The question is about characterising the sets $S(K)$ of primes which split completely in a given galoisian extension $K|\mathbb{Q}$. Do recent results such as Serre's modularity conjecture (as proved by Khare-Wintenberger), or certain cases of the Fontaine-Mazur conjecture (as proved by Kisin), have anything to say about such subsets, beyond what Class Field Theory has to say ?

I'll now introduce some terminology and recall some background.

Let $\mathbb{P}$ be the set of prime numbers. For every galoisian extension $K|\mathbb{Q}$, we have the subset $S(K)\subset\mathbb{P}$ consisting of those primes which split (completely) in $K$. The question is about characterising such subsets; we call them galoisian subsets.

If $T\subset\mathbb{P}$ is galoisian, there is a unique galoisian extension $K|\mathbb{Q}$ such that $T=S(K)$, cf. Neukirch (13.10). We say that $T$ is abelian if $K|\mathbb{Q}$ is abelian.

As discussed here recently, a subset $T\subset\mathbb{P}$ is abelian if and only if it is defined by congruences. For example, the set of primes $\equiv1\pmod{l}$ is the same as $S(\mathbb{Q}(\zeta_l))$. "Being defined by congruences" can be made precise, and counts as a characterisation of abelian subsets of $\mathbb{P}$.

Neukirch says that Langlands' Philosophy provides a characterisation of all galoisian subsets of $\mathbb{P}$. Can this remark now be illustrated by some striking example ?

Addendum (28/02/2010) Berger's recent Bourbaki exposé 1017 arXiv:1002.4111 says that cases of the Fontaine-Mazur conjecture have been proved by Matthew Emerton as well. I didn't know this at the time of asking the question, and the unique answerer did not let on that he'd had something to do with Fontaine-Mazur...

  • $\begingroup$ It doesn't answer the question, but such sets are studied in mathi.uni-heidelberg.de/~wingberg/agwingberg/paperwingberg/…. $\endgroup$
    – user19475
    Jan 14, 2010 at 19:45
  • $\begingroup$ The fact that a galoisian set of primes numbers $T$ uniquely determines the finite galoisian extension $K|\mathbf{Q}$ such that $T=S(K)$ goes back to M. Bauer, Über einen Satz von Kronecker, Arch. der Math. u. Phys. (3) 6, 218--219 (1903). $\endgroup$ Jan 1, 2012 at 12:44

2 Answers 2


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 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.

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    $\begingroup$ Such a concrete illustrative example is exactly what I wanted. Many thanks! $\endgroup$ Jan 20, 2010 at 6:37
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    $\begingroup$ A related and naive question: instead of answering when a prime p splits, has people use a similar technique to say anything about when the Frobenius $Frob_p$ has certain orders (e.g. generating the full Galois group)? $\endgroup$
    – Anonymous
    Feb 28, 2010 at 17:25
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    $\begingroup$ Given a finite Galois ext. of number fields $L/K$, if we let $\rho_i:Gal(L/K) \to GL_{n_i}(\mathbb C)$ run over the irreps. of $Gal(L/K)$, then a prime $v$ that is unramified in $L$ splits completely in $L$ if and only if $\rho_i(Frob_v)$ is trivial for each $i$, if and only if the Satake parameter associated to the local factor at $v$ in the (conjecturally) associated cuspidal automorphic rep'n $\pi_i$ of $GL_{n_i}(\mathbb A_K)$ is trivial for each $i$. So we get a finite list of cuspidal automorphic rep's $\pi_i$ whose behaviour at unramified primes $v$ controls their splitting in $L$. $\endgroup$
    – Emerton
    Jul 24, 2010 at 0:19
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    $\begingroup$ Its worth adding that this example goes back at least to Shimura, in "A reciprocity law in non-solvable extensions", J. Reine Angew. Math. v. 221 (1966) pp. 209–220. I have heard stories that Hecke was looking at this automorphic form L-function at the same time Artin was looking at the same example from the viewpoint of Artin L-functions, each without being aware of the other. $\endgroup$
    – Stopple
    Mar 21, 2011 at 16:24
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    $\begingroup$ Actually, this example appeared in F. van der Blij: Binary quadratic forms of discriminant -23, Indagationes Mathematicae, 1952. $\endgroup$ May 23, 2018 at 22:23

In order to add my bit to the already rich content on this site, here is a nice family of galoisian extensions $K_l|{\bf Q}$ with group ${\rm GL}_2({\bf F}_l)$ (indexed by primes $l\neq5$) for which a ``reciprocity law'' can be written down explicitly. I've come across this family recently while writing an expository article.

Let $E$ be the elliptic curve (over $\bf Q$) of conductor $11$ defined by $y^2+y=x^3-x^2$, with associated modular form $$ \eta_{1^2,11^2}=q\prod_{k>0}(1-q^k)^2(1-q^{11k})^2=\sum_{n>0}c_nq^n. $$ Let $K_l={\bf Q}(E[l])$, which is thus galoisian over $\bf Q$ and unramified at every prime $p\neq11,l$.

One can deduce from cor.1 on p.308 of Serre (Inventiones 1972) that for every prime $l\neq5$, the representation $$ \rho_{E,l}:{\rm Gal}(K_l|{\bf Q})\rightarrow{\rm GL}_2({\bf F}_l) $$ we get upon choosing an ${\bf F}_l$-basis of $E[l]$ is an isomorphism; cf. the online notes on Serre's conjecture by Ribet and Stein. Shimura did this for $l\in[9,97]$ (Crelle 1966).

Suppose henceforth that $l$ is a prime $\neq5$ and that $p$ is a prime $\neq11,l$. The characteristic polynomial of $\rho_{E,l}({\rm Frob}_p)\in{\rm GL}_2({\bf F}_l)$ is $$ T^2-\bar c_pT+\bar p\in{\bf F}_l[X]. $$ The prime $p$ splits completely in $K_l$ if and only if ${\rm Frob}_p=1$ in ${\rm Gal}(K_l|{\bf Q})$, which happens if and only if $$ \rho_{E,l}({\rm Frob}_p)=\pmatrix{1&0\cr0&1}. $$ If so, then $p,c_p\equiv1,2\pmod l$ but not conversely, for the matrix $\displaystyle\pmatrix{1&1\cr0&1}$ also has the characteristic polynomial $T^2-\bar2T+\bar1$. But these congruences on $p,c_p$ do rule out an awful lot of primes as not splitting completely in $K_l$.

In summary, we have the following ``reciprocity law" for $K_l$ : $$ \hbox{($p$ splits completely in $K_l$)} \quad\Leftrightarrow\quad E_p[l]\subset E_p({\bf F}_p), $$ where $E_p$ is the reduction of $E$ modulo $p$. Indeed, reduction modulo $p$ identifies $E[l]$ with $E_p[l]$ and the action of ${\rm Frob}_p$ on the former space with the action of the canonical generator $\varphi_p\in{\rm Gal}(\bar{\bf F}_p|{\bf F}_p)$ on the latter space. To say that $\varphi_p$ acts trivially on $E_p[l]$ is the same as saying that $E_p[l]$ is contained in the ${\bf F}_p$-rational points of $E_p$. The analogy with the multiplicative group $\mu$ is perfect: $$ \hbox{($p\neq l$ splits completely in ${\bf Q}(\mu[l])$)} \quad\Leftrightarrow\quad \mu_p[l]\subset \mu_p({\bf F}_p) $$ ($\Leftrightarrow l|p-1\Leftrightarrow p\equiv1\pmod l$), where $\mu_p$ is not the $p$-torsion of $\mu$ but the reduction of $\mu$ modulo $p$.

I requested Tim Dokchitser to compute the first ten $p$ which split completely in $K_7$, and his instantaneous response was 4831, 22051, 78583, 125441, 129641, 147617, 153287, 173573, 195581, and 199501.

It is true that all this (except the list of these ten primes) was known before Serre's conjecture was proved (2006--9) or even formulated (1973--87), but I find this example a very good illustration of the kind of reciprocity laws it provides.

I hope you enjoyed it as much as I did.

  • $\begingroup$ From the reciprocity law we also have that "splits completely" implies $p-c_p+1 = 0 \pmod{l^2}$. Is there a way to understand this from the modular form and representation? $\endgroup$ Jan 3, 2011 at 15:25
  • $\begingroup$ Perhaps one should look at the representation on the $l$-adic Tate module. $\endgroup$ Jan 4, 2011 at 6:39

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