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\title{Counting Discriminants of Number Fields}
\author{Henri Cohen, Francisco Diaz y Diaz and Michel Olivier\\
Laboratoire A2X, U.M.R. 5465 du C.N.R.S.,\\
Universit\'e Bordeaux I,\\
351 Cours de la Lib\'eration,\\
33405 TALENCE Cedex, FRANCE}
%\keywords{relative extension, class field theory, discriminant}
%\email{\{cohen,diaz,olivier\}@math.u-bordeaux.fr}\\
\maketitle
\begin{abstract}
This paper is intended as a summary of the conjectures and results
obtained recently on the asymptotic and exact counting of relative and 
absolute extensions of number fields. No proofs are given.
\end{abstract}

\smallskip

\centerline{Version of \today}

\smallskip

\section{General Conjectures and Results}

Let $\ov{\Q}$ be a fixed algebraic closure of $\Q$. Let $K\subset\ov{\Q}$ be a number field, and
let $G$ be a transitive permutation group on $n$ letters. We consider the set ${\cal F}_n(G)$ of all 
extensions $L/K$ of degree $n$ with $L\subset\ov{\Q}$ such that the Galois group of the Galois
closure of $L/K$ is isomorphic to $G$. We write
$$N_{K,n}(G,X)=|\{L\in{\cal F}_n(G),\ |\N(\gd(L/K))|\le X\}\;,$$
where $\gd(L/K)$ denotes the relative ideal discriminant, $\N$ the absolute norm (so that of course
$|\N(\gd(L/K))|=|d(L)|$, the absolute value of the discriminant of $L$). The aim of this paper
is to give results and conjectures on exact and asymptotic values of these quantities,
without proof. It is very easy to generalize the results to the case where the behavior
of a \emph{finite} number of places of $K$ in the extension $L/K$ are specified, but we will not
do this. It is also usually possible to give additional main terms and rather good error terms
instead of asymptotic formulas, but we will not do this either.

A number of general conjectures on the subject has been made by several authors. In view of
available data and theorems, it now seems reasonable to formulate the following precise
conjectures (see for example \cite{Coh2} and \cite{Mal}).

\begin{conjecture}\label{con1}
\begin{enumerate}\item For each number field $K$ and transitive group $G$ on
$n$ letters as above, there exist three strictly positive constants $a_K(G)$, $b_K(G)$ and
$c_K(G)$ such that $$N_{K,n}(G,X)\sim c_K(G)\,X^{a_K(G)}(\log X)^{b_K(G)}\;.$$
\item Furthermore, the constant $a_K(G)$ should not depend on $K$ (so will be denoted by $a(G)$)
and should be a rational number satisfying $0<a(G)\le1$, and $b_K(G)$ should be a nonnegative
integer, equal to $0$ if $a(G)=1$.
\item If $G$ is a primitive transitive group, we should have $a(G)<1$ except when $G\isom S_n$,
in which case we should have $a(S_n)=1$.
\item On the contrary, if $n$ is composite (so that there exist imprimitive groups $G$), there
exists at least one imprimitive transitive group $G$ such that $a(G)=1$.
\item The total number of extensions $L/K$ in $\ov{\Q}$ of degree $n$ and absolute discriminant
bounded by $X$ should be asymptotic to $c_K\,X$ for some positive constant $c_K$.
\end{enumerate}\end{conjecture}

An even more precise version of this conjecture concerning the value of 
$a(G)$ has been made by G.~Malle \cite{Mal} as follows.

\begin{definition} For any element $g\in S_n$ different from the identity, define the index 
$\ind(g)$ of $g$ by the formula $$\ind(g)=n-|\text{orbits of }g|\;.$$ 
We define the index $i(G)$ of a transitive subgroup $G$ of $S_n$ by the formula 
$$i(G)=\min_{g\in G,\ g\neq1}\ind(g)\;.$$
\end{definition}

Examples: \begin{enumerate}\item The index of a transposition is equal to $1$, and this is the
lowest possible index for a nonidentity element. It follows that $i(S_n)=1$.
\item If $G$ is an Abelian group, and if $\ell$ is the smallest prime divisor of $|G|$, then
it is easy to show that $i(G)=|G|(1-1/\ell)$.\end{enumerate}

\begin{conjecture} (Malle). For any transitive subgroup $G$ of $S_n$, we should have $a(G)=1/i(G)$.
\end{conjecture}

It can be shown that the statements (2), (3), and (4) about $a(G)$ in
Conjecture \ref{con1} follow from Malle's conjecture.

The following results give support to the conjecture (see \cite{Wri},
\cite{CoDiOl2}, \cite{CoDiOl3}).

\begin{theorem}\begin{enumerate}\item (Wright). Malle's conjecture is
true for all Abelian groups $G$.
\item (CDO). Malle's conjecture is true for $n\le4$ (we will give more
precise results below) and for $G=D_\ell$, for $\ell$ prime.
\end{enumerate}
\end{theorem} 

In addition, Malle gives partial results towards the statement that his 
conjecture is compatible with direct products and with wreath products.

Malle does not give a conjecture for the value of the exponent of the 
logarithm $b_K(G)$ (in fact he does not even conjecture such a precise
form). However, in the Abelian case, the same paper of Wright proves the 
following theorem:

\begin{theorem} (Wright). Let $G$ be an Abelian group, and let $\ell$ be
the smallest prime divisor of $|G|$ (so that $a(G)=(|G|(1-1/\ell))^{-1}$).
Denote by $B_\ell(G)$ the number of elements of $G$ of order $\ell$, 
which will be of the form $\ell^k-1$ for some positive $k$.
Then we have $$b_K(G)=\dfrac{B_\ell(G)}{[K(\zeta_\ell):K]}-1\;.$$
\end{theorem}

The paper of Wright also claims an explicit expression for the constant
$c_K(G)$, but although it is a finite product of adelic integrals, as
far as the authors are aware, it has not been computed explicitly by
this method apart from the case where $G$ is of order $2$. We have
computed it by other means in more general situations.

Finally, concerning statement (4) of the general conjecture, Malle
proves the following theorem:

\begin{theorem} If $n$ is a composite number divisible by either $2$ or
$3$, there exists an imprimitive transitive subgroup $G$ of $S_n$ such
that $a(G)=1$.
\end{theorem}

He of course conjectures that this remains true for any composite $n$,
not only those divisible by $2$ and $3$. In particular, if the main
conjecture is true, this shows that, for composite $n$, the proportion 
of $S_n$-extensions among all extensions of degree $n$ is strictly less
than $1$. Thanks to our results, this is now a theorem for $n=4$.

This is in complete opposition with the situation for
\emph{polynomials}, where Hilbert's irreducibility theorem shows that
``almost all'' polynomials of degree $n$ have Galois group $S_n$.

\section{Results in Small Degrees}

\subsection{$G=C_2$}

$$N_{K,2}(C_2,X)\sim c_K(C_2)\,X\text{\quad with}$$
$$c_K(C_2)=\dfrac{1}{2^{r_2}}\dfrac{\zeta_K(1)}{\zeta_K(2)}\;.$$
Here $r_2=r_2(K)$ is the number of complex places of $K$, and
$\zeta_K(1)$ is a useful abuse of
notation for the residue of $\zeta_K(s)$ at $s=1$.

This very simple result deserves to be better known, but its proof is
not simple (a few pages). It is due (using Shintani's theory of zeta
functions of prehomogeneous vector spaces, see \cite{Shi1}, \cite{Shi2},
\cite{Wri-Yuk}) to Datskowsky and Wright \cite{Dat-Wri}, and was
recently reproved by the authors using Kummer theory. Of course, in particular
$$c_{\Q}(C_2)=\dfrac{1}{\zeta(2)}=\dfrac{6}{\pi^2}\;.$$

Using some nontrivial series manipulation \cite{Coh3}, one can for
example compute exactly (3 weeks CPU time)
$$N_{\Q,2}(C_2,10^{25})=6079271018540266286517795\;.$$

\subsection{$G=C_3$}

$$N_{K,3}(C_3,X)\sim\begin{cases}c_K(C_3)\,X^{1/2}&\text{\quad if $\zeta_3\notin K$}\\
c_K(C_3)\,X^{1/2}\log X&\text{\quad if $\zeta_3\in K$\;.}\end{cases}$$
Here, when $\zeta_3\notin K$, setting $K_z=K(\zeta_3)=K(\sqrt{-3})$, we have

\begin{align*}c_K(C_3)&=\dfrac{1}{2\cdot 3^{r_1+r_2-1}}\dfrac{\zeta_{K_z}(1)}{\zeta_K(2)}
\prod_{\leg{K_z}{\p}=1}\left(1-\dfrac{2}{\N\p(\N\p+1)}\right)\\
&\phantom{=}\kern-6pt\prod_{\leg{K_z}{\p}=0}\kern-5pt\left(1+\dfrac{1}{\N\p+1}-\dfrac{1}{\N\p^{(e(\p)+1)/2}}\right)
\prod_{\substack{\leg{K_z}{\p}=-1\\\p\mid3}}\kern-10pt\left(1+\dfrac{2}{\N\p}-\dfrac{2}{\N\p^{e(\p)/2}}\right)\;.
\end{align*}
In the above, $\leg{K_z}{\p}=-1,0\text{ or }1$ means that $\p$ is inert,
ramified or split in the quadratic extension $K_z/K$, and
$e(\p)=e(\p/3)$ is the absolute ramification index of a prime ideal
dividing 3. Note that, of course, in the second product the condition
$\leg{K_z}{\p}=0$ implies that $\p\mid3$ (it is in fact equivalent to
$e(\p)$ being odd).

On the other hand, if $\zeta_3\in K$ we have the much simpler formula
$$c_K(C_3)=\dfrac{1}{4\cdot 3^{r_2}}\zeta_K(1)^2\prod_{\p}\left(1+\dfrac{2}{\N\p}\right)\left(1-\dfrac{1}{\N\p}\right)^2\;.$$
In particular, 
$$c_{\Q}(C_3)=\dfrac{11\sqrt3}{36\pi}\prod_{p\equiv1\pmod6}\left(1-\dfrac{2}{p(p+1)}\right)\;.$$
The general result is apparently new, but the result for $K=\Q$ is due
to H.~Cohn \cite{Cohn}.

Using some nontrivial series manipulation \cite{Coh3}, one can for
example compute exactly
$$N_{\Q,3}(C_3,10^{37})=501310370031289126\;.$$

\subsection{$G=S_3$}

$$N_{K,3}(S_3,X)\sim c_K(S_3)\,X\text{\quad with}$$
$$c_K(S_3)=\left(\dfrac23\right)^{r_1-1}\left(\dfrac16\right)^{r_2}\dfrac{\zeta_K(1)}{\zeta_K(3)}=\dfrac{2^{r_1-r_2-1}}{3^{r_1+r_2-1}}\dfrac{\zeta_K(1)}{\zeta_K(3)}\;.$$
In particular $$c_{\Q}(S_3)=\dfrac{1}{\zeta(3)}\;.$$
This last result is the beautiful and difficult result of 
Davenport--Heilbronn \cite{Dav-Hei1}, \cite{Dav-Hei2}, but the
general result is much deeper and is due to Datskowsky and Wright 
\cite{Dat-Wri}. Our methods enable us to find the correct power of $X$
and $\log X$ ($1$ and $0$ respectively), but give a dreadful
expression for the constant $c_K(S_3)$.

Using the algorithmic methods of K.~Belabas \cite{Bel}, based on the
Heilbronn--Davenport theory, one can for example compute exactly
$$N_{\Q,3}(S_3,10^{11})=81414013239\;.$$

\subsection{$G=C_4$}

$$N_{K,4}(C_4,X)\sim c_K(C_4)\,X^{1/2}\;.$$
We have not had time to compute the constant $c_K(C_4)$ in the general
case. For $K=\Q$, we have
$$c_{\Q}(C_4)=\dfrac{3}{\pi^2}\left(\left(1+\dfrac{\sqrt2}{24}\right)\prod_{p\equiv1\pmod4}\left(1+\dfrac{2}{p^{1/2}(p+1)}\right)-1\right)\;.$$
This result is due in principle to A.~Baily \cite{Bai}, but there are 
computational mistakes in his paper.

Using some nontrivial series manipulation \cite{Coh3}, one can for
example compute exactly
$$N_{\Q,4}(C_4,10^{30})=122051516492357\;.$$

\subsection{$G=V_4=C_2\times C_2$}

$$N_{K,4}(V_4,X)\sim c_K(V_4)\,X^{1/2}\log^2X\text{\quad with}$$
\begin{align*}c_K(V_4)&=\dfrac{1}{3\cdot4^{r_2+2}}\zeta_K(1)^3\prod_{\p\nmid2}\left(1+\dfrac{3}{\N\p}\right)\left(1-\dfrac1{\N\p}\right)^3\\
&\phantom{=}\prod_{\p\mid2}\left(1+\dfrac{4}{\N\p}+\dfrac{1}{\N\p^2}-\dfrac{1}{\N\p^{e(\p)+1}}-\dfrac{1}{\N\p^{e(\p)+2}}\right)\left(1-\dfrac1{\N\p}\right)^3\;,\end{align*}
where here $e(\p)=e(\p/2)$ is the absolute ramification index of a prime
ideal $\p$ over 2. In particular, 
$$c_{\Q}(V_4)=\dfrac{23}{960}\prod_p\left(1+\dfrac3p\right)\left(1-\dfrac1p\right)^3\;.$$
This last result is due in principle to A.~Baily \cite{Bai}, but once
again there are computational mistakes in his paper.

Using some nontrivial series manipulation \cite{Coh3}, one can for
example compute exactly
$$N_{\Q,4}(V_4,10^{35})=6894524058812256194\;.$$

\subsection{$G=D_4$}

$$N_{K,4}(D_4,X)\sim c_K(D_4)\,X\text{\quad with}$$
$$c_K(D_4)=\sum_{[k:K]=2}\dfrac{1}{2^{r_2(k)}\N(\gd(k/K))^2}
\dfrac{\zeta_k(1)}{\zeta_k(2)}\;.$$
In particular,
$$c_{\Q}(D_4)=\dfrac6{\pi^2}\sum_D\dfrac{2^{-r_2(D)}}{D^2}\dfrac{L_D(1)}{L_D(2)}\;,$$
where the sum runs over all (positive or negative) discriminants of
quadratic fields, $r_2(D)=r_2(\Q(\sqrt D))$, and $L_D(s)$ is the
Dirichlet $L$-series attached to the character $\leg{D}n$. Note that we
do not know how to compute this sum in any other way than the naive
method, hence we know only about 7 or 8 decimals.

Using genus theory in a suitable way, combined with series manipulation,
one can for example compute exactly
$$N_{\Q,4}(D_4,10^{17})=10465196820067560\;.$$
All the above results are due to the authors (see \cite{CoDiOl4}).
Since $D_4$ quartic extensions of $\Q$ form a positive proportion of all 
quartic number fields, it is remarkable that one is able to compute
$N_{\Q,4}(D_4,X)$ for $X=10^{17}$: the published tables only go up to $10^6$. It
should be emphasized at this point that all our methods for counting
exactly are algorithmic, and that if the number of fields is reasonable,
we can just as easily construct \emph{tables} of extensions, see for
example \cite{CoDiOl2}.

\subsection{$G=A_4$}

$$N_{K,4}(A_4,X)\sim\begin{cases}c_K(A_4)\,X^{1/2}\log X&\text{\quad if $\zeta_3\notin K$}\\
c_K(A_4)\,X^{1/2}\log^2 X&\text{\quad if $\zeta_3\in K$\;.}\end{cases}$$
Here $c_K(A_4)$ is a complicated explicit constant. To give an example
of the complexity, we have in particular
$$c_{\Q}(A_4)=\lim_{N\to\infty}\dfrac{1}{3\log2\zeta(3)}\sum_{\substack {K_3\\ N<f(K_3)\le 2N}}\dfrac{h(K_3)R(K_3)c_2(K_3)c_r(K_3)}{f(K_3)^2}P(K_3)$$
with
$$P(K_3)=\prod_{p\text{ split in }K_3}\dfrac{(1+3/p)(1-1/p)^2}{1+1/p+1/p^2}\;,$$
where $K_3$ ranges over all cyclic cubic extensions of $\Q$ up to isomorphism
(which can easily be described explicitly), $f(K_3)$, 
$h(K_3)$, $R(K_3)$ denote the conductor, class number and regulator of $K_3$,
$$c_r(K_3)=\prod_{p\mid f(K_3)}\dfrac{1}{1+1/p+1/p^2}$$
and $c_2(K_3)=11/2$ if $2$ is inert in $K_3$, while $c_2(K_3)=23/5$ if $2$ is
totally split in $K_3$.

Using this method, we can also compute exactly (in 3 weeks of CPU time)
$$N_{\Q,4}(A_4,10^{13})=5278424\;.$$
The reason that it takes so long to compute is the computation of the
class number and regulator for all the necessary cyclic cubic fields $K_3$.

These results are due to the authors \cite{CoDiOl3}.

\subsection{$G=S_4$}

$$N_{K,4}(S_4,X)\sim c_K(S_4)\,X\;,$$
where $c_K(S_4)$ is an even more complicated explicit constant similar
to the one above (except that the sum will be over all noncyclic cubic
extensions of $K$ instead of cyclic cubic extensions, and there will be
no limit). We do not even give the expression for $c_{\Q}(S_4)$, although it is
quite similar to $c_{\Q}(A_4)$.

Using an extension of Shintani's method requiring a huge amount of work
and computations, it seems that A.~Yukie will soon be able to prove the
same result for $S_4$-extensions, but with a much nicer
expression for the constant $c_{K}(S_4)$ (exactly as Datskowsky and
Wright's result gave a nice expression for $c_K(S_3)$), see \cite{Yuk}.
The expression that he finds (which is not quite proved yet) is
$$c_{K,4}(S_4)=2\left(\dfrac{37}{96}\right)^{r_1}\left(\dfrac1{24}\right)^{r_2}\prod_{\p}\left(1+\dfrac1{\N\p^2}-\dfrac1{\N\p^3}-\dfrac1{\N\p^4}\right)\;,$$
so that in particular
$$c_{\Q}(S_4)=\dfrac{37}{48}\prod_p\left(1+\dfrac1{p^2}-\dfrac1{p^3}-\dfrac1{p^4}\right)\;.$$

Using our method, we can compute exactly 
$$N_{\Q,4}(S_4,10^7)=6541232\;.$$

\section{More General Results}

Our methods enable us to attack essentially all solvable groups. For
example, Wright's result gives for a prime number $\ell$
$$N_{K,\ell}(C_\ell,X)\sim c_K(C_\ell)\,X^{1/(\ell-1)}\,(\log
X)^{(\ell-1)/[K_z:K]-1}$$
for some constant $c_K(C_\ell)$, but we have computed this constant in a
completely explicit manner, see \cite{CoDiOl5}, \cite{CoDiOl6}.

We have also shown that (for $\ell$ an odd prime)
$$N_{K,\ell}(D_\ell,X)\sim c_K(D_\ell)\,X^{2/(\ell-1)}\;,$$
where the exponent is in agreement with Malle's conjecture.

\begin{thebibliography}{14}
\bibitem{Bai} A.~Baily, {\it On the density of discriminants of quartic
fields\/}, J. reine angew. Math.~{\bf 315} (1980), 190--210.
\bibitem{Bel} K.~Belabas, A fast algorithm to compute cubic fields,
Math.~Comp.~{\bf 66} (1997), 1213--1237.
\bibitem{CoDiOl1} H.~Cohen, F.~Diaz y Diaz and M.~Olivier,
Density of number field discriminants, in preparation.
\bibitem{CoDiOl2} H.~Cohen, F.~Diaz y Diaz and M.~Olivier,
Construction of tables of quartic fields using Kummer theory,
Proceedings ANTS IV, Leiden (2000), Lecture Notes in Computer Science 
{\bf 1838}, Springer-Verlag, 257--268.
\bibitem{CoDiOl3} H.~Cohen, F.~Diaz y Diaz and M.~Olivier,
On the density of discriminants of quartic number fields, in preparation.
\bibitem{CoDiOl4} H.~Cohen, F.~Diaz y Diaz and M.~Olivier,
Enumeratic quartic dihedral extensions, submitted.
\bibitem{CoDiOl5} H.~Cohen, F.~Diaz y Diaz and M.~Olivier,
Densit\'e des discriminants des extensions cycliques de degr\'e premier, 
C.~R.~Acad.~Sci.~Paris {\bf 330} (2000), 61--66.
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