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\begin{document}

\title{Lecture 1}

\author{Mustafa Arslan}

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

\date{January 19, 2000}


\email{arslan@math.lsu.edu}
\maketitle

\section{Local Fields}
\subsection{$\Z_p$ and $\Q_p$}

\begin{Def}
Given a collection of sets and maps
\begin{equation}
\begin{array}{ccccccc}
  &&& \phi_3 &&  \phi_2 & \\
\cdots& \rightarrow& X_3& \rightarrow& X_2& \rightarrow& X_1 \\
\end{array}
\end{equation}
The subset
$$\dirlim X_\bullet=\{ (\ldots,x_3,x_2,x_1):\
x_n=\phi_{n+1}(x_{n+1})\ \}$$
of $\prod X_j$ is called {\em direct limit} of the projective
system $(X_n,\phi_n)$.
\end{Def}

If each $X_j$ is a topological space, and each $\phi _{j}$ are
continuous, we give
$$\dirlim X_\bullet\subset\prod X_j$$
the subspace topology, where $\prod X_j$ has the product topology
\footnote{Remember, the open sets of product topology on $\prod
X_j$ are of the form
$$U_{j_1}\times\cdots\times U_{j_n}\times\prod_{j\neq j_i} X_j$$
where $U_{j_i}$ are open in $X_{j_i}$.}

In particular if each $X_j$ is finite, with the discrete topology, then
$\prod X_j$ is compact\footnote{This is well-known Tychonoff theorem}.
\begin{lem}
If each $X_j$ is finite with discrete topology then $\dirlim
X_\bullet$ is a
closed subset of $\prod X_j$.
\end{lem}
\begin{proof}
If $p=(\ldots,p_3,p_2,p_1)$ is not in $\dirlim X_j$ then
$$\phi_{n+1}(p_{n+1})\neq p_n.$$
for some $n$ and hence
$$\left( \{p_{n+1}\}\times\{p_n\}\times\prod_{j\neq n,n+1} X_m\right)$$
is an open neighborhood of $p$ not intersecting with $\dirlim X_\bullet$.
\end{proof}
\begin{exa}
Let $p$ be a prime integer and $X_j=\Z/p^j\Z$ and let the maps
$\phi_j$ are
defined by
$$\begin{array}{cccc}
\phi_j:&\Z/p^j\Z&\rightarrow&\Z/p^{j-1}\Z\\
&[d]_{p^j}&\mapsto&[d]_{p^{j-1}}
\end{array}
$$
\end{exa}
\begin{Def}
The direct limit $\Z_p=\dirlim \Z_\bullet$ of the above example is
called
the {\em ring of p-adic integers}
\end{Def}

Note that $\Z_p$ is a compact topological ring.
In general for a ring $A$ and for an ideal $I$, $\dirlim A/I^n$ is
called
$I-adic$ completion of $A$.

\begin{lem}
\label{L:I.1.5}
For all $m\leq n$ there is a short exact sequence
\begin{equation}
\begin{array}{ccccccccc}
&&&p^n&&&&&\\
0&\rightarrow&\Z/p^{m-n}\Z&\rightarrow&\Z/p^m\Z&\rightarrow
&\Z/p^n\Z&\rightarrow&0
\end{array}
\end{equation}
where the first map is $[d]_{p^{m-n}}\mapsto[p^nd]_{p^m}$ and the
second map is
$[e]_{p^m}\mapsto[e]_{p^n}$
\end{lem}
\begin{proof}
Let $[d_1]_{p^{m-n}}$ and $[d_2]_{p^{m-n}}$ represent two distinct
elements
in $\Z/p^{m-n}\Z$. This means that
$$d_1-d_2\nequiv 0\ (Mod\ p^{m-n})$$
Hence we have
$$p^nd_1-p^nd_2\nequiv 0\ (Mod\ p^m)$$
and this proves the injectivity of the map $p^n$ i.e.
exactness at the first place. The surjectivity of the map
$[e]_{p^m}\mapsto[e]_{p^n}$ (hence the exactness at $\Z/p^n\Z$)
is obvious. Hence it remains to show exactness
at $\Z/p^m\Z$. Clearly;
$$[d]_{p^{m-n}}\mapsto[p^nd]_{p^m}\mapsto[p^n]_{p^n}=0$$
and if $[e]_{p^n}=0$ in $\Z/p^n\Z$ then $[e/p^n]_{p^{m-n}}=0$ and this
shows the exactness at $\Z/p^m\Z$ and so the sequence (2) is exact.
\end{proof}
\begin{pro}
\label{P:I.1.6}
The sequence
\begin{equation}
\begin{array}{ccccccccc}
&&&p^n&&\varepsilon_n&&& \\
0&\rightarrow&\Z_p&\rightarrow&\Z_p&\rightarrow&\Z/p^n\Z&\rightarrow&0
\end{array}
\end{equation}
is exact; where $\varepsilon_n$ is the induced projection map.
\end{pro}
\end{document}