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\begin{document}

\title{Lecture 14}

\author{Michael Broome}

\address{Louisiana State University \\
Department of Mathematics \\
Baton Rouge, LA 70803, USA}

\date{\today}

\email{broome@math.lsu.edu}

\maketitle

\section{Section 6:Nonarchimedean Absolute Values}
\begin{lem}
\label{L:VI.6.4}
Let $(K,|\,|_v)$ be complete with respect to a discrete valuation.  Let $\pi$ be a prime element. Let $S\subset O_v$ be any coset representatives for $k_v=O_v/\mathfrak{p}_v.$  Then every element $a\in O_v$ has a unique expansion.  See below.
\[
a=\sum_{n=0}^{\infty}a_n\pi^n\qquad, where\,\, a_n\in S.
\]
Conversely, every such infinite series converges to an element of $O_v$.
\end{lem}
\begin{proof}
The proof is the same as in Proposition 1.5 with the appropriate substitutions.
\end{proof}


\begin{pro}
\label{P:VI.6.1}
Let $(K,|\,|_v)$ be a nonarchimedean valued field.  Then $K$ is locally compact $\iff$ the following are true:
\begin{itemize}
\item[(1)]$K$ is complete.
\item[(2)]$\Gamma$ is discrete.
\item[(3)]$k_v$, the residue field, is finite.
\end{itemize}
\end{pro}
\begin{proof}
($\Rightarrow$)Suppose $K$ is locally compact.\\
$\qquad$ To show (1), use the fact that if every point has a compact neighborhood, then the space is complete.  Therefore, $K$ is complete.\\
To show (2), we will use Proposition 3.2, that is, the ultrametric inequality.  Suppose $x\in\mathfrak{p}_v$.  Therefore, $|1+x|=1$ because $|x|<1$.  So, we have the following diagram:\\
\[
\begin{matrix}
K^{\times}&\stackrel {|\,|_v}{\longrightarrow}&\R_{+}^{\times}\\
&\searrow&\uparrow\\
&&K^{\times}/(1+\mathfrak{p_v})\\
\end{matrix}
\]
Note that $(1+\mathfrak{p_v})$ is a subgroup of $K^{\times}$ and it is also open.  Thus, the natural quotient topology on $K^{\times}/(1+\mathfrak{p_v})$ is discrete.  Since $\R_{+}^{\times}$ is the homomorphic image of $K^{\times}/(1+\mathfrak{p_v})$, it is also discrete.  Also, we know that $\Gamma_v\subset\R_{+}^{\times}$, which implies that $\Gamma_v$ must also be discrete. \\
To show (3), note that $O_v$ is closed and bounded and hence compact.  Also, $\mathfrak{p}_v\subset O_v$ and $\mathfrak{p}_v$ is open.  Thus, $O_v/\mathfrak{p}_v$ is discrete because $\mathfrak{p}_v$ is open in the natural topology, and $O_v/\mathfrak{p}_v$ is compact since it is the image of a compact space $O_v$ defined by the map:
\[
O_v\longrightarrow O_v/\mathfrak{p}_v
\] 
($\Leftarrow$) We will show that $O_v$ is compact, which implies that it is locally compact.  Note that $|\,|_v$ will make $O_v$ into a metric space, and for a metric space, compactness is equivalent to sequential compactness.  That is, there is a subsequence that converges.\\
Let $a_j\in O_v$.  By Lemma 6.4, we have the following expansion of $a_j$:
\[
a_j=\sum_{n=0}^{\infty}a_{jn}\pi^n,\qquad where\, a_{jn}\in S.
\]
Note that elements in $S$ are coset representatives for $k_v=O_v/\mathfrak{p}_v$, and since $k_v$ is finite by assumption, $S$ must be a finite set also.  Because $S$ is finite, there exists $a_0^*$ which occurs infinitely often as a constant coefficient of an $a_j$.  Now pick out the $a_j$ for which $a_{j0}=a_0^*$.  This is a subsequence of the original sequence.  Thus, we now have the following:
\[
a_j=a_0^*+a_{j1}\pi+a_{j2}\pi^2+\cdots
\]
Now we will concentrate on the coefficient of the $\pi$ term.  Again, by the above argument, there exists an $a_1^*$ which occurs infinitely often as $a_{j1}$ in this subsequence.  Next, extract a subsequence from the new sequence by selecting the $a_j$ for which $a_{j1}=a_1^*$.  The new subsequence is the following:
\[
a_j=a_0^*+a_1^*\pi+a_{j2}\pi^2+a_{j3}\pi^3+\cdots
\]
Continue this process.  We will get a subsequence that is of the form:
\[
a^*=a_0^*+a_1^*\pi+a_2^*\pi^2+a_3^*\pi^3+\cdots
\]
Note that $a^*$ is a limit point of the original sequence $a_j$, which implies that $O_v$ is sequentially compact.  Therfore, $O_v$ is compact and thus locally compact. 
\end{proof}

\section{Section 7:Algebraic Extensions of Complete Discrete Valuations}
Consider the following scenario:  we have $(K,|\,|_v)$ is a complete, discrete, nonarchimedean valued field.  Let $L/K$ be a finite extension field.  By Theorem 4.3,there is a unique way to extend the absolute value on $K$ to an absolute value on $L$ ,say $|\,|_w$.  In fact, we have the following formula:
\[
|x|_w=|N_{L/K}(x)|^{1/n},\qquad where\, n=[L:K]
\]
We also have the following diagram:
\[
\begin{matrix}
K&\subset&L\\
\cup&&\cup\\
O_v&\subset&O_w\\
\cup&&\cup\\
\mathfrak{p}_v&\subset&\mathfrak{p}_w
\end{matrix}
\] 
Also, we have the following relations for the above sets:
\begin{itemize}
\item[(1)]$K\cap O_w=O_v$.
Note that $O_w=\{x\in L:|x|_v\leq 1\}$.  So if we restrict to $K$, the absolute value on $L$ will restrict to the absolute value on $K$.  So the result will hold.
\item[(2)]$O_v\cap\mathfrak{p}_w=\mathfrak{p}_v$\\
Note that $\mathfrak{p}_w=\{x\in L:|x|_w<1\}$.  So if we restrict to $O_v$, the absolute value on $L$ will restrict to the absolute value on $K$.  Thus, the result holds.
\item[(3)]$\Gamma_v\subset\Gamma_w$\\
The result follows from the fact that the norm on $K$ is a subset of the norm on $L$.
\item[(4)]$k_v\subset k_w$\\
Consider the following diagram:
\[
\begin{matrix}
O_v&\hookrightarrow&O_w\\
&\stackrel{\varphi}{\searrow}&\downarrow\\
&&O_w/\mathfrak{p}_w=k_w\\
\end{matrix}
\]
,where $\varphi$ is the natural homomorphism.  Note that the kernel of $\varphi$ is \\$O_v\cap\mathfrak{p}_w=\mathfrak{p}_v$ by (2).  This along with the diagram implies that 
\[
k_v=O_v/\mathfrak{p}_v=O_v/ker(\varphi)\hookrightarrow O_w/\mathfrak{p}_w=k_w.
\]
Therefore, the result holds.
\end{itemize}

\begin{Def}
\label{D:VI.7.0}
The ramification index is the integer \[e=e(w|v)=[\Gamma_w:\Gamma_v].\]  Note that this is finite.
\end{Def}

\begin{Def}
\label{D:VI.7.1}
The residual degree is the index \[f=f(w|v)=[k_w:k_v]=dim_{k_v}(k_w).\]  Note that this number is also finite by the next lemma.
\end{Def}

\begin{lem}
\label{L:VI.7.1}
$f\leq n=[L:K]$
 \end{lem}
\bibliographystyle{amsplain}



\providecommand{\bysame}{\leavevmode\hbox to3em{\hrulefill}\thinspace}
\begin{thebibliography}{10}

\bibitem{sL}
S. Lang, \emph{Algebraic Number Theory}, Second Edition, Graduate  
Texts in Mathematics, {\bf 110}, Springer - Verlag, 1994.

\end{thebibliography}

\end{document}