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\newcommand{\cft}{Class Field Theory}
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\begin{document}
%topmatter
\title [An Overview of Class Field Theory]{An Overview of Class Field Theory}
\author {Thomas R. Shemanske}
\address{Department of Mathematics, Dartmouth College, Hanover, New Hampshire 03755}
\email {Thomas.Shemanske@dartmouth.edu}
\date{\today}
%endtopmatter
\maketitle
\section[intro]{{\bf Introduction}} In these notes, we try to give a reasonably
simple exposition on the question of what is \cft. We strive more for an intuitive
discussion rather than complete accuracy on all points. A great deal of what
follows has been lifted without proper reference from the two very informative
papers by Garbanati and Wyman
\cite{Garbanati},\cite{Wyman}. The questions which we shall pose and try to
answer in the next section are:
\begin{enumerate}
\item What is \cft?
\item What are the goals of \cft?
\item What are the main results of \cft\ over $\Q$?
\end{enumerate}
\section[Origins]{\bf{The Origins of \cft}} In examining the work of Abel,
Kronecker (1821 -- 1891) observed that certain abelian extensions of imaginary
quadratic number fields are generated by the adjunction of special values of
automorphic functions arising from elliptic curves. For example, if $K$ is an
imaginary quadratic number field and ${\mathfrak A} = {\mathbb Z}\omega_1 +
{\mathbb Z}\omega_2$ is an ideal of $K$ with Im$(\omega_1/\omega_2) > 0$, then
$K(j(\omega_1/\omega_2))$ is an abelian extension of $K$, where $j$ is the
modular function.
Kronecker wondered whether all abelian extensions of $K$ could be obtained in
this manner (Kronecker's Jugendtraum). This leads to the question of ``finding"
all abelian extensions of number fields. Kronecker conjectured and Weber (1842
-- 1913) proved:
\begin{nnthm} [Kronecker--Weber (1886--1887)] \label{thm:KronWeber}
Every abelian extension of
$\Q$ is contained in a cyclotomic extension of $\Q$.
\end{nnthm}
To Kronecker and Weber, \cft\ was the task of finding all abelian extensions,
and of finding a generalization of Dirichlet's theorem on primes in arithmetic
progressions which is valid in number fields.
Hilbert saw that \cft\ is much more --- that it is the theory of abelian
extensions. In his famous address to the ICM in Paris in 1900, Hilbert posed
numerous questions two of which are the focus of the endeavors in \cft.
\pagebreak[3]
\begin{Ventry}{Hilbert's 12th:}
\item[Hilbert's 9th:] To develop the most general reciprocity law in an
arbitrary number field, generalizing Gauss' law of quadratic reciprocity.
\begin{Ventry}{---}
\item[---] For abelian extensions, this is the Artin reciprocity law
\item[---] For non-abelian extensions, the question is still open and one
cannot expect an answer similar to the one in the abelian case. In particular,
``congruence conditions will not suffice".
\end{Ventry}
\item[Hilbert's 12th:] Generalize Kronecker's Jugendtraum.
\end{Ventry}
\subsection{\bf What is a reciprocity law?} Let $f \in \Z[X]$ be monic and
irreducible, and let $K_f$ be the splitting field of $f$ over $\Q$. Then
$K_f/\Q$ is a finite Galois extension. Let $p \in \Z$ be a prime and
$\F = \Z/p\Z$. Reducing $f$ mod $p$ gives a polynomial $f_p \in \F[X]$.
If $f_p$ factors into distinct linear factors over $\F$ then we say that $f$
{\it splits completely modulo $p$}.
Define $Spl(f) = \{\,p\in \Z \mid f \text{ splits completely modulo }
p\,\}$. With finite exceptions, $Spl(f) = \{\,p\in \Z \mid p \text{ splits
completely in } K_f\,\}$ via the Dedekind-Kummer theorem (see \S4).
By a reciprocity law, we intend a means by which to describe the factorization of
$f_p$ as a function of $p$, or somewhat less demanding, a ``rule" which
determines which primes belong to $Spl(f)$.
First, why is this of interest? In response, we have the Inclusion theorem:
\begin{nnthm} [Inclusion Theorem] Let $f$, $g$ be irreducible polynomials in
$\Z[X]$ with splitting fields $K_f$, $K_g$ respectively. Then
$K_f \supset K_g$ if and only if $Spl(f) \subset^* Spl(g)$.
\end{nnthm}
\noindent Here $\subset^*$ means with finitely many exceptions.
Thus $K_f = K_g$ if and only if $Spl(f) =^* Spl(g)$, that is the set $Spl(f)$
captures the Galois extension.
\begin{proof} $(\Rightarrow)$ This direction is straightforward. If $p \in
Spl(f)$ then $e(K_f/\Q) = 1$ and $f(K_f/\Q)=1$. Since $K_f \supset K_g$ and $e$
and $f$ are multiplicative in towers, we have $e(K_g/\Q) = 1$ and
$f(K_g/\Q)=1$, and hence $p \in Spl(g)$.
$(\Leftarrow)$ This direction follows from the Tchebotarev density
theorem.
\end{proof}
We give an example.
\ms
\noindent{\bf Example.} Let $p$ be a prime $p\equiv 1\pmod4$, $f(X) = X^2 -
p$, and $g(X) = X^p -1$. Then $K_f = \Q(\sqrt p)$ and $K_g = \Q(\zeta_p)$
where $\zeta_p$ is a primitive $p$-th root of unity. Since $p\equiv 1\pmod4$,
we have $K_f \subset K_g$. We must show that $Spl(f) \supset^* Spl(g)$. It is
well-known that a prime $q \in Spl(g)$ (i.e., $q$ splits completely in
$\Q(\zeta_p)$) iff $q \equiv 1\pmod p$ and $q \in Spl(f)$ (i.e., $q$ splits
completely in $\Q(\sqrt p)$) iff $\legendre q p = 1$ (via the
Dedekind-Kummer theorem). Clearly any prime $q$ satisfying $q \equiv 1\pmod p$
satisfies $\legendre q p = 1$, hence $Spl(f) \supset Spl(g)$.
\bs
Another theorem of great importance is the
\begin{nnthm} [Abelian Polynomial Theorem] The set $Spl(f)$ can be described
by congruences with respect to a modulus depending only on $f$ ($K_f$) if and
only if $K_f$ is an abelian extension of $\Q$.
\end{nnthm}
\noindent The Artin reciprocity law is a precise version of ($\Leftarrow$), and
($\Rightarrow$) says that ``congruence conditions" will not suffice to
characterize a reciprocity law for non-abelian extensions.
\ms
\noindent {\bf Examples}:
\begin{enumerate}
\item Let $p \in \Z$ be an odd prime, and
consider the quadratic polynomial $f(X) = X^2 - q$ where $q$ is an odd prime.
Then modulo $p$, three things can happen:
%\renewcommand{\theenumi}{\alph{enumi}}
%\renewcommand{\labelenumi}{(\theenumi)}
\begin{enumerate}
\item $f_p(X) = l(X)^2$, linear $l(X)$
\item $f_p(X) = l_1(X)l_2(X)$ distinct linear factors ($f$ splits
completely modulo $p$)
\item $f_p$ is irreducible.
\end{enumerate}
\noindent (a) occurs iff $x^2 \equiv q\pmod p$ has one solution iff $p = q$.
\noindent (b) occurs iff $x^2 \equiv q\pmod p$ has two solutions iff
$\legendre{q}{p} = +1$.
\noindent (c) occurs iff $x^2 \equiv q\pmod p$ has no solutions iff
$\legendre{q}{p} = -1$.
To determine for which $p$ the congruence $x^2 \equiv q\pmod p$
is solvable, is apriori an infinite problem. On the other hand, it is one from
which the traditional form of quadratic reciprocity rescues us.
Suppose $q = 17$ in the above example. Then $\legendre{17}{p} = +1$
iff
$\legendre{p}{17} = +1$ iff $p \equiv 1,2,4,8,9,13,15,16\pmod {17}$.
Thus $p \in Spl(x^2 - 17)$ iff (with finite exceptions)
$p \equiv 1,2,4,8,9,13,15,16\pmod {17}$.
\item Next consider the cyclotomic polynomials, $\Phi_n$. Let
$\zeta$ be a primitive $n^{th}$ root of unity and $\Phi_n$ the irreducible
polynomial of
$\zeta$ over $\Q$. We know that the degree of $\Phi_n$ is $\phi(n)$ and that
$x^n - 1 = \prod_{d \mid n} \Phi_d$.
To describe $Spl(\Phi_n)$ we need to answer which primes split completely in
$K_{\Phi_n} = \Q(\zeta)$. If $p \nmid n$ then for any prime of
$K_{\Phi_n}$ lying above $p$, we know $e = 1 $ and $fg = \phi(n)$. Moreover,
$f$ is determined by the relation that it is the smallest positive integer such
that $p^f \equiv 1\pmod n$. Thus $p$ splits completely in $K_{\Phi_n}$ iff
$f=1$, hence $p \in Spl(\Phi_n)$ iff (wfe) $p \equiv 1\pmod n$, characterizing
$Spl(\Phi_n)$ by congruence conditions.
\end{enumerate}
Let us loosely define the {\it arithmetic of a number field $K$} to be the study
of the ideals of $K$ and the quotient rings determined by the ideals of $K$ as
well as the study of the ideal class group and groups isomorphic to subgroups
or quotient groups of the ideal class group.
\ms
\noindent {\bf Goals of \cft}:
\begin{enumerate}
\item Describe all finite abelian extensions of $K$ in terms of the
arithmetic of $K$.
\item Canonically realize $Gal(L/K)$ in terms of the arithmetic of $K$
whenever $Gal(L/K)$ is abelian.
\item Describe the decomposition of a prime ideal from $K$ to $L$ in terms
of the arithmetic of $K$ whenever $L/K$ is abelian (i.e., provide a reciprocity
law).
\end{enumerate}
\subsection{\bf Summary of \cft\ over $\Q$.}
\noindent Notation: $\Q_m = \Q(e^{2\pi i/m})$. We may assume that $m \not\equiv
2(4)$. For if $m\equiv 2\pmod4$ with $m = 2m_0$, then we easily observe that
$-e^{2\pi i/{m_0}}$ is a primitive $m$th root of unity, and hence that
$\Q_m = \Q_{m_0}$. Over
$\Q$, the Kronecker-Weber Theorem motiviates the following definition:
\begin{nndefinition}
Let $L/\Q$ be a finite abelian extension. A positive integer $m$ is called a
{\it defining modulus} or an {\it admissible modulus of $L$} if $L \subset
\Q_m$. Such an $m$ exists by the Kronecker-Weber theorem. The {\it conductor} of
$L$, ${\mathfrak f}_L$, is the smallest admissible modulus of $L$.
\end{nndefinition}
\noindent {\bf Examples:}
\begin{enumerate}
\item $L = \Q_m$. Then ${\mathfrak f}_L = m$, since $\Q_m \subset \Q_n$ implies
that $\Q_m = \Q_m \cap \Q_n = \Q_{(m,n)}$ implies that $m \mid n$.
\item Let $L$ be the maximal real subfield of $\Q_m$. Then $L = \Q(\zeta +
\zeta^{-1})$ where $\zeta = e^{2 \pi i /m}$ (it is the fixed
field of complex conjugation). Note that if $m = 3,4$, then $L = \Q$.
For $m \ge 5$, ${\mathfrak f}_L = m$. For $m = 3,4$, ${\mathfrak f}_L = 1$.
\item $L = \Q(\sqrt d)$, $d$ square-free integer, $|d| > 1$.
Then $${\mathfrak f}_L = |\text{disc}(L)| =
\begin{cases}
|d|& \text{if $d\equiv 1\pmod 4$}\\
|4d|& \text{if $d\equiv 2,3\pmod 4$}.
\end{cases}$$
\end{enumerate}
To gain some feeling for why the last example holds, recall that if $L = \Q_p$
($p$ an odd prime), then disc$(L) = (-1)^{\frac{p-1}2} p^{p-2}$ is the square
of an integer in
$\mathcal O_L$, thus
$$\Q\Big(\sqrt{(-1)^{\frac{p-1}2} p}\Big)
\subset
\Q_p.$$
It follows that for a prime $p$
$$\Q(\sqrt p) \subset
\begin{cases}
\Q_p&\text{if $p \equiv 1\pmod 4$}\\
\Q_{4p}& \text{if $p \equiv 3 \pmod 4$}\\
\Q_8&\text{if $p = 2$}.
\end{cases}$$
\noindent Moreover, if $d = \pm 2^\nu p_1 p_2 \cdots p_r$ is squarefree, then
$\Q(\sqrt d) \subset
\Q(\sqrt{2^\nu},\sqrt{p_1},\sqrt{p_2},\ldots,\sqrt{p_r}) \subset
\Q(\zeta_{4\cdot2^\nu})\Q(\zeta_{p_1},\zeta_{p_2},\ldots,\zeta_{p_r},\zeta_{4})
= \Q(\zeta_{4d})$.
\begin{nnthm} Let $L/\Q$ be a finite abelian extension, and $m$ an
admissible modulus of $L$. Then ${\mathfrak f}_L \mid m$.
\end{nnthm}
\begin{proof} $L \subset \Q_m \cap \Q_{{\mathfrak f}_L} = \Q_{({\mathfrak
f}_L,m)}$ which implies ${\mathfrak f}_L \mid m$.
\end{proof}
\bs
Let $L$ be an abelian extension of $\Q$, and let $m$ be an admissible modulus of
$L$. Then $L \subset \Q_m$. Let $a \in \Z$ with $(a,m) =
1$, and denote by $\llegendre{L}{a}$ the Artin symbol, the
automorphism of $L$ obtained by restricting to the field $L$ the automorphism of
$\Q_m$ determined by ($\zeta
\mapsto \zeta^a$). Then the Artin map is the homomorphism
$$\llegendre{L}{*}:(\Z/m\Z)^\times \to Gal(L/\Q).$$
\noindent The Artin map is onto since every automorphism of $L$ extends to one of
$\Q_m$ which has the above form. Denote the kernel of $\llegendre{L}{*}$
by $I_{L,m}$. Identifying $(\Z/m\Z)^\times$ with $Gal(\Q_m/\Q)$, we see that
$I_{L,m}$ is identified with $Gal(\Q_m/L)$, so under the Galois correspondence (see
diagram below),
$L$ is the fixed field of the subgroup $I_{L,m}$ of $(\Z/m\Z)^\times$.
$$
\begin{diagram}
\node{\Q_m}\arrow{e}\arrow{s,-}
\node{\{1\}}\arrow{w}\arrow{s,-}\\
\node{L}\arrow{e}\arrow{s,-}
\node{I_{L,m}}\arrow{w}\arrow{s,-}\\
\node{\Q}\arrow{e}
\node{(\Z/m\Z)^\times}\arrow{w}
\end{diagram}
$$
\noindent This information is summarized in the
\begin{nnthm} [Artin Reciprocity] Let $L/\Q$ be a finite abelian
extension with defining modulus $m$. Then the following sequence is exact:
$$1 \rightarrow I_{L,m} \hookrightarrow (\Z/m\Z)^\times \rightarrow Gal(L/\Q)
\rightarrow 1.$$
\end{nnthm}
\noindent Thus, the Artin map induces an isomorphism between $Gal(L/\Q)$ and
$(\Z/m\Z)^\times/I_{L,m}$ thus canonically realizing $Gal(L/\Q)$ in terms of the
arithmetic of $\Q$. In particular, this
says that every abelian extension is given in terms of the arithmetic of $\Q$,
and so realizes one of the primary goals of \cft.
\bs
As a special case, if $L$ is a quadratic extension of $\Q$ contained in
$\Q_m$, then $Gal(L/\Q)$ is isomorphic to $\{\pm 1\}$, and identifying the
isomorphic groups, the Artin map essentially can be defined by $\legendre L,a =
\legendre a m$. To make clearer what we mean, we examine some typical cases in
the examples below.
\ms
\noindent {\bf Examples}:
\begin{enumerate}
\item Let $p$ be an prime $p \equiv 1\pmod4$. Then $\Q(\sqrt p) \subset \Q_p$.
If $L = \Q(\sqrt p)$, then since $[L:\Q] = 2$, $I_{L,p}$ is a subgroup of
index two in $(\Z/p\Z)^\times$. Since $(\Z/p\Z)^\times$ is cyclic, there is a
unique such subgroup, namely the squares (or quadratic residues) mod $p$.
For any $a$ prime to $p$, $\legendre L a = \pm1$, and
$I_{L,p}$ is the kernel of $\legendre L *$. With $I_{L,p}$ identified as the
group of squares mod $p$ and $Gal(L/\Q)$ identified with $\{\pm1\}$, it is clear
that $\legendre L a = \legendre a p$.
\item A considerably more complicated example is $L = \Q(\sqrt 7)$.
Clearly the conductor of $L$ is 28, so take $m = 28$ in the setup above.
Here we will see that $\legendre L a$ is almost $\legendre a {28}$.
The only real difficulty in interpreting the quadratic residue symbol
$\legendre a 2$, so we digress for a moment.
Recall that $\legendre a 2$ is defined by
$$\legendre a 2 =
\begin{cases}
1&\text{if $a \equiv 1\pmod 8$}\\
-1&\text{if $a \equiv 5\pmod 8$}\\
0&\text{otherwise}.
\end{cases}$$
In particular, if $a$ is squarefree and $p$ is any prime, then $\legendre a p$ is
1, $-1$ or 0 depending upon whether $p$ splits, is inert, or ramifies in $\Q(\sqrt
a)$. The difficulty we encounter is that if $a \equiv 3\pmod 4$, then
$\legendre a 4 = \legendre a 2^2 \ne \legendre {a^2} 2$, the first expression
equalling zero, while the last equals 1.
To continue, let $\zeta_m = e^{2\pi i/m}$, and consider the tower of fields below.
$$
\begin{diagram}
\node[2]{\Q(\zeta_{28})}
\arrow{sw,-} \arrow{s,-} \arrow{se,-}\\
\node{\Q(\zeta_4)}\arrow{s,=}
\node{\Q(\sqrt {-1},\sqrt {-7})}\arrow{sw,-}\arrow{s,-}\arrow{se,-}
\node{\Q(\zeta_7)}\arrow{s,-}\\
\node{\Q(\sqrt{-1})}\arrow{se,-}
\node{\Q(\sqrt 7)}\arrow{s,-}
\node{\Q(\sqrt{-7})}\arrow{sw,-}\\
\node[2]{\Q}
\end{diagram}
$$
By the Galois correspondence, there is a corresponding lattice of groups.
$$
\begin{diagram}
\node[2]{H(\zeta_{28})}
\arrow{sw,-} \arrow{s,-} \arrow{se,-}\\
\node{H(\zeta_4)}\arrow{s,=}
\node{H(\sqrt {-1},\sqrt {-7})}\arrow{sw,-}\arrow{s,-}\arrow{se,-}
\node{H(\zeta_7)}\arrow{s,-}\\
\node{H(\sqrt{-1})}\arrow{se,-}
\node{H(\sqrt 7)}\arrow{s,-}
\node{H(\sqrt{-7})}\arrow{sw,-}\\
\node[2]{H(1)}
\end{diagram}
$$
Here we set the notation by putting $H(1) = (\Z/28\Z)^\times$ (and
$H(\zeta_{28}) = \{1\}$). Then for example,
$H(\sqrt{-7})$ is the subgroup of $(\Z/28\Z)^\times$
corresponding to $Gal(\Q(\zeta_{28})/\Q(\sqrt{-7}))$.
Our purpose is to calculate $I_{L,{28}}$ where $L = \Q(\sqrt 7)$, and to compare
the values of $\legendre L a$ with those of $\legendre a {28}$. The subgroup
$I_{L,{28}}$ will simply be $H(\sqrt 7)$.
If we consider the tower $\Q \subset \Q(\sqrt{-7}) \subset \Q(\zeta_7)$, then
as a subgroup of $(\Z/7\Z)^\times$, $\Q(\sqrt{-7})$ corresponds to the subgroup
of quadratic residues mod 7 (as in example 1), that is to $\{1, 2, 4\}$. Modulo
28 (i.e. $a \equiv 1,2,4\pmod 7$ and $a\equiv 1,3\pmod4$), this yields
$H(\sqrt{-7}) = \{1, 9, 11, 15, 25, 23\}\subset (\Z/28\Z)^\times$.
The tower $\Q \subset \Q(\sqrt{-1}) = \Q(\zeta_4)$ is degenerate yielding the
trivial subgroup of $(\Z/4\Z)^\times$ corresponding to $\Q(\sqrt{-1})$, or
$\{a | a \equiv 1\pmod4\}$. Modulo 28 (i.e., $a \equiv 1\pmod4$ and
$a \not\equiv 0\pmod7$), this yields
$H(\sqrt{-1}) = \{1, 5, 9, 13, 17, 25\}$.
As $\Q(\sqrt{-1},\sqrt{-7})$ is the compositum of $\Q(\sqrt{-1})$ and
$\Q(\sqrt{-7})$, Galois theory tells us that
$H(\sqrt{-1},\sqrt{-7}) = H(\sqrt{-1}) \cap H(\sqrt{-7}) = \{ 1, 9, 25\}$.
For the record, we note that $\legendre a {28} = +1$ if and only if
$(a,28) = 1$ and $\legendre a 7 = \legendre a 4$, which is true if and only if
$a \equiv 1,2,4\pmod7$ and $a\equiv 1\pmod4$. Note that since
$\legendre a 4 = \legendre a 2^2$, $\legendre a 4$ is never equal to $-1$.
Thus $\{a | \legendre a {28} = 1\} = \{ 1, 9, 25\}$, and is not equal to
$H(\sqrt 7)=I_{L,{28}}$ which has order 6. It is now a trivial matter to deduce
that $H(\sqrt 7)= \{1, 3, 9, 19, 25, 27\}$.
\item To handle more general examples like $L = \Q(\sqrt{\pm35})$, we need only
consider one of the two tower of fields below and use the techniques of the
preceding examples.
If $L = \Q(\sqrt{-35})$, we consider the tower
$$
\begin{diagram}
\node[2]{\Q(\zeta_{35})}
\arrow{sw,-} \arrow{s,-} \arrow{se,-}\\
\node{\Q(\zeta_7)}\arrow{s,-}
\node{\Q(\sqrt {-7},\sqrt {5})}\arrow{sw,-}\arrow{s,-}\arrow{se,-}
\node{\Q(\zeta_5)}\arrow{s,-}\\
\node{\Q(\sqrt{-7})}\arrow{se,-}
\node{\Q(\sqrt{-35})}\arrow{s,-}
\node{\Q(\sqrt{5})}\arrow{sw,-}\\
\node[2]{\Q}
\end{diagram}
$$
\noindent whereas if $L = \Q(\sqrt{35})$, we consider the tower
$$
\begin{diagram}
\node[2]{\Q(\zeta_{140})}
\arrow{sw,-} \arrow{s,-} \arrow{se,-}\\
\node{\Q(\zeta_{28})}\arrow{s,-}
\node{\Q(\sqrt {7},\sqrt {5})}\arrow{sw,-}\arrow{s,-}\arrow{se,-}
\node{\Q(\zeta_5)}\arrow{s,-}\\
\node{\Q(\sqrt{7})}\arrow{se,-}
\node{\Q(\sqrt{35})}\arrow{s,-}
\node{\Q(\sqrt{5})}\arrow{sw,-}\\
\node[2]{\Q}
\end{diagram}
$$
\noindent and proceed as in the previous examples.
\end{enumerate}
\bs
To continue our investigation of class fields, we have the following theorem
which gives information about the conductor of an abelian extension.
\begin{nnthm} [Conductor--Ramification Theorem] If $L$ is a finite
abelian extension of $\Q$, then a prime $p$ of $\Q$ ramifies in $L$ if and only
if $p \mid {\mathfrak f}_L$.
\end{nnthm}
\begin{nncor} If $L \neq \Q$ is a finite abelian extension of $\Q$, then at
least one prime $p$ ramifies in $L$.
\end{nncor}
\begin{proof} Since $L \neq \Q$, $L \nsubseteq \Q_1 = \Q$, hence ${\mathfrak f}_L
> 1$, and so is divisible by at least one prime.
\end{proof}
\noindent For contrast, we have the result of Minkowski that a prime $p$ of $\Q$
ramifies in a number field $L$ if and only of $p \mid \text{disc}(L)$. This
says that for abelian extensions, there should be a connection between the
conductor and the discriminant (see the conductor-discriminant formula below).
\begin{nnthm}[Decomposition Theorem] Let $m$ be a defining modulus
of $L$. If $p \nmid m$ (in particular $p$ is unramified) then the order of
$pI_{L,m}$ in $(\Z/m\Z)^\times/I_{L,m}$ is $f$, the residue class degree.
\end{nnthm}
\noindent
Notice that this generalizes the theorem about the decomposition of primes in
cyclotomic fields. If we choose $L = \Q_m$, then $I_{L,m} = 1$, and we are
reduced to talking about the order of $p$ in $(\Z/m\Z)^\times$.
\noindent Let $m = {\mathfrak f}_L$ in the above theorem. Since $efg = [L:\Q]$,
\begin{align*}
p \in Spl(L/\Q) &\Leftrightarrow e=1,\ f=1\\
&\Leftrightarrow p \nmid {\mathfrak f}_L,\ p\in I_{L,{\mathfrak f}_L}
\end{align*}
the first condition because $e = 1$ and the second because $f = 1 $ via the
Decomposition theorem.
If $I_{L,{\mathfrak f}_L} = \{a_1, \ldots a_s\}$ with $a_i \in \Z$ and
$(a_i,{\mathfrak f}_L) = 1$, then $p \in Spl(L/\Q) \Leftrightarrow
p \equiv a_i\pmod {{\mathfrak f}_L}$ for some $i$. This acomplishes the goal of
describing $Spl(L/\Q)$ in terms of congruence conditions, and hence the
decomposition of primes in terms of the arithmetic of $\Q$.
\subsection{\bf Duality}
Let $X_m$ denote the character group of $(\Z/m\Z)^\times$. That is
$\chi \in X_m$ implies that $\chi:(\Z/m\Z)^\times \to {\mathbb C}^\times$ is a
homomorphism.
\begin{nndefinition} We say that $d$ is a {\it defining modulus} for $\chi$ if $a
\equiv 1\pmod d$ implies that $\chi(a) = 1$. The {\it conductor} of $\chi$,
denoted
${\mathfrak f}_\chi$, is the smallest defining modulus for $\chi$.
\end{nndefinition}
\noindent If $m$ is a defining modulus for a finite abelian extension $L$, let
$$X_{L,m} = \{\chi \in X_m | \chi(h) = 1 \hbox{ for all } h \in I_{L,m}\}$$
\noindent Recall that $I_{L,m}$ is the subgroup of
$ (\Z/m\Z)^\times \cong Gal(\Q_m/\Q) $ corresponding to the subfield $L
\subset \Q_m$ via the Galois correspondence. That is, $I_{L,m} \cong
Gal(\Q_m/L)$, and from duality we see that
$$X_{L,m}\cong Gal(\Q_m/L)^\perp \cong
\widehat{Gal(\Q_m/\Q)/Gal(\Q_m/L)} \cong \widehat{Gal(L/\Q)}.$$
\noindent Finally, we have the
\begin{nnthm}[Conductor--Discriminant Formula] Let $m$ be an
admissible modulus for a finite abelian extension $L$ of $\Q$. Then
$${\mathfrak f}_L = \text{lcm}\{{\mathfrak f}_\chi | \chi \in X_{L,m}\}$$
and
$$|\text{disc}(L)| = \prod_{\chi \in X_{L,m}}{\mathfrak f}_\chi.$$
\end{nnthm}
\noindent In particular, ${\mathfrak f}_L \mid \text{disc}(L)$, and so we always
have the tower of fields:
$$\Q \subset L \subset \Q_{{\mathfrak f}_L} \subset \Q_{|\text{disc}(L)|}.$$
\section{{\bf Global \cft}}
In order to generalize \cft\ to ground fields other than $\Q$, several issues
need to be addressed:
\begin{enumerate}
\item The Kronecker-Weber theorem is valid only for ground field $\Q$, so we need
a new notion of admissible modulus (a very deep theorem).
\item We need to handle all the infinite primes.
\item We need a generalized notion of congruence.
\item With what shall we replace $(\Z/m\Z)^\times$ and $\Q_m$?
\end{enumerate}
\bs
Let $\M$ be a modulus and let $\M_0$ denote its finite part. For a number
field $K$, let $I_K^\M$ denote the group of fractional ideals of $K$ relatively
prime to $\M_0$. Let
$$K_{\M,1} = \{\,\alpha \in K^\times \mid
\alpha \equiv 1\ (\bmod\,^* \M)\,\}.$$
\noindent Recall that $\alpha \equiv 1\ (\bmod\,^* \M)$ means that
\begin{align*}
\text{ord}_\P(\alpha -1) &\ge \text{ord}_\P(\M_0) \quad \text{ for all }\ \P \mid
\M_0\quad \text{ and}\\
\alpha &> 0 \text{ at each real prime dividing $\M$}
\end{align*}
Let $R_\M = \{ \alpha\O_K \mid \alpha \in K_{\M,1}\,\}$. $R_\M$ is called the
{\it ray mod $\M$}. Let $C_\M = I_K^\M/R_\M$, the {\it ray class group}.
Special cases are familiar. If $\M = 1$, then the ray class group $C_1$ is just
the ideal class group of the field $K$. If $K = \Q$ and $\M = mp_\infty$, where
$m$ is a positive integer, then $C_\M \cong (\Z/m\Z)^\times$.
Let $L/K$ be a Galois extension and let $\M$ be a $K$-modulus. Define
$I_L^\M = I_L^{\M_0\O_L}$ and
\begin{align*}
L_{\M,1} = \{\,\alpha \in L^\times \mid &
\alpha \equiv 1\ (\bmod\,^* \M_0\O_L)\\
&\text{and where $\alpha > 0$ at each real prime of $L$}\\
&\text{dividing a real prime occuring in $\M$}\,\}.
\end{align*}
\noindent Finally let $R_{L,\M} = \{ \alpha\O_L \mid \alpha \in L_{\M,1}\,\}$, and
$C_{L,\M} = I_L^\M/R_{L,\M}$.
Recall that the norm of an ideal relative to a Galois extension $L/K$ is
defined as follows: If $\P$ is a prime of $K$ and $\PP$ is a prime of $L$
lying above $\P$ with inertial degree $f$, then we define the norm of $\PP$ to
be
$\N(\PP) = \P^f$. We extend the definition of the norm to the group of
fractional ideals by multiplicativity. Note that when $K = \Q$,
$\N(\PP) = \P^f = p^f\Z$ for the prime $p\Z = \PP\cap \Z$, while the absolute
norm of $\PP$ is equal the cardinality of the residue class field $\O_L/\PP$
which is $p^f$, so this definition provides a natural generalization of the
absolute norm.
One can show that $\N(R_{L,\M}) \subset R_\M$, and so the definition of the
norm can be extended to $C_{L,\M}$ by defining
$\N({\mathfrak A}R_{L,\M}) = \N({\mathfrak A}) R_\M$. Put
$$I_{L/K,\M} = \N(C_{L,\M}) < C_\M.$$
\noindent For example, if $K=\Q$ and $L \subset \Q_m$ (i.e. $m$ is an admissible
modulus of $L$), then we have the diagram:
$$
\begin{diagram}
\node{\Q_m}\arrow{e}\arrow{s,-}
\node{\{1\}}\arrow{w}\arrow{s,-}\\
\node{L}\arrow{e}\arrow{s,-}
\node{I_{L,m}}\arrow{w}\arrow{s,-}\\
\node{\Q}\arrow{e}
\node{(\Z/m\Z)^\times}\arrow{w}
\end{diagram}
$$
If we let $\M = mp_\infty$, then it can be shown that
$I_{L/K,\M} \cong I_{L,m}$. Notice also that $C_\M \cong (\Z/m\Z)^\times$ and
that $[C_\M : I_{L/\Q,\M}] = [L:\Q]$ by the Galois correspondence. Generalizing
this fact, we have the deep theorem:
\begin{nnthm} Let $\M$ be a $K$-modulus and $L/K$ an abelian extension
of number fields. Then there exists a unique $K$-modulus $\f$ such that
$[C_\M : I_{L/K,\M}] = [L:K]$ iff $\f \mid \M$.
\end{nnthm}
\noindent The unique modulus $\f$ is called the {\it conductor} of $L/K$ and any
$K$-modulus divisible by $\f$ is called an {\it admissible} modulus of $L/K$.
This is
not a very intuitive theorem because we don't have something natural like the
Kronecker-Weber theorem with which to define the conductor. The theorem is
proved in two steps. The first inequality to be established was that $[C_\M :
I_{L/K,\M}] \le [L:K]$ for all moduli $\M$. This was done by Weber (1897-8).
It is now known as the ``second inequality". In 1920, Tagaki showed that $[C_\M
: I_{L/K,\M}] \ge [L:K]$ for some modulus $\M$, now known as the ``first
inequality".
\noindent One can also show that
$${\mathfrak f}_{L/\Q} =
\begin{cases}
(f_L)& \text{if $L \subset \R$}\\ (f_L)p_\infty&\text{ if $L \not\subset \R$}
\end{cases}$$
Now we need an analog of the cyclotomic fields and the Kronecker-Weber
theorem.
\begin{nnthm}[Existence] Given a $K$-modulus $\M$ and a subgroup
$I_\M$ of the ray class group $C_\M$, there exists a unique abelian
extension $L/K$ such that
\begin{enumerate}
\item $\M$ is an admissible modulus of $L/K$
\item $I_{L/K,\M} = \N(C_{L,\M}) = I_\M$\qquad or
\item The kernel of the Artin map $I_K^\M \to Gal(L/K)$ is $H_\M$ where
$I_\M = H_\M/R_\M.$
\end{enumerate}
\end{nnthm}
\noindent $L$ is called the {\it class field} of the subgroup $I_\M$. When $I_\M=
R_\M$, the trivial subgroup, the class field $L$ is called the {\it ray class
field} and is denoted $K(R_\M)$. If $K=\Q$ and $\M = mp_\infty$, then $K(R_\M) =
\Q_m$, that is the cyclotomic fields are the ray class fields for the moduli
$\M = mp_\infty$. The ray class field for the modulus $\M = m$ is the maximal
real subfield of $\Q_m$.
We have two more important theorems:
\begin{nnthm}Given an abelian extension $L/K$, there exists a
$K$-modulus $\M$ so that $L \subset K(R_\M)$.
\end{nnthm}
As a consequence, we recover the Kronecker-Weber
theorem.
\begin{nnthm}Given an abelian extension $L/K$, the conductor $\f$ is the
``smallest" $K$-modulus $\M$ such that $L \subset K(R_\M)$. Moreover, $\M$ is
an admissible modulus of $L/K$ iff $L \subset K(R_\M)$.
\end{nnthm}
Thus it follows that every abelian extension of $K$ is a subfield of a ray class
field for $K$. We have classified the abelian extensions, but we have not
constructed them. More later.
We have the following generalization of Dirichlet's theorem on primes in
arithmetic progressions.
\begin{nnthm} Let $I_\M$ be a subgroup of the ray class group $C_\M$.
Then $I_\M = H_\M/R_\M$ where $R_\M \subset H_\M \subset I_K^\M$. Then there
are an infinite number of primes in each coset of $I_K^\M/H_\M$. In fact, the
primes in the coset have density $1/[I_K^\M\,:\,H_\M]$.
\end{nnthm}
\noindent If $K = \Q$ and $\M = mp_\infty$, then $C_\M \cong (\Z/m\Z)^\times$.
If we choose $H_\M = R_\M$, then $I_K^\M/H_\M = C_\M \cong (\Z/m\Z)^\times$,
and we have recovered the Dirichlet theorem over $\Q$.
It can be shown that if $\M$ is an admissible modulus for an abelian extension of
number fields $L/K$, then the Artin map, $\llegendre{L/K} *$, is trivial on
the ray mod $\M$, $R_\M$, and so the definition of the Artin map can be
extended to the ray class group, $C_\M$.
In 1927, Artin proved
\begin{nnthm}[Artin Reciprocity] Let $L/K$ be an abelian
extension of number fields, and let $\M$ be an admissible modulus of $L/K$.
Then the following sequence is exact:
$$1 \rightarrow I_{L/K,\M}= \N(C_{L/K,\M})
\hookrightarrow C_\M \rightarrow Gal(L/K) \rightarrow 1.$$
\end{nnthm}
\begin{nncor}Let $\M$ be an admissible modulus of the abelian
extension $L/K$. Then $L \subset K(R_\M)$, $Gal(K(R_\M)/K) \cong C_\M$ and
$Gal(K(R_\M)/L) \cong I_{L/K,\M}$.
\end{nncor}
\begin{nndefinition} The Hilbert class field of a number field $K$ is the
ray class field $K(R_1)$, and will be denoted $\K$.
\end{nndefinition}
From above we see that $Gal(\K/K) \cong C_1$, the ideal class group of $K$.
Thus much work is done in trying to understand subfields of $\K$ to help
understand the structure of the ideal class group.
\begin{nnthm}
A $K$-prime $\P$ ramifies in $L$ iff $\P\mid\f$.
\end{nnthm}
\begin{nnthm}
The Hilbert class field $\K$ is the maximal
abelian unramified extension of $K$.
\end{nnthm}
\begin{proof} Since $\K = K(R_1)$, ${\mathfrak f}_{\K/K} = (1)$. If $L$ is an
unramified extension of $K$, then $\f=(1)$ by the above theorem. Since $(1)$
is an admissible modulus for $L/K$, we have $L \subset K(R_1) = \K$.
\end{proof}
\begin{nnthm}
Each fractional ideal of $K$ becomes principal in $\K$.
\end{nnthm}
\noindent This does not say that $\K$ has class number one. Instead it suggests
the ``class tower problem". Let $K_0 = K$ and $K_i = \tilde K_{i-1}$ for
$i\ge1$. Does there exist a $j$ such that $K_j = K_{j-1}$? This would imply
that the class number of $K_{j-1}$ equals 1. Golod and Shafarevich (1964)
showed that any imaginary quadratic field $\Q(\sqrt{-d})$ where $d$ is a
positive integer divisible by at least six primes has an infinite class field
tower.
\section{{\bf Equivalence of the reciprocity laws.}}
We consider the case of a prime $p$, $p \equiv 1 \ (\bmod\, 4)$, and $q$ an odd
prime, $q \ne p$.
Gauss' law says that $\legendre p q = \legendre q p$.
Wyman asks for a rule which describes the primes $q$ which split completely in
$\Q(\sqrt p)$.
Artin says that
$$1 \rightarrow H \hookrightarrow I_\Q^{pp_\infty} \rightarrow
Gal(\Q(\sqrt p)/\Q) \rightarrow 1$$
\noindent for an appropriately defined subgroup $H$ is exact where the map
$I_\Q^{pp_\infty} \rightarrow Gal(\Q(\sqrt p)/\Q)$ is the Artin map.
\noindent Consider the diagram:
$$
\begin{diagram}
\node{L=\Q_p} \arrow{s,-}\arrow{e}
\node{\{1\}}\arrow{s,-}\arrow{w}\\
\node{K= \Q(\sqrt p)}\arrow{s,-}\arrow{e}
\node{Gal(\Q_p/\Q(\sqrt p))}\arrow{s,-}\arrow{w}\\
\node{\Q}\arrow{e}\node{Gal(\Q_p/\Q)}\arrow{w}
\end{diagram}
$$
The Artin map which we need to consider is $\llegendre {K/\Q} * $.
However, from the properties of the Frobenius automorphism, we know that
$\llegendre {K/\Q} * = \llegendre {L/\Q} * \bigg|_K$, so we compute
$\llegendre {L/\Q} *$ instead.
\begin{nnlem} Let $m > 0$, $q$ a prime with $q\nmid m$, and ${\mathcal Q}$ a
prime of $\Q_m$ lying above $q$. Then the $m$-th roots of unity are distinct
modulo ${\mathcal Q}$.
\end{nnlem}
\begin{proof} Let $\zeta_m$ be a primitve $m$-th root of unity.
Then
$$ X^m -1 = \prod_{j=0}^{m-1} (X - \zeta_m^j)$$
\noindent implies
$$ X^{m-1} + \cdots + X + 1 = \prod_{j=1}^{m-1} (X - \zeta_m^j)$$
\noindent and hence
$$ m = \prod_{j=1}^{m-1} (1 - \zeta_m^j).$$
If $\zeta_m^j \equiv \zeta_m^k \ (\bmod\, {\mathcal Q})$, then
$(1-\zeta_m^l) \equiv 0 \ (\bmod\, {\mathcal Q})$ for some $l$, hence
$(m,{\mathcal Q})
\ne 1$. Since ${\mathcal Q}$ is prime, we have ${\mathcal Q}\mid m\O$ and hence
$m\O \subset {\mathcal Q}$. Thus $m \in {\mathcal Q}\cap \Z = q\Z$ which implies
$q\mid m$, a contradiction.
\end{proof}
\noindent Now let $\sigma:I_\Q^{mp_\infty} \to Gal(\Q_m/\Q)$ be the Artin map, and
denote by $\sigma_a$ the automorphism $\sigma(a) \in Gal(\Q_m/\Q)$
characterized by $\sigma_a(\zeta_m) = \zeta_m^a$. $\sigma_q$
is the element of the Galois group, $Gal(\Q_m/\Q)$, which satisfies
$\sigma_q(x) \equiv x^q \ (\bmod\, {\mathcal Q})$ for all $x \in \Z[\zeta_m]$. In
particular, $\sigma_q(\zeta_m) \equiv \zeta_m^q \ (\bmod\, {\mathcal Q})$, and
since every automorphism of $Gal(\Q_m/\Q)$ is characterized by $\tau(\zeta_m) =
\zeta_m^a$ for some $a$, we have by the lemma that
$\sigma_q(\zeta_m) = \zeta_m^q$.
\begin{nnthm} The following are equivalent:
\begin{enumerate}
\item $\legendre p q = 1$.
\item $\displaystyle{X^2 - X + \frac{1-p}4 \equiv 0 \ (\bmod\, q)}$ is
solvable.
\item $q$ splits in $\displaystyle{\Q(\sqrt p) =
\Q\Big(\frac{1+\sqrt p}{2}\Big)}$.
\item $\legendre {K/\Q}q = 1$.
\item $\legendre q p = 1$.
\end{enumerate}
\end{nnthm}
\begin{proof} The equivalence of (1) and (2) can be seen directly. If
$\legendre p q = 1$, then $p \equiv \alpha^2 \ (\bmod\, q)$ for some
$\alpha \in \Z$. $\displaystyle{X\equiv \frac{1+\alpha}{2} \ (\bmod\, q)}$
solves
$\displaystyle{X^2 - X + \frac{1-p}{4} \equiv 0 \ (\bmod\, q)}$. Conversely,
if
$\displaystyle{X^2 - X + \frac{1-p}{4} \equiv (X-\alpha)(X-\beta) \ (\bmod\, q)}$,
then
$(\alpha - \beta)^2 \equiv p \ (\bmod\, q)$, hence $\legendre p q = 1$.
Note that this is really pretty obvious if we think of
$\alpha,\beta \equiv \frac{1 \pm \sqrt p} {2}$, the real roots of the
quadratic.
The equivalence of (2) and (3) is a consequence of the Dedekind-Kummer theorem.
\begin{nnthm}[Dedekind-Kummer] Let $A$ be a Dedekind domain with
quotient field $K$, let $E/K$ be a finite separable extension, and let $B$ be
the integral closure of $A$ in $E$. Suppose that $B = A[\alpha]$ for some
$\alpha \in E$ and let $f(X)$ be the irreducible polynomial for $\alpha$ over
$K$. Let $\P$ be a prime ideal of $A$. Let $\overline{f(X)}$ denote the
reduction of $f(X)$ modulo $\P$. Suppose
$$\overline{f(X)} = \overline{P_1(X)}^{e_1}\cdots\overline{P_g(X)}^{e_g}$$
is the factorization of $f(X)$ modulo $\P$ into powers of distinct monic
irreducible polynomials in $(A/\P)[X]$. Let $P_i(X) \in A[X]$ be a monic
polynomial in $A[X]$ which reduces mod $\P$ to $\overline{P_i(X)}$. Let
$\PP_i$ be the ideal of $B$ generated by $\P$ and $P_i(\alpha)$. Then $\PP_i$
is a prime ideal of $B$ lying above $\P$, $e_i$ is the ramification index, the
$\PP_i$'s are distinct, and
$$\P B = \PP_1^{e_1}\cdots\PP_g^{e_g}$$
is the factorization of $\P$ in $B$.
\end{nnthm}
\noindent We merely note that the roots of $X^2 - X + \frac{1-p}{4}$ are
$\frac{1\pm\sqrt p}{ 2}$ which generate the ring of integers of $\Q(\sqrt
p)$. For the converse, observe that if $X^2 - X + \frac{1-p}{4}$ was
irreducible mod $q$, then
$q$ would be inert in $\Q(\sqrt p)$.
The equivalence of (3) and (4) is an elementary property of the Frobenius
automorphism.
The equivalence of (4) and (5) is where the fun is. Recall that for any
integer $a$ not divisible by $p$, we denote by $\sigma_a$ the automorphism
$\llegendre {L/\Q} a$ of $Gal(\Q_p/\Q)$. From above, we know that
$\sigma_a(\zeta_p) = \zeta_p^a$, and from elementary properties of the Frobenius
that $\displaystyle{\sigma_a|_K = \llegendre{K/\Q} a}$.
The map $a \leftrightarrow \sigma_a$ gives the isomorphism between
$(\Z/p\Z)^\times$ and $Gal(\Q_p/\Q)$. Consider the diagram modified from above:
$$
\begin{diagram}
\node{L=\Q_p} \arrow{s,-}\arrow{e}
\node{\{1\}}\arrow{s,-}\arrow{w}\\
\node{K= \Q(\sqrt p)}\arrow{s,-}\arrow{e}
\node{\{\text{squares}\}}\arrow{s,-}\arrow{w}\\
\node{\Q}\arrow{e}\node{(\Z/p\Z)^\times}\arrow{w}
\end{diagram}
$$
\noindent By the Galois correspondence,
$$Gal(\Q(\sqrt p)/\Q) \cong Gal(\Q_p/\Q)/Gal(\Q_p/\Q(\sqrt p))
\cong (\Z/p\Z)^\times/\{\text{squares}\}$$
\noindent Now (4) is true if and only if
$\sigma_q|_K = 1$ in $Gal(\Q(\sqrt p)/\Q)$, hence if and only if
$\sigma_q \in Gal(\Q_p/\Q(\sqrt p))$, hence
under the correspondence above if and only if $q \in \{\text{squares}\}$ if
and only if (5).
\end{proof}
\section{{\bf Examples of Hilbert Class Fields}}
\begin{enumerate}
\item $K = \Q$. Then $\tilde K = \Q$ since any proper extension
of
$\Q$ is ramified (as a consequence of Minkowski's bound on the discriminant).
\item $K = \Q(\sqrt{-15})$. Then $\tilde K = \Q(\sqrt{-3},\sqrt 5)$.
To see this we need to do a little work. Let $L = \Q(\sqrt{-3},\sqrt 5)$ and
consider the tower of fields:
$$
\begin{diagram}
\node[2]{\Q(\sqrt{-3},\sqrt 5)}
\arrow{sw,-} \arrow{s,-} \arrow{se,-}\\
\node{\Q(\sqrt{-15})}\arrow{se,-}
\node{\Q(\sqrt 5)}\arrow{s,-}
\node{\Q(\sqrt{-3})}\arrow{sw,-}\\
\node[2]{\Q}
\end{diagram}
$$
First, we show that $L/K$ is an unramified extension of fields. Consider the
infinite primes first. Since both primes of $K$ are complex, there can be no
ramification from $K$ to $L$ at the infinite places. Observe that
$\Delta_{K/\Q} = -15$ hence 3 and 5 are the only primes which ramifiy in $K$.
It is then clear that 3 and 5 ramify in $L$. Moreover, these are the only
finite primes $p$ of $\Q$ which ramify in $L$, since if $p \ne
3,5$ is prime, then (by checking discriminants) $p$ is unramified in both
$\Q(\sqrt{-3})$ and $\Q(\sqrt 5)$, and hence in the compositum $L$. Thus the
only primes which can ramify from $K$ to $L$ are the primes in $K$ lying above 3
and 5.
Consider a prime $\PP$ of $L$ lying above 3. Note that since $L/\Q$ is Galois
(it is the compositum of Galois extensions), it doesn't really matter which
prime $\PP$ we choose. Let $\P = \PP \cap K$, and $\P' = \PP \cap \Q(\sqrt
5)$. We know that
$$e(\PP/3) = e(\PP/\P)e(\P/3) = e(\PP/\P')e(\P'/3)$$
\noindent and that $e(\P'/3) = 1$, $e(\P/3) = 2$, and that
$e(\PP/\P') \le [L\,:\,\Q(\sqrt 5)] = 2$. This implies that
$e(\PP/\P) = 1$. This together with an analogous argument for the prime 5
shows us that $L/K$ is an unramified (necessarily abelian) extension. Thus
$L \subset \tilde K$.
Out of the study of Dirichlet $L$-series come various analytic formulae for the
class number of number fields (see Borevich and Shafarevich for example).
The significance is that $Gal(\tilde K/K)$ is isomorphic to the ideal class
group of $K$, and hence $[\tilde K\,:\,K] = h_K$.
If $K = \Q(\sqrt
{-d})$ with $d >2$ and the conductor ${\mathfrak f}_K$ of $K$ (in the old sense -- ignoring
the infinite prime) is odd, then
$$h_K = \frac{1}{2 - \legendre 2 d}\cdot
\sum_{\substack{0