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\title{Math 129: Algebraic Number Theory\\
{\bf Lecture 14: Galois Representations, Absolute Values}}
\author{William Stein}
\date{Thursday, March 23, 2004}
\begin{document}
\maketitle 

We start with a quick review of last time:

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 $p\in\Z$.  
The {\em decomposition group} of $\p$ is the subgroup
$$
  D_\p = \{\sigma \in G : \sigma(\p)=\p\} \leq G.
$$
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 $[G:D_\p]=g$.
(This clarifies the key point that was confusing me last time.)

We proved:
\begin{lemma}
The decomposition subgroups $D_\p$ corresponding to primes $\p$
lying over a given $p$ are all conjugate in $G$.
\end{lemma}

\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{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}

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

There is a natural reduction homomorphism
\begin{equation}\label{eqn:red}
  \vphi:D_\p \to \galff,
\end{equation}
and we proved last time that it is surjective.  The key
point was that the reduction of the characteristic polynomial
of a certain lift of a generator of $\F_\p$ was in $\F_p[x]$.
Now, back to our story....

\section{The Inertia Group}
\begin{definition}[Inertia Group]
The {\em inertia group} is the kernel $I_\p$ of $D_\p\to\galff$.
\end{definition}
Thus we have 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}

\edit{Put this back in here in book:
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 (\ref{eqn:red})
is a canonical isomorphism $D_\p \isom \Gal(\F_\p/\F_p)$.  \edit{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 Major 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
$$R_p(x) = x^{\deg(F_p)}\cdot F_p(1/x) = 1 + \cdots +
\det(\Frob_p)\cdot x^{\deg(F_p)}$$ 
be the polynomial obtain
by reversing the order of the coefficients of $F_p$.  
Following E.~Artin, set
\begin{equation}\label{eqn:artin}
L(\rho,s) = \prod_{p\text{ unramified}}
\frac{1}{R_p(p^{-s})}.
\end{equation}
(Note: Please change what's in the handout for day13 to (\ref{eqn:artin}).)
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 some way to analytically continue
$L(\rho,s)$ to the whole complex plane, except possibly at $1$.  
(A standard fact from complex analysis is that this analytic
continuation must be unique.)
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 $\GL_2$}), which played
a crucial roll in Wiles's proof of Fermat's Last Theorem.  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.

\section{Absolute Values}
We will now move onto something different: absolute values,
completions, local fields, adeles, and ideles.  A common technique in
mathematics is to ``reduce to the local case'' in order to gain a much
clearer understanding of what is going on, and the methods we will
introduce next, which are basic to much modern number theory,
 allow us to do that for number fields.

Fix a field $K$ for the rest of this section.
\begin{definition}[Absolute Value]
An {\em absolute value} on $K$ is a map $\absval{\cdot}: K\to \R$
such that there is some $\alpha>0$ such that for all $a,b\in K$,
we have
\begin{itemize}
\item $\absval{a}\geq 0$ and $\absval{a}=0$ if and only
if $a=0$,
\item $\absval{a}\cdot\absval{b} = \absval{ab}$, and
\item $\absval{a+b}^{\alpha} \leq \absval{a}^{\alpha} + \absval{b}^{\alpha}$.
\end{itemize}
\end{definition}

\begin{example}
Suppose $\sigma:K\hra \C$.  Let $\absval{a} = |\sigma(a)|$, where
$|\sigma(a)|$ is the usual absolute value on $\C$.   

On any field $K$, the {\em trivial absolute value} is the one
given by 
$$
  \absval{a} = \begin{cases} 0 & \qquad\text{if }a=0\\
  1&\qquad\text{if }a\neq 0.\end{cases}
$$
\end{example}

\begin{definition}[Places]
Two absolute values $\absval{\cdot}_1$ and $\absval{\cdot}_2$ on a
field $K$ are {\em equivalent} if there exists a $c>0$ such that
$\absval{\cdot}_1 = \absval{\cdot}_2^c$.  A {\em place} for a field
$K$ is an equivalence class of absolute values on $K$.
\end{definition}

Recall that a metric space is a set $X$ equipped with a metric
$d:X\times X\to \R_{\geq 0}$, which is a function such that there is
an $\alpha$ so that for all
$x,y,z\in X$ we have $d(x,x)=0$, $d(x,y)=d(y,x)$, $d(x,z)^\alpha\leq
d(x,y)^\alpha+d(y,z)^\alpha$, and $d(x,y)=0\implies x=y$.
A metric on a space induces a topology, in which a basis of open
sets is the collection of all open balls
$$
 B(a,r) = \{x\in K : d(x,a)<r\}.
$$

If $\absval{\cdot}$ is an absolute value on a field~$K$, then
$d(x,y)=\absval{x-y}$ is a metric on~$K$, and hence $\absval{\cdot}$
induces a topology on~$K$.  For example, the trivial absolute
value induces the discrete topology, since every point set
is open, as $B(a,\frac{1}{2}) = \{a\}$.

\begin{proposition}
Two valuations $\absval{\cdot}_1$ and $\absval{\cdot}_2$
are equivalent if and only if they induce the same topology on $K$.
\end{proposition}
\begin{proof}
$(\Longrightarrow)$ First suppose 
$\absval{\cdot}_1$ and $\absval{\cdot}_2$ are
equivalent.  Then there is a $c>0$ such that 
$\absval{\cdot}_1 = \absval{\cdot}_2^c$.
Thus
\begin{align*}
  B_1(a,r) &= \{x \in K : \absval{x-a}_1 < r\}\\
   &=\{x \in K : \absval{x-a}_2 < r^{1/c}\} = B_2(a,r^{1/c}).
\end{align*}
Thus the collection of all open balls for $\absval{\cdot}_1$
is exactly the same as the set of open balls for $\absval{\cdot}_2$,
hence the topologies are the same.

$(\Longleftarrow)$
Next suppose $\absval{\cdot}_1$ and $\absval{\cdot}_2$ induce
the same topology on $K$.  To deduce that the two absolute
values are equivalent, we prove a lemma.
\begin{lemma}\label{lem:absvalconv}
For any sequence $\{x_n\}$ in~$K$ we
have 
$$
  \absval{x_n}_1 \to 0 \iff \absval{x_n}_2 \to 0.
$$
\end{lemma}
\begin{proof}
It suffices to prove that if $\absval{x_n}_1\to 0$
then $\absval{x_n}_2\to 0$, since the proof of the
other implication is the same. 
Let $\eps>0$.  The topologies induced by the two absolute
values are the same, so $B_2(0,\eps)$ can be covered by
open balls $B_1(a_i,r_i)$.  One of these open balls
$B_1(a,r)$ contains~$0$, and using the triangle inequality
we see that there is $\eps'>0$ such that
$$
  B_1(0,\eps') \subset B_1(a,r)\subset B_2(0,\eps).
$$
Since $\absval{x_n}_1\to 0$, there exists $N$ such
that for $n\geq N$ we have $\absval{x_n}_1 <\eps'$.
For such $n$, we have $x_n\in B_1(0,\eps')$, so $x_n\in B_2(0,\eps)$,
so $\absval{x_n}_2<\eps$.  Thus $\absval{x_n}_2\to 0$.
\end{proof}
If $x\in K$ and $i=1,2$, then $\absval{x^n}_i \to 0$
if and only if $\absval{x}_i^n\to 0$, which is the
case if and only if $\absval{x}_i<1$.   Thus 
Lemma~\ref{lem:absvalconv} implies that 
$\absval{x}_1<1$ if and only if $\absval{x}_2<1$.

If there is no nonzero $x\in K$ with $\absval{x}_1<1$,
then $\absval{x}_1=1$ for all $x\neq 0$ (otherwise
$\absval{x}_1>1 \implies \absval{1/x}_1<1$), so 
$\absval{\cdot}_1$ induces the discrete topology. 
Since $\absval{\cdot}_1$ and $\absval{\cdot}_2$
induce the same topology, $\absval{\cdot}_2$ also
induces the discrete topology. If $\absval{\cdot}_2$
is nontrivial then there exists a nonzero $x\in K$
with $\absval{x}_2<1$.  Then $\absval{x^n}_2\to 0$,
so for all $\eps>0$ there exists $N$ such that for $n\geq N$
we have $x^n\in B(0,\eps)$.  In the discrete topology every
point is open, so there exists $\eps>0$ such that
$B(0,\eps)=\{0\}$.  But then $x^n=0$ for $n$ sufficiently
large, hence $x=0$, a contradiction.  This completes
the proof of the proposition when $\absval{\cdot}_1$
is trivial.

We may assume there is a nonzero $x_0\in K$ with $\absval{x_0}_1<1$.
As mentioned above $\absval{x_0}_2<1$ as well.  We will use this $x_0$
to show that $\absval{\cdot}_1$ and $\absval{\cdot}_2$ are equivalent.
Let $\gamma>0$ be the real number such that $\absval{x_0}_2 =
\absval{x_0}_1^{\gamma}$, so $\gamma =
\log(\absval{x_0}_2)/\log(\absval{x_0}_1)$.  For any nonzero $x\in K$
with $\absval{x}_1<1$, let~$\lambda$ be such that
$\absval{x}_1=\absval{x_0}_1^{\lambda}$.  
We now prove that $\absval{x}_2 = \absval{x_0}_2^{\lambda}$.
If $a/b$ is any nonzero
rational number with $a/b>\lambda$, then 
$$
\absval{\frac{x_0^a}{x^b}}_1 = \frac{\absval{x_0}_1^a}{\absval{x}_1^b}
= \frac{\absval{x_0}_1^a}{\absval{x_0}_1^{\lambda b}} =
\absval{x_0}_1^{a-\lambda b} < 1
$$
(since $a-\lambda b>0$ and $\absval{x_0}_1<1$), so 
$\absval{x_0^a/x^b}_2<1$ also.  Hence
$\absval{x_0}_2^a < \absval{x}_2^b$, so
$\absval{x_0}_2^{a/b}<\absval{x}_2$.
Likewise, if $a/b$ is a nonzero rational number with $a/b<\lambda$,
arguing as above we see that $\absval{x_0}_2^{a/b}>\absval{x}_2$.
Combining these two facts, we see that $\absval{x}_2 =
\absval{x_0}_2^{\lambda}$.  Thus
$$
\absval{x}_2 = \absval{x_0}_2^{\lambda} = \absval{x_0}_1^{c\lambda}
  = \absval{x}_1^c,
$$
hence $\absval{\cdot}_1$ is equivalent to $\absval{\cdot}_2$,
as claimed.
\end{proof}

Assume henceforth that $K$ has characteristic $0$.
\begin{definition}[Archimedean]
An absolute value is {\em archimedean} if
there is an integer $n\in K$ such that $\absval{n}>1$.
An absolute value is {\em non-archimedean} if it
is not archimedean.
\end{definition}

\begin{proposition}
Let $K$ be a number field.  The archimedean valuations
are all of the form $\absval{a} = |\sigma(a)|^c$, where
$\sigma:K\hra \C$ is an embedding, $c>0$, and $|\cdot|$
is the usual absolute value on $\C$.
\end{proposition}
\begin{proof}\end{proof}

\begin{proposition}
Suppose $\absval{\cdot}$ is a non-archimedean valuation
on a number field $K$.  Then there is a prime $\p$
of $\O_K$ and a constant $C>1$ such that 
$$\absval{a}=C^{-\ord_{\p}(a)},$$
where $\ord_{\p}(a)$ is the largest power
of $\p$ that divides the ideal $a\O_K$.  (Note
that when $a=0$, $\absval{a} =C^{-\infty}=0$, as expected.)
\end{proposition}

\section{Completions and Local Fields}

Completion of $K$ at $\p$.  Hensel's Lemma.  Weak Approximation.

%\bibliography{biblio}
\end{document}
