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\title{Math 129: Algebraic Number Theory\\
{\bf Lecture 13: Galois Extensions}}
\author{William Stein}
\date{Thursday, March 8, 2004}
\begin{document}
\maketitle 

\section{The Decomposition and Inertia Groups}
Suppose $K$ is a number field that is Galois over $\Q$ with
group $G=\Gal(K/\Q)$.
Fix a prime $\p\subset \O_K$ lying over $p\in\Z$.  
\begin{definition}[Decomposition group]\label{def:decomp}
The {\em decomposition group} of $\p$ is the subgroup
$$
  D_\p = \{\sigma \in G : \sigma(\p)=\p\} \leq G.
$$
\end{definition}
(Note: The decomposition group is called the ``splitting group''
in Swinnerton-Dyer.  Everybody I know calls it the decomposition
group, so we will too.)
Recall that $G$ acts on the set of primes $\p$ lying
over $p$.  Thus the decomposition group is the stabilizer in $G$
of $\p$.  The orbit-stabilizer theorem implies that $[G:D_\p]$
equals the orbit of $\p$, which we proved last time\edit{ref}
equals the number $g$ of primes lying over $p$, so $[G:D_\p]=g$.

\begin{lemma}
The decomposition subgroups $D_\p$ corresponding to primes $\p$
lying over a given $p$ are all conjugate in $G$.
\end{lemma}
\begin{proof}
We have $\tau(\sigma(\tau^{-1}(\p))) = \p$ if and only if
$\sigma(\tau^{-1}(\p)) = \tau^{-1}\p$.  Thus $\tau\sigma\tau^{-1}\in D_p$
if and only if $\sigma\in D_{\tau^{-1}\p}$, so 
$\tau^{-1}D_p\tau = D_{\tau^{-1}\p}$.   The lemma now follows
because, as we proved
before\edit{reference}, $G$ acts transitively on the set of
$\p$ lying over~$p$.
\end{proof}

The decomposition group is extremely useful because it allows us
to see the extension $K/\Q$ as a tower of extensions, such that at
each step in the tower we understand well the splitting behavior
of the primes lying over~$p$.  Now might be a good time to glance
ahead at Figure~\ref{fig} on page~\pageref{fig}.

We characterize the fixed field of $D=D_\p$ as follows.
\begin{proposition}\label{prop:nosplit}
The fixed field $K^D$ of $D$
$$K^D=\{a \in K : \sigma(a) = a\text{ for all }
\sigma \in D\}$$
is the smallest subfield $L\subset K$ such that $\p\cap L$
does not split in~$K$ (i.e., $g(K/L)=1$).
\end{proposition}
\begin{proof}
First suppose $L=K^D$, and note that by Galois theory $\Gal(K/L)\isom
D$, and by the theorem we proved on Tuesday\edit{ref}, the group $D$
acts transitively on the primes of $K$ lying over $\p\cap L$.  One of
these primes is $\p$, and $D$ fixes $\p$ by definition, so there is
only one prime of $K$ lying over $\p\cap L$, i.e., $\p\cap L$ does not
split in~$K$.  Conversely, if $L\subset K$ is such that $\p\cap L$
does not split in~$K$, then $\Gal(K/L)$ fixes $\p$ (since it is the only
prime over $\p\cap L$), so $\Gal(K/L)\subset D$, hence $K^D\subset L$.
\end{proof}
Thus $p$ does not split in going from $K^D$ to $K$---it does some
combination of ramifying and staying inert.  To fill in more of
the picture, the following proposition asserts that $p$ splits
completely and does not ramify in $K^D/\Q$.
\begin{proposition}\label{prop:noresidue}
Let $L=K^D$ for our fixed prime $p$ and Galois extension $K/\Q$.
Let $e=e(L/\Q),f=f(L/\Q),g=g(L/\Q)$ be for $L/\Q$ and $p$. 
Then $e=f=1$ and $g=[L:\Q]$, i.e., $p$ does not ramify and splits 
completely in~$L$.  Also $f(K/\Q)=f(K/L)$ and $e(K/\Q)=e(K/L)$.
\end{proposition}
\begin{proof}
As mentioned right after Definition~\ref{def:decomp}, the
orbit-stabilizer theorem implies that $g(K/\Q)=[G:D]$, and 
by Galois theory $[G:D]=[L:\Q]$.
Thus
\begin{align*}
 e(K/L)\cdot f(K/L) &= [K:L]
   =[K:\Q]/[L:\Q] \\
&= \frac{e(K/\Q)\cdot f(K/\Q) \cdot g(K/\Q)}{[L:\Q]}
   = e(K/\Q)\cdot f(K/\Q).
\end{align*}
Now $e(K/L)\leq e(K/\Q)$ and $f(K/L)\leq f(K/\Q)$, so 
we must have $e(K/L)=e(K/\Q)$ and $f(K/L)=f(K/\Q)$.  
Since $e(K/\Q)=e(K/L)\cdot e(L/\Q)$ and $f(K/\Q)=f(K/L)\cdot f(L/\Q)$,
the proposition follows.
\end{proof}

\newcommand{\galff}{\Gal(\F_\p/\F_p)}
\newcommand{\autff}{\Aut(\F_\p/\F_p)}

\subsection{Galois groups of finite fields}\label{sec:galoisfinite}
Each $\sigma\in D=D_\p$ acts in a well-defined
way on the finite field $\F_\p = \O_K/\p$, so we obtain
a homomorphism 
$$
  \vphi:D_\p \to \galff.
$$
We pause for a moment and derive a few basic properties of $\galff$,
which are in fact general properties of Galois groups for finite fields. 
Let $f=[\F_\p:\F_p]$.

The group $\autff$ contains the element $\Frob_p$ defined by
$$\Frob_p(x) = x^p,$$ because $(xy)^p = x^p y^p$  and
 $$(x+y)^p = x^p + px^{p-1}y + \cdots + y^p \con
x^p+y^p\pmod{p}.$$ By a homework problem\edit{cite}, the group
$\F_\p^*$ is cyclic, so there is an element $a\in\F_\p^*$ of order
$p^f-1$, and $\F_\p=\F_p(a)$.  Then $\Frob_p^n(a) = a^{p^n} = a$ if
and only if $(p^f-1)\mid p^n-1$ which is the case preciselywhen
$f\mid n$, so the order of $\Frob_p$ is~$f$.  Since the order of the
automorphism group of a field extension is at most the degree of the
extension, we conclude that $\autff$ is generated by $\Frob_p$.  Also,
since $\autff$ has order equal to the degree, we conclude that
$\F_\p/\F_p$ is Galois, with group $\galff$ cyclic of order $f$
generated by $\Frob_p$.  (Anther general fact: Up to isomorphism
there is exactly one finite field of each degree.  Indeed, if there
were two of degree $f$, then both could be characterized as the set of
roots in the compositum of $x^{p^f}-1$, hence they would be equal.)

\subsection{The Exact Sequence}\label{sec:exactseq}
As mentioned above, there is a natural reduction homomorphism
$$
  \vphi:D_\p \to \galff.
$$
\begin{theorem}
The homomorphism $\vphi$ is surjective.
\end{theorem}
\begin{proof}
Let $\tilde{a} \in \F_\p$ be an element such that $\F_\p = \F_p(a)$. 
Lift $\tilde{a}$ to an algebraic integer $a\in \O_K$, and let
$f=\prod_{\sigma\in D_p}(x-\sigma(a))\in K^D[x]$ be the characteristic polynomial of~$a$ over $K^D$.
Using Proposition~\ref{prop:noresidue} we see that~$f$ 
reduces to the minimal polynomial $\tilde{f}=\prod (x-\tilde{\sigma(a)})\in \F_p[x]$
of $\tilde{a}$ (by the Proposition the coefficients of $\tilde{f}$
are in $\F_p$, and $\tilde{a}$ satisfies $\tilde{f}$, and the
degree of $\tilde{f}$ equals the degree of the minimal polynomial
of $\tilde{a}$).   The roots of $\tilde{f}$ are of the form 
$\tilde{\sigma}(a)$, and
the element $\Frob_p(a)$ is also a root of
$\tilde{f}$, so it is of the form $\tilde{\sigma(a)}$.
We conclude that the generator $\Frob_p$ of $\galff$ is
in the image of $\vphi$, which proves the theorem. 
\end{proof}

\begin{definition}[Inertia Group]
The {\em inertia group} is the kernel $I_\p$ of $D_\p\to\galff$.
\end{definition}
Combining everything so far,  we find an exact sequence of groups
\begin{equation}\label{eqn:exact}
   1 \to I_\p \to D_\p \to \galff\to 1.
\end{equation}
The inertia group is a measure of how $p$ ramifies in $K$.
\begin{corollary}
We have $\# I_\p= e(\p/p)$, where $\p$ is a prime of $K$ over $p$.
\end{corollary}
\begin{proof}
The sequence (\ref{eqn:exact}) implies that 
$\#I_\p = \#D_\p/f(K/\Q)$.  Applying Propositions~\ref{prop:nosplit}--\ref{prop:noresidue}, we have
$$\#D_\p = [K:L] = \frac{[K:\Q]}{g} = \frac{efg}{g} = ef.$$
Dividing both sides by $f=f(K/\Q)$ proves the corollary.
\end{proof}

We have the following characterization of $I_\p$.
\begin{proposition}\label{prop:charip}
Let $K/\Q$ be a Galois extension with group $G$, let $\p$ be a prime lying
over a prime $p$.  Then 
$$
I_\p = \{\sigma\in G : \sigma(a) = a\pmod{\p}\text{ for all } a\in\O_K\}.
$$
\end{proposition}
\begin{proof}
By definition $I_\p = 
\{\sigma\in D_\p : \sigma(a) = a\pmod{\p}\text{ for all } a\in\O_K\}$,
so it suffices to show that if $\sigma\not\in D_\p$, then there
exists $a\in\O_K$ such that $\sigma(a)=a\pmod{\p}$.  
If $\sigma\not\in D_\p$, we have $\sigma^{-1}(\p)\neq \p$, so
since both are maximal ideals, there exists $a\in\p$ with $a\not\in\sigma^{-1}(\p)$,
i.e., $\sigma(a)\not\in\p$.  Thus $\sigma(a)\not\con a\pmod{\p}$.
\end{proof}

Figure~\ref{fig} is a picture of the
splitting behavior of a prime $p\in\Z$.
\begin{figure}\label{fig}
\includegraphics[width=\textwidth]{splitting.eps}
\caption{The Splitting of Behavior of a Prime in a Galois Extension}
\end{figure}

\section{Frobenius Elements}
Suppose that $K/\Q$ is a finite Galois extension with group $G$ and
$p$ is a prime such that $e=1$ (i.e., an unramified prime).  Then 
$I=I_\p=1$ for any $\p\mid p$, so the map $\vphi$ of Section~\ref{sec:exactseq}
is a canonical isomorphism $D_\p \isom \Gal(\F_\p/\F_p)$.  By
Section~\ref{sec:galoisfinite}, the group $\Gal(\F_\p/\F_p)$ is
cyclic with canonical generator $\Frob_p$.  
The {\em Frobenius element} corresponding to $\p$ is
$\Frob_\p\in D_\p$. It is the unique element of $G$ such that for all
$a\in\O_K$ we have 
$$
  \Frob_\p(a)\con a^p\pmod{\p}.
$$
(To see this argue as in the proof of Proposition~\ref{prop:charip}.)
Just as the primes $\p$ and decomposition groups $D$ are all
conjugate, the Frobenius elements over a given prime are conjugate.
\begin{proposition}
For each $\sigma \in G$, we have 
$$
 \Frob_{\sigma\p} = \sigma\Frob_\p\sigma^{-1}.
$$
In particular, the Frobenius elements lying over a given
prime are all conjugate.\edit{reference to theorem that all primes conjugate.}
\end{proposition}
\begin{proof}
Fix $\sigma\in G$. For any $a\in\O_K$ we have
$\Frob_\p(\sigma^{-1}(a)) - \sigma^{-1}(a) \in \p$. 
Multiply by $\sigma$ we see that 
$\sigma\Frob_\p(\sigma^{-1}(a)) - a \in \sigma\p$,
which proves the proposition.
\end{proof}

Thus the conjugacy class of $\Frob_\p$ in $G$ is a well defined
function of~$p$.  For example, if $G$ is abelian, then $\Frob_\p$ does
not depend on the choice of $\p$ lying over $p$ and we obtain a well
defined symbol $\kr{K/\Q}{p} =\Frob_\p\in G$ called the {\em Artin
symbol}.  It extends to a map from the free abelian
group on unramified primes to the group $G$ (the fractional ideals of
$\Z$).  Class field theory (for $\Q$) sets up a natural bijection
between abelian Galois extensions of $\Q$ and certain maps from
certain subgroups of the group of fractional ideals for $\Z$.  We have
just described one direction of this bijection, which associates to an
abelian extension the Artin symbol (which induces a homomorphism).
The Kronecker-Weber theorem asserts that the abelian extensions of
$\Q$ are exactly the subfields of the fields $\Q(\zeta_n)$, as $n$
varies over all positive integers.  By Galois theory there is a
correspondence between the subfields of $\Q(\zeta_n)$ (which has
Galois group $(\Z/n\Z)^*$) and the subgroups of $(\Z/n\Z)^*$.  Giving
an abelian extension of $\Q$ is {\em exactly the same} as giving an
integer~$n$ and a subgroup of $(\Z/n\Z)^*$. Even more importantly, the
reciprocity map $p\mapsto \kr{Q(\zeta_n)/\Q}{p}$ is simply $p\mapsto
p\in (\Z/n\Z)^*$.  This is a nice generalization of quadratic
reciprocity: for $\Q(\zeta_n)$, the $efg$ for a prime $p$ depends in a
simple way on nothing but $p\mod n$.

\section{Galois Representations and a Conjecture of Artin}
The Galois group $\Gal(\Qbar/\Q)$ is an object of central importance
in number theory, and I've often heard that in some sense number
theory is the study of this group.  A good way to study a group is to
study how it acts on various objects, that is, to study its
representations.

Endow $\galq$ with the topology which has as a basis of open neighborhoods
of the origin the subgroups $\Gal(\Qbar/K)$, where $K$ varies
over finite Galois extensions of $\Q$. (Note: This is {\bf not} the
topology got by taking as a basis of open neighborhoods the collection
of finite-index normal subgroups of $\galq$.)
Fix a positive integer $n$ and let $\GL_n(\C)$ be the group of
$n\times n$ invertible matrices over $\C$ with the discrete topology.

\begin{definition}
A {\em complex  $n$-dimensional representation} of $\galq$
is a continuous homomorphism
$$
  \rho:\galq\to \GL_n(\C).
$$
\end{definition}
For $\rho$ to be continuous means that there is a finite Galois
extension $K/\Q$ such that $\rho$ factors through $\Gal(K/\Q)$:
$$\xymatrix{ {\galq}\ar[rr]^{\rho}\ar[dr]& &{\GL_n(\C)}\\
&{\Gal(K/\Q)}\ar[ur]_{\rho'}}
$$ 
For example, one could take $K$ to be
the fixed field of $\ker(\rho)$.  (Note that continous implies that
the image of $\rho$ is finite, but using Zorn's lemma one can show
that there are homomorphisms $\galq\to\{\pm 1\}$ with finite image
that are not continuous, since they do not factor through the Galois
group of any finite Galois extension.)

Fix a Galois representation $\rho$ and a finite Galois extension
$K$ such that $\rho$ factors through $\Gal(K/\Q)$.
For each prime $p\in\Z$ that is not ramified in $K$, there is an
element $\Frob_\p\in\Gal(K/\Q)$ that is well-defined up to conjugation
by elements of $\Gal(K/\Q)$.  This means that $\rho'(\Frob_p)\in \GL_n(\C)$
is well-defined up to conjugation.  Thus the characteristic polynomial
$F_p\in\C[x]$ is a well-defined invariant of $p$ and $\rho$.  Let
$R_p(x) = x^{\deg(F_p)}\cdot F_p(1/x)$ be the polynomial obtain
by reversing the order of the coefficients of $F_p$.  
Following E.~Artin, let $n=[K:\Q]$ and set
$$
L(\rho,s) = \prod_{p\text{ unramified}}
\frac{1}{R_p(p^{-s})}.
$$
We view. $L(\rho,s)$ as a function of a single complex variable $s$.
One can prove that $L(\rho,s)$ is holomorphic on some right
half plane, and extends to a meromorphic function on all $\C$.
\begin{conjecture}[Artin]
The $L$-series of any continuous representation $\Gal(\Qbar/\Q)\to\GL_n(\C)$
is an entire function on all $\C$, except possibly at $1$.
\end{conjecture}
This conjecture asserts that there is a way to analytically continue
$L(\rho,s)$ to the whole complex plane, except possibly at $1$.  The
simple pole at $s=1$ corresponds to the trivial representation (the
Riemann zeta function), and if $n\geq 2$ and $\rho$ is irreducible,
then the conjecture is that $\rho$ extends to a holomorphic function
on all $\C$.

The conjecture follows from class field theory for $\Q$ when
$n=1$. When $n=2$ and the image of $\rho$ in $\PGL_2(\C)$ is a
solvable group, the conjecture is known, and is a deep theorem of
Langlands and others (see {\em Base Change for $\GL_2$}).  When $n=2$
and the projective image is not solvable, the only possibility is that
the projective image is isomorphic to the alternating group $A_5$.
Because $A_5$ is the symmetric group of the icosahedron, these
representations are called {\em icosahedral}.  In this case Joe
Buhler's Harvard Ph.D. thesis gave the first example, there is a whole
book (Springer Lecture Notes 1585, by Frey, Kiming, Merel, et al.),
which proves Artin's conjecture for 7 icosahedral representation (none
of which are twists of each other).  Kevin Buzzard and I (Stein)
proved the conjecture for 8 more examples.  Subsequently, Richard
Taylor, Kevin Buzzard, and Mark Dickinson proved the conjecture for an
infinite class of icosahedral Galois representations (disjoint from
the examples).  The general problem for $n=2$ is still open, but
perhaps Taylor and others are still making progress toward it.


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