at 19:47
Joel -- an afterthought -- I think that this question was raised explicitly way way before Langlands. There are exercises in Cassels-Froehlich about the primes that split completely in an (arbitrary) finite extension of number fields (proving e.g. that these primes determine the extension up to iso etc etc), plus some sort of assertion that beyond abelian situations [so perhaps implicitly allowing solvable cases] no-one has a clue how to classify these primes. I know C-F was after Langlands but one can easily imagine that the question was much older... –  user30035 Mar 21 '13 at 20:05
There is Shimura's 1966 Crelle paper "A reciprocity law in non-solvable extensions". –  paul garrett Mar 21 '13 at 21:11
PS Keith isn't it just something like $p=x^2+27y^2$? My source for all of this was my memory of the first few sections of Cox's book. When I get into the office my new source will be Cox's book and perhaps I'll have to retract some assertions :-/ –  user30035 Mar 22 '13 at 7:39
$S_3$ is solvable. I thought you wanted nonsolvable examples. –  David Speyer Mar 22 '13 at 11:44
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1 Answer

This is a very interesting question and would make an excellent topic for a doctoral thesis in the history of mathematics. I will interpret the question as

Which pre-Langlands results, problems, and theories --- apart from what is easily deducible from the theory of $\;\mathrm{GL}_1$ (from Gauß to Tate) --- can now be considered a part of the Langlands programme ?

There is nothing original in my answer : everything is gleaned from the writings of Langlands, Serre and Weil. I may have misrepresented some of their words, and in any case our future doctoral candidate will have to delve deeper into the original sources.

Fricke & Klein (1912) observe that the modular curve $X_0(11)$ of level $\Gamma_0(11)$ is defined by the equation $\sigma^2=1-20\tau+56\tau^2-44\tau^3$.

Hasse (193?) asks a doctoral student (Pierre Humbert) to prove that the $L$-function of an elliptic curve $E$ over $\mathbf{Q}$ (defined as the product over various primes $p$ of the $\zeta$-function of $E$ modulo $p$) is entire and satisfies a functional equation. Humbert sagely decides to work on quadratic forms with Siegel instead.

Weil (1951) asks in his report Sur la théorie du corps de classes for a galoisian interpretation of the whole idèle class group of a number field (as opposed to the quotient of the said group by the connected component of the identity), analogous to the galoisian interpretation in the function field case. See http://mathoverflow.net/questions/41318 in this regard.

Weil (1952) shows that certain elliptic curves with complex multiplications (such as $y^2=x^4+1$) are modular.

Deuring (1953--1957) proves (following a suggestion by Weil) that all elliptic curves with complex multiplications are modular.

Eichler (1954) proves that the $L$-function of $X_0(N)$ is essentially the product of Hecke $L$-functions attached to cuspidal eigenforms of weight $2$ and level $N$. This was generalised by Shimura (1958) and completed by Igusa (1959).

Taniyama (1955) asks at the Tokyo-Nikko conference a somewhat imprecise question which some interpret as implying that one can prove Hasse's conjecture for $E$ by showing that $E$ is modular.

Shimura (1966) explicitly determines the reciprocity law for the splitting of rational primes in the number field obtained by adjoining the $l$-torsion ($l$ prime) of the Fricke curve $X_0(11)$ in terms of the coefficient $c_l$ of $q^l$ in the modular form $$ q\prod_{n>0}(1-q^n)^2(1-q^{11n})^2 $$ (but only for $l<100$ for which he could check that the mod-$l$ representation is surjective).

Weil (1967) proves that if an elliptic curve over $\mathbf{Q}$ is modular, then it has to be modular of level equal to its conductor, and assigns the Übungsaufgabe to the interested reader to show that every elliptic curve over $\mathbf{Q}$ is indeed modular.

Around this time Langlands wrote a letter to Weil and changed the world.

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Dear Chandan, thanks for writing this very instructive answer. However, somehow this is not exactly the kind of answer I was looking for. As you say very clearly, you are answering the question "what pre-Langlands cases of reciprocity are now integrated into the Langlands program ?", and you offer a long list of interesting examples. But I wonder: are there any case of reciprocity result which at this point remain completely foreign to the Langlands program? Very possibly, there aren't any, and if this is so your answer is the best possible. However, I had the hope when asking this ... –  Joël Mar 28 '13 at 19:08
... a question that someone would be able to give such an example. I believe it might help if I give some completely imaginary example of what I was hoping for. (Phony) Theorem: Let $P_n(x)$ the polynomial $n! (1 + x + x^2/2 + x^3/3! + \dots + x^n /n!)$ which is known to be of Galois group $S_n$ if (in particular) $n$ is odd. Then a $P_n(x)$ is still irreducible modulo $p$ if and only if $1/(p+1)$ is an energy level of the molecule $C_n H_{2n+2}$. This is proved using first Dessin d'enfants then quantum field theory in that 1997 paper of such and such. –  Joël Mar 28 '13 at 19:17
In other words: with the Langlands program, we have taken the habit of thought that reciprocity must be expressed in terms of modular of automorphic forms. For example, to get info of how a minimal polynomial for the field of $l$-torsion points of $X_0(11)$, we look at the coefficients of some explicit modular form, as you recall. But are there case where we know a law of reciprocity for some polynomial where the conditions on $p$ does not apparently have anything to do with modular forms or automorphic forms? –  Joël Mar 28 '13 at 19:23
@Joël. Something can be deduced by transcendental number theoretical methods (Chudnovski, André, Bost) applied to the Grothendieck-Katz conjecture on differential equations, although this is certainly much weaker than a reciprocity. For example, one can prove without any Galois theory that an irreducible polynomial of degree $>1$ with integer coefficients can't have a root modulo almost all prime numbers, even modulo too high a (weak form of) density of prime numbers. (One proves that the power series $(1+x)^\alpha$ would be an algebraic function, hence $\alpha\in\mathbf Q$.) –  ACL Mar 28 '13 at 23:27
Dear Joël, with this clarification of what exactly you want, I would once again recommend Philippe Satgé's paper from 1977 numdam.org/numdam-bin/fitem?id=AIF_1977__27_4_1_0 about which I asked a question here some time ago mathoverflow.net/questions/119305/…. –  Chandan Singh Dalawat Mar 29 '13 at 0:33
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