I'm an undergraduate following Neukirch's Algebraic Number Theory; Please do not assume much more than chapters $1$ and $2$ of this book to answer. The topics covered are: algebraic number fields, behaviour of ideals, prime splitting, $p$-adic numbers, valuations in general, … .
So, to the question:
There is this famous theorem of number theory, called Gauss' Quadratic Reciprocity, which gives us a criterion for deciding whether a equation $x^{2}\equiv p\pmod{q}$, for given primes $p,q$ has solutions.
That's what Quadratic Reciprocity was for me: a computational tool (an interesting one).
I'm about to begin Class Field Theory and I'm been told that I'll study generalizations of this Gauss' law, namely the so called Reciprocity Map (I think due to Artin). That been said, It would be nice to be able to look at the classic Quadratic Reciprocity from a new point a view that makes it very explicit what aspects of it we are generalizing. Or perhaps just a way of look at the classic result as more than a computational tool would be nice already.
There is a part of Neukirch's book where he discusses the quadratic reciprocity (way before CFT), and he says that "the law of decomposition in the cyclotomic field provides the proper explanation to Gauss' Reciprocity Law".
The proof itself, however, seems a bit technical and I'm not getting it.
Can someone help me out? Thanks.
Best Answer
There are a few ways to generalize quadratic reciprocity. Let me describe for you how class field theory over $\mathbb{Q}$ works.
Let $K$ be a finite abelian extension of $\mathbb{Q}$ of degree $n$, Galois group $\Gamma$. One goal of class field theory is the following: give a general rule describing how prime numbers $p$ decompose into a product of primes in $\mathcal O_K$. It's cumbersome to give a succinct rule which describes how all primes split, so the usual workaround is to give a rule which applies to all primes except those in a finite $S$, usually containing all the ramified primes.
This generalizes quadratic reciprocity in the following way: let $K = \mathbb{Q}(\beta)$ for some $\beta \in \mathcal O_K$, and let $f \in \mathbb{Z}[X]$ be the minimal polynomial of $f$ over $\mathbb{Q}$. For almost all primes $p$, you have $(\mathcal O_K)_{(p)} = \mathbb{Z}_{(p)}[\beta]$, and so you can apply the following criterion: if you factor $f$ into a product of irreducibles $p_1^{e} \cdots p_g^{e}$ over $\mathbb{Z}/p\mathbb{Z}[X]$, then $e$ will be the ramification index of $p$ in $K$, the degree $f$ of any of the $p_i$ will be the inertia, and $p$ will split into $\frac{n}{ef}$ primes.
So splitting of primes is analogous to factoring certain polynomials over $\mathbb{Z}/p\mathbb{Z}[X]$. In particular, quadratic reciprocity deals with determining for which primes $p$ the polynomial $X^2 - q \in \mathbb{Z}/p\mathbb{Z}$ is irreducible for a fixed prime $q$ (possibly $q$ is a negative prime). In other words, quadratic reciprocity answers the question of which primes $p$ split in $\mathbb{Q}(\sqrt{q})$, or more generally which primes split in a quadratic extension of $\mathbb{Q}$.
Now, here is how class field theory gives you an algorithm for determining (with a finite number of exceptions) how primes split in $K$, or at least implies the existence of such an algorithm. Given a prime number $p$, fix a prime $\mathfrak p$ of $K$ lying over $p$. Remember that for $p$ unramified in $K$, the decomposition group $$\Gamma_{p} = \{ \sigma \in \Gamma : \sigma \mathfrak p = \mathfrak p \}$$ is cyclic, and it has a particularly nice generator, commonly denoted $(p, K/\mathbb{Q})$. It does not depend on the choice of $\mathfrak p$, because $K/\mathbb{Q}$ is abelian. It is called the Frobenius at $p$. The order $f = f_p$ of $(p,K/\mathbb{Q})$ is the inertial degree of $p$ in $K$, and $p$ splits into $\frac{n}{f}$ primes in $K$.
So given a divisor $g$ of $n$, we want an algorithm for determining which primes $p$ split into $g$ primes in $K$, i.e. which primes satisfy $g = \frac{n}{f_p}$.
The nonconstructive part of the proof is the Kronecker-Weber theorem. It says that there exists some integer $m$ such that $K$ is contained in $\mathbb{Q}(\zeta_m)$. Suppose we have found such an $m$. Let $L = \mathbb{Q}(\zeta_m)$, $G = \textrm{Gal}(L/\mathbb{Q})$, and $H = \textrm{Gal}(L/K)$. There is a canonical identification of $G$ with the group $(\mathbb{Z}/m\mathbb{Z})^{\ast}$, and under this identification we can think of $H$ as a subgroup of $(\mathbb{Z}/m\mathbb{Z})^{\ast}$. For $p$ not dividing $m$, the Frobenius $(p,L/\mathbb{Q})$ has the effect $\zeta_m \mapsto \zeta_m^p$. In particular, $(p,L/\mathbb{Q})$ can be identified with $p$ modulo $m$, and the inertial degree of $p$ in $L$ is then the multiplicative order of $p$ modulo $m$.
In general, if $\phi: A \rightarrow B$ is a homomorphism of finite groups, and $a \in A$, then the order of $\phi(a)$ is the smallest number $f$ such that $a^f \in \textrm{Ker } \phi$. Now, you can check that the restriction of $(p,L/\mathbb{Q})$ to an automorphism of $K$ is exactly $(p,K/\mathbb{Q})$. Therefore, the inertial degree of $(p,K/\mathbb{Q})$ is the smallest number $f$ such that $p^f$ is congruent modulo $m$ to a member of $H$.
So to summarize, here is how to determine how primes split in $K$:
Find an $m$ such that $K \subseteq \mathbb{Q}(\zeta_m)$.
Identify the Galois group of $\mathbb{Q}(\zeta_m)$ with $(\mathbb{Z}/m\mathbb{Z})^{\ast}$, and identify $H = \textrm{Gal}(\mathbb{Q}(\zeta_m)/\mathbb{Q})$ as a subgroup of $(\mathbb{Z}/m\mathbb{Z})^{\ast}$.
With the exception of the prime factors of $m$, the primes $p$ which split into $g$ factors in $K$ are exactly those primes for which $p+H$ has order $\frac{n}{g}$ in $G/H$.