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\begin{document}
\title{Lecture 17?}
\maketitle

\begin{pro}
\label{7.2}
 Let $(K,\nu)$ be a complete, valued field with respect to a discrete
valuation $\nu$. Let
 $(L, \omega)$ be a finite extension.  Then $\ok{L}=\ok{\omega}$
 is the integral closure of $\ok{K}$ in $L$.
\end{pro}
\begin{proof}
Let $\tilde{O}_{K}$ be the integral closure of $\ok{K}$ in $L$. We
first show that $\tilde{O}_{K} \subseteq \ok{L}$.

Let $x \in \tilde{O_{K}}.$  Then $x$ satisfies an equation of
integral dependence $$x^{n} + a_{1}x^{n-1} + \cdots + a_{n} = 0,
\quad a_{i} \in \ok{K}.$$ Suppose $x \notin \ok{L}$.  Then $\omega
(x) = -k < 0$, for some $k$.  Then $$ \omega (x^{n} + a_{1}x^{n-1}
+ \cdots + a_{n}) = \max_{i}\{ \omega (x^{n}), \omega(
a_{i}x^{n-i})\} = \omega (x^{n}) = -nk,$$ as $ \omega (a_{i}) \le
0,$ for every $i$. But  $$ \omega (x^{n} + a_{1}x^{n-1} + \cdots +
a_{n}) = \omega (0) = +\infty,$$ a contradiction. Therefore, $x
\in \ok{L}$.

We now show that $\ok{L} \subseteq \tilde{O}_{K}$. Let $x \in
\ok{L}$, we show that $\ok{K}[x]$ is  finitely generated. Clearly,
$\ok{K}[x] \subseteq \ok{L}.$  And, as seen previously, $\ok{L}
\cong (\ok{K})^{n}$, where $ [\ok{L}:\ok{K}]=n.$ Therefore, as
$\ok{K}$ is Noetherian, $\ok{L} \cong(\ok{K})^{n}$ is Noetherian.
But $\ok{L}$ Noetherian implies that every sub-module is finitely
generated.  In particular, $\ok{K}[x]$ is finitely generated,
implying that $x$ is integral over $\ok{K}$ and thus
$x\in\tilde{O}_{K}.$
\end{proof}
\bigskip

\begin{lem}
\label{7.1}
A unique factorization domain (UFD) is integrally
closed in its fraction field.
\end{lem}

\begin{proof}
Let $R$ be a UFD and $K$ its field of fractions.  Let $x \in K.$
Then $x$ satisfies an equation of integral dependence $$x^{n} +
a_{1}x^{n-1} + \cdots + a_{n} = 0, \quad a_{i} \in R.$$  As $R$ is
a UFD, we can write $x=\frac{p}{q}$, where $p,q \in K$ and
$(p,q)=1$.  Substituting, we have $$\left(\frac{p}{q}\right)^{n} +
a_{1}\left(\frac{p}{q}\right)^{n-1} + \cdots + a_{n} = 0,$$ and
multiplying by $q^{n},$ $$p^{n} + a_{1}p^{n-1}q + \cdots +
a_{n}q^{n} = 0.$$ Thus, solving for $p^{n}$, $$
p^{n}=q(-a_{1}p^{n-1} - \cdots - a_{n}q^{n-1}),$$ we have that
$p|q$, a contradiction to $(p,q)=1$. Hence, $R$ is integrally
closed in its field of fractions.
\end{proof}
\bigskip

\section{Absolute Values of Algebraic Number Fields}
\label{8}

\begin{thm}
\label{8.1} Let $K$ be an algebraic number field  and $\nu$ a
non-trivial absolute value on $K$.Then
\begin{itemize}
    \item [$(i)$] If $\nu$ is archimedean, then it is equivalent
    to either a real or a complex embedding.
    \item [$(ii)$] If $\nu$ is non-archimedean, then $\nu$ is
    equivalent to the $\mfr{p}$-adic valuation of a prime ideal
    $\mfr{p} \subseteq \ok{K}$. Furthermore, every such prime
    defines a non-archimedean valuation of $K$.
\end{itemize}
\end{thm}

\begin{proof}
Let $\hat{K}_{\nu}$ be the completion of $K$ with respect to
$\nu$.
 $(i)$  As $\nu$ is archimedean, Theorem 5.2 implies that
    $\hat{K}_{\nu}= \R$ or $\C$.  But restricting the extension,
    we get the usual absolute value on $\R$ or $\C$ respectively.
    Thus, $\nu$ is equivalent to a real or complex embedding.

    $(ii)$ If $\nu$ is non-archimedean, define $$\mfr{p}_{\nu} =
    \{x\in \ok{K}: |x|_{\nu} < 1 \}. $$  It is easily verified
    that this is a prime ideal of $\ok{K}$.  Now, by Ostrowski's
    theorem $$ \nu|_{\Q} \mbox{ is non-archimedean } \iff
    \nu|_{\Q} = \nu_{p}, \mbox{ the p-adic valuation }.$$ Thus,
    $\nu$ is an extension of the p-adic valuation to $K$, that is,
    a $\mfr{p}$-adic valuation, proving the result.
\end{proof}

\begin{lem}
Let $K$ be an algebraic number field and $\nu$ a non-archimedean
valuation of $K$.  Then $$x\in \ok{K} \implies |x|_{\nu} \le 1.$$
\end{lem}

\begin{proof}
As $x \in \ok{K}$, $x$ satisfies an equation of integral
dependence $$x^{n} + a_{1}x^{n-1} + \cdots + a_{n} = 0, \quad
a_{i} \in \Z.$$ If $\nu (x) <0$, say $\nu (x) =-k$, then
$$\nu(x^{n} + a_{1}x^{n-1} + \cdots + a_{n})= \max_{i}\{ \nu
(x^{n}), \nu( a_{i}x^{n-i})\} = \nu (x^{n}) = -nk.$$  Then
$$-nk=\nu(x^{n} + a_{1}x^{n-1} + \cdots + a_{n})=\nu (0) =
+\infty,$$ a contradiction.  Therefore $\nu (x) \ge 0$.  But $\nu
(x) \ge 0 \iff |x|_{\nu} \le 1$, which gives the result.
\end{proof}

\end{document}