show/hide this revision's text 3 added 20 characters in body

I think it is easiest to illustrate the role of the Langlands program (i.e. non-abelian class field theory) in answering this question by giving an example.

E.g. consider the Hilbert class field $K$ of $F := {\mathbb Q}(\sqrt{-23})$; this is a degree 3 extension abelian extension of $F$, and an $S_3$ extension of $\mathbb Q$. (It is the splitting field of the polynomial $x^3 - x - 1$.)

The 2-dimensional representation of $S_3$ thus gives a representation $\rho:Gal(K/{\mathbb Q}) \hookrightarrow GL_2({\mathbb Q}).$
A prime $p$ splits in $K$ if and only if $Frob_p$ is the trivial conjugacy class in $Gal(K{\mathbb Q})$, if and only if $\rho(Frob_p)$ is the identity matrix, if and only if trace $\rho(Frob_p) = 2$. (EDIT: While $Frob_p$ is a 2-cycle, resp. 3-cycle, if and only if $\rho(Frob_p)$ has trace 0, resp. -1.)

Now we have the following reciprocity law for $\rho$: there is a modular form $f(q)$, in fact a Hecke eigenform, of weight 1 and level 23, whose $p$th Hecke eigenvalue gives the trace of $\rho(Frob_p)$. (This is due to Hecke; the reason that Hecke could handle this case is that $\rho$ embeds $Gal(K/{\mathbb Q})$ as a dihedral subgroup of $GL_2$, and so $\rho$ is in fact induced from an abelian character of the index two subgroup $Gal(K/F)$.)

In this particular case, we have the following explicit formula:

$$f(q) = q \prod_{n=1}^{\infty}(1-q^n)(1-q^{23 n}).$$

If we expand out this product as $f(q) = \sum_{n = 1}^{\infty}a_n q^n,$ then we find that $trace \rho(Frob_p) = a_p$ (for $p \neq 23$), and in particular, $p$ splits completely in $K$ if and only if $a_p = 2$. (For example, you can check this way that the smallest split prime is $p = 59$; this is related to the fact that $59 = 6^2 + 23 \cdot 1^2$.). (EDIT: While $Frob_p$ has order $2$, resp. 3, if and only if $a_p =0$, resp. $-1$.)

So we obtain a description of the set of primes that split in $K$ in terms of the modular form $f(q)$, or more precisely its Hecke eigenvalues (or what amounts to the same thing, its $q$-expansion).

The Langlands program asserts that an analogous statement is true for any Galois extension of number fields $E/F$ when one is given a continuous representation $Gal(E/F) \hookrightarrow GL_n(\mathbb C).$ This is known when $n = 2$ and either the image of $\rho$ is solvable (Langlands--Tunnell) or $F = \mathbb Q$ and $\rho(\text{complex conjugation})$ is non-scalar (Khare--Wintenberger--Kisin). In most other contexts it remains open.
show/hide this revision's text 2 added 191 characters in body

I think it is easiest to illustrate the role of the Langlands program (i.e. non-abelian class field theory) in answering this question by giving an example.

E.g. consider the Hilbert class field $K$ of $F := {\mathbb Q}(\sqrt{-23})$; this is a degree 3 extension abelian extension of $F$, and an $S_3$ extension of $\mathbb Q$. (It is the splitting field of the polynomial $x^3 - x - 1$.)

The 2-dimensional representation of $S_3$ thus gives a representation $\rho:Gal(K/{\mathbb Q}) \hookrightarrow GL_2({\mathbb Q}).$
A prime $p$ splits in $K$ if and only if $Frob_p$ is the trivial conjugacy class in $Gal(K{\mathbb Q})$, if and only if $\rho(Frob_p)$ is the identity matrix, if and only if trace $\rho(Frob_p) = 2$. (EDIT: While $Frob_p$ is a 2-cycle, resp. 3-cycle, if and only if $\rho(Frob_p)$ has trace 0, resp. -1.)

Now we have the following reciprocity law for $\rho$: there is a modular form $f(q)$, in fact a Hecke eigenform, of weight 1 and level 23, whose $p$th Hecke eigenvalue gives the trace of $\rho(Frob_p)$. (This is due to Hecke; the reason that Hecke could handle this case is that $\rho$ embeds $Gal(K/{\mathbb Q})$ as a dihedral subgroup of $GL_2$, and so $\rho$ is in fact induced from an abelian character of the index two subgroup $Gal(K/F)$.)

In this particular case, we have the following explicit formula:

$$f(q) = q \prod_{n=1}^{\infty}(1-q^n)(1-q^{23 n}).$$

If we expand out this product as $f(q) = \sum_{n = 1}^{\infty}a_n q^n,$ then we find that $trace \rho(Frob_p) = a_p$ (for $p \neq 23$), and in particular, $p$ splits completely in $K$ if and only if $a_p = 2$. (For example, you can check this way that the smallest split prime is $p = 59$; this is related to the fact that $59 = 6^2 + 23 \cdot 1^2$.)1^2$.). (EDIT: While $Frob_p$ has order $2$, resp. 3, if and only if $a_p =0$, resp. $-1$.)

So we obtain a description of the set of primes that split in $K$ in terms of the modular form $f(q)$, or more precisely its Hecke eigenvalues (or what amounts to the same thing, its $q$-expansion).

The Langlands program asserts that an analogous statement is true for any Galois extension of number fields $E/F$ when one is given a continuous representation $Gal(E/F) \hookrightarrow GL_n(\mathbb C).$ This is known when $n = 2$ and either the image of $\rho$ is solvable (Langlands--Tunnell) or $\rho(\text{complex conjugation})$ is non-scalar (Khare--Wintenberger--Kisin). In most other contexts it remains open.
show/hide this revision's text 1

I think it is easiest to illustrate the role of the Langlands program (i.e. non-abelian class field theory) in answering this question by giving an example.

E.g. consider the Hilbert class field $K$ of $F := {\mathbb Q}(\sqrt{-23})$; this is a degree 3 extension abelian extension of $F$, and an $S_3$ extension of $\mathbb Q$. (It is the splitting field of the polynomial $x^3 - x - 1$.)

The 2-dimensional representation of $S_3$ thus gives a representation $\rho:Gal(K/{\mathbb Q}) \hookrightarrow GL_2({\mathbb Q}).$
A prime $p$ splits in $K$ if and only if $Frob_p$ is the trivial conjugacy class in $Gal(K{\mathbb Q})$, if and only if $\rho(Frob_p)$ is the identity matrix, if and only if trace $\rho(Frob_p) = 2$.

Now we have the following reciprocity law for $\rho$: there is a modular form $f(q)$, in fact a Hecke eigenform, of weight 1 and level 23, whose $p$th Hecke eigenvalue gives the trace of $\rho(Frob_p)$. (This is due to Hecke; the reason that Hecke could handle this case is that $\rho$ embeds $Gal(K/{\mathbb Q})$ as a dihedral subgroup of $GL_2$, and so $\rho$ is in fact induced from an abelian character of the index two subgroup $Gal(K/F)$.)

In this particular case, we have the following explicit formula:

$$f(q) = q \prod_{n=1}^{\infty}(1-q^n)(1-q^{23 n}).$$

If we expand out this product as $f(q) = \sum_{n = 1}^{\infty}a_n q^n,$ then we find that $trace \rho(Frob_p) = a_p$ (for $p \neq 23$), and in particular, $p$ splits completely in $K$ if and only if $a_p = 2$. (For example, you can check this way that the smallest split prime is $p = 59$; this is related to the fact that $59 = 6^2 + 23 \cdot 1^2$.)

So we obtain a description of the set of primes that split in $K$ in terms of the modular form $f(q)$, or more precisely its Hecke eigenvalues (or what amounts to the same thing, its $q$-expansion).

The Langlands program asserts that an analogous statement is true for any Galois extension of number fields $E/F$ when one is given a continuous representation $Gal(E/F) \hookrightarrow GL_n(\mathbb C).$ This is known when $n = 2$ and either the image of $\rho$ is solvable (Langlands--Tunnell) or $\rho(\text{complex conjugation})$ is non-scalar (Khare--Wintenberger--Kisin). In most other contexts it remains open.