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In the following theorems and lemmas we will have:
\begin{itemize}
  \item $\O_K$ a \gls{dedd}
  \item $K = \Frac(\O_K)$
  \item $L / K$ finite separable
  \item $\O_L$ the \gls{intc} of $\O_K$ in $L$
    (which is a \gls{dedd} by \cref{thm_10_2}).
\end{itemize}

\begin{fclemma}[]
  \label{lemma_10_4}
  % Lemma 10.4
  Assuming:
  - $0 \neq x \in \O)K$
  Then:
  \[
    (x) = \prod_{\substack{p \neq 0 \\ \text{prime
    ideal}}} p^{\ddv_p(x)}
  .\]
\end{fclemma}

\begin{proof}
  $x \O_{K, \localp{(p)}} = (p \O_{K,
  (p)})^{\ddv_p(x)}$ by definition of $\ddv_p(x)$.

  Lemma follows from property of \gls{locon}
  \[
    I = J \iff I \O_{K, (p)} = J \O_{K, (p)}
  \]
  for all prime ideals $p$.
\end{proof}

\begin{notation*}
  \glssymboldefn{mover}%
  $P \le \O_L$, $p \le \O_K$ non-zero prime
  ideals. We write $P \mid p$ if $p \O_L =
  P_1^{e_1} \cdots P_r^{e_r}$ and $P \in \{P_1,
  \ldots, P_r\}$ ($e_i > 0$, $P$ distinct).
\end{notation*}

\begin{fcthm}[]
  \label{thm_10_5}
  % Theorem 10.5
  Assuming:
  - $\O_K$, $\O_L$, $K$, $L$ as usual
  - for $p$ a non-zero prime ideal of $\O_K$, we
  write $p\O_L P_1^{e_1} \cdots P_r^{e_r}$
  Then:
  the \glspl{absval} on $L$ extending
  $|\bullet|_p$ (up to equivalence) are precisely
  $|\bullet|_{P_1}, \ldots, |\bullet|_{P_L}$.
\end{fcthm}

\begin{proof}
  By \cref{lemma_10_4} for any $0 \neq x \in \O_K$
  and $i = 1, \ldots, r$ we have $v_{P_i}(x) = e_i
  v_p(x)$. Hence, up to equivalence,
  $|\bullet|_{P_i}$ extends $|\bullet|_p$.

  Now suppose $|\bullet|$ is an \gls{absval} on
  $L$ extending $|\bullet|_p$. Then $|\bullet|$ is
  bounded on $\Zbb$, hence is \gls{nonarch}.
  Let $R = \{x \in L \st |x| \le 1\} \le L$ be the
  \gls{valr} for $L$ with respect to $|\bullet|$.
  Then $\O_K \subseteq R$, and since $R$ is
  \gls{intlyc} in $L$ (\cref{lemma_6_8}), we have
  $\O_L \subseteq R$. Set
  \begin{align*}
    P
    &\defeq \{x \in \O_L \st |x| < 1\} \\
    &= m_R \cap \O_L
  \end{align*}
  (where $m_R$ is the maximal ideal of $R$).

  Hence $P$ a prime ideal in $\O_L$. It is
  non-zero since $p \subseteq P$. Then $\O_{L,
  (p)} \subseteq R$, since $s \in \O_L \setminus P
  \implies |s| = 1$.

  But $\O_{L, (p)}$ is a \gls{dvr}, hence a
  maximal subring of $L$, so $\O_{L, (p)} = R$.
  Hence $|\bullet|$ is \gls{nequ} to
  $|\bullet|_p$. Since $|\bullet|$ extends
  $|\bullet|_p$, $P \cap \O_K = p$ so $P_1^{e_1}
  \cdots P_r^{e_r} \subseteq P$, so $P = P_i$ for
  some $i$.
\end{proof}

Let $K$ be a number field. If $\sigma : K \to
\Rbb, \Cbb$ is a real or complex embedding, then
$x \mapsto |\sigma(x)|_\infty$ defines an
\gls{absval} on $K$ (Example Sheet 2) denoted
$|\bullet|_\sigma$.

\begin{corollary}[]
  \label{coro_10_6}
  % Corollary 10.6
  Let $K$ be a number field with ring of integers
  $\O_K$. Then any \gls{absval} on $K$ is
  \gls{nequ} to either
  \begin{enumerate}[(i)]
    \item $|\bullet|_p$ for some non-zero prime
      ideal of $\O_K$.
    \item $|\bullet|_\sigma$ for some $\sigma : K
      \to \Rbb, \Cbb$.
  \end{enumerate}
\end{corollary}

\begin{proof}
  \textbf{Case 1:} $|\bullet|$ \gls{nonarch}. Then
  $|\bullet| \big|_\Qbb$ is equivalent to
  $|\bullet|_p$ for some prime $p$ by
  \nameref{thm_7_9}. \cref{thm_10_5} gives that
  $|\bullet|$ is \gls{nequ} to $|\bullet|_p$ for
  some $\pideal \subseteq \O_K$ a prime ideal with
  $\pideal \mover p$.

  \textbf{Case 2:} $|\bullet|$ \gls{arch}. See
  Example Sheet 2.
\end{proof}

\subsection{Completions}

$\O_K$ a \gls{dedd}, $L / K$ a finite separable
extension.

Let $\pideal \subseteq \O_K$, $P \subseteq \O_L$
be non-zero prime ideals with $P \mover \pideal$.

\glssymboldefn{completep}%
We write $K_{\pideal}$ and $L_P$ for the
completions of $K$ and $L$ with respect to the
\glspl{absval} $|\bullet|_{\pideal}$ and
$|\bullet|_P$ respectively.

\begin{lemma}
  \label{lemma_10_7}
  % Lemma 10.7
  \phantom{}
  \begin{enumerate}[(i)]
    \item The natural $\pi_P : L \otimes_K
      K_{\pideal} \to L_P$ is surjective.
    \item $[L_P : K_P] \le [L : K]$.
  \end{enumerate}
\end{lemma}

\begin{proof}
  Let $M = LK_{\pideal} = \Im(\pi_P) \subseteq
  L_P$.

  Write $L = K(\alpha)$ then $M =
  K_{\pideal}(\alpha)$. Hence $M$ is a finite
  extension of $K_{\pideal}$ and $[M :
  K_{\pideal}] \le [L : K]$. Moreover $M$ is
  complete (\cref{thm_6_1}) and since $L
  \subseteq M \subseteq L_P$, we have $M = L_P$.
\end{proof}

\begin{fclemma}[Chinese remainder theorem]
  \label{lemma_10_8}
  % Lemma 10.8
  Assuming:
  - $R$ a ring
  - $I_1, \ldots, I_n \subseteq R$ ideals
  - $I_i + I_j = R$ for all $i \neq j$
  Then:
  \begin{enumerate}[(i)]
    \item $\bigcap_{i = 1}^n = \prod_{i = 1}^n
      I_i$ ($=I$ say).
    \item $R / I \cong \prod_{I = 1}^n R / I_i$.
  \end{enumerate}
\end{fclemma}

\begin{proof}
  Example Sheet 2.
\end{proof}

\begin{theorem}[]
  \label{thm_10_9}
  % Theorem 10.9
  The natural map
  \[
    L \otimes_K K\completep\pideal \to \prod_{P \mover
    \pideal} L_P
  \]
  is an isomorphism.
\end{theorem}

\begin{proof}
  Write $L = K(\alpha)$ and let $f(X) \in K[X]$ be
  the minimal polynomial of $\alpha$. Then we have
  \[
    f(X) = f_1(X) \cdots f_r(X) \in K\completep\pideal[X]
  \]
  where $f_i(X) \in K\completep\pideal[X]$ are
  distinct irreducible (separable). Since $L \cong
  K[X] / f(X)$,
  \[
    L \otimes_K K\completep\pideal \cong
    K\completep\pideal[X] / f_i(X) \cong \prod_{i
    = 1}^{r} K\completep\pideal[X] / f_i(X)
  .\]
  Set $L_i = K\completep\pideal[X] / f_i(X)$ a
  finite extension of $K\completep\pideal$.
  Then $L_i$ contains both $K\completep\pideal$
  and $L$ (use $K[X] / f(x) \to
  K\completep\pideal[X] / f_i(X)$ injective since
  morphism of fields). Moreover $L$ is dense
  inside $L_i$ (approximate coefficients of
  $K\completep\pideal[X] / f_i(X)$ with an element
  of $K[X] / f_i(X)$).

  The theorem follows from the following three
  claims:
  \begin{enumerate}[(1)]
    \item $L_i \cong L_P$ for some prime $P$ of
      $\O_L$ dividing $\pideal$.
    \item Each $P$ appears at most once.
    \item Each $P$ appears at least once.
  \end{enumerate}
  Proof of claims:
  \begin{enumerate}[(1)]
    \item Since $[L_i : K\completep\pideal] <
      \infty$, there is a unique \gls{absval} on
      $L_i$ extending $|\bullet|_{\pideal}$.
      \cref{thm_10_5} gives us that $|\bullet|
      \big|_L$ is \gls{nequ} to $|\bullet|_P$ for
      some $P \mover \pideal$. Since $L$ is dense
      in $L$ and $L_i$ is complete, we have $L_i
      \cong L_P$.
    \item Suppose $\varphi : L_i \to L_j$ is an
      isomorphism preserving $L$ and
      $K\completep\pideal$; then
      \[
        \varphi : K\completep\pideal[X] / f_i(X)
        \to K\completep\pideal[X] / f_i(X)
      \]
      takes $x$ to $x$ and hence $f_i = f_i$.
    \item By \cref{lemma_10_7}, the natural map
      $\pi_P : L \otimes_K K\completep\pideal
      \to L_P$ is surjective for any prime $P
      \mover \pideal$.

      Since $L_P$ is a field, $\pi_P$ factors
      through $L_i$ for some $i$, and hence $L_i
      \cong L_P$ by surjectivity of $\pi_P$.
      \qedhere
  \end{enumerate}
\end{proof}