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% 31/10/2024 12PM

\begin{fcthm}[]
  % Theorem 8.5
  Assuming:
  - $K$ is a \gls{lf}
  Then:
  $K$ is the completion of a \gls{gf}.
\end{fcthm}

\begin{proof}
  \textbf{Case 1:} $\absval|\bullet|$ is
  \gls{arch}. Then $\Rbb$ is the completion of
  $\Qbb$, and $\Cbb$ is the completion of
  $\Qbb(i)$ (with respect to
  $\absval|\bullet|_\infty$).

  \textbf{Case 2:} $\absval|\bullet|$
  \gls{nonarch}, \gls{eqc}. Then $K \cong
  \Fbb_q((t))$ is the completion of $\Fbb_q(t)$
  with respect to the $t$-adic valuation.

  \textbf{Case 3:} $\absval|\bullet|$
  \gls{nonarch} \gls{mixc}. Then $K =
  \Qp(\alpha)$, with $\alpha$ a root of a monic
  irreducible polynomial $f(X) \in \Zp[X]$. Since
  $\Zbb$ is dense in $\Zp$, we choose $g(X) \in
  \Zbb[X]$ as in \cref{prop_8_4}. Then $K =
  \Qbb(\beta)$ with $\beta$ a root of $g(X)$.
  Since $\Qbb(\beta)$ dense in $\Qp(\beta) = K$,
  and $K$ is complete, we must have that $K$ is
  the completion of $\Qbb(\beta)$.
\end{proof}

\newpage
\part{Dedekind domains}

\newpage
\section{Dedekind domains}

\begin{fcdefn}[Dedekind domain]
  \glsnoundefn{dedd}{Dedekind domain}{Dedekind
  domains}%
  % Definition 9.1
  A \emph{Dedekind domain} is a ring $R$ such that
  \begin{enumerate}[(i)]
    \item $R$ is a Noetherian integral domain.
    \item $R$ is \gls{intlyc} in $\Frac(R)$.
    \item Every non-zero prime ideal is maximal.
  \end{enumerate}
\end{fcdefn}

\begin{example*}
  \phantom{}
  \begin{itemize}
    \item The ring of integers in a number field
      is a \gls{dedd}.
    \item Any PID (hence a \gls{dvr}) is a
      \gls{dedd}.
  \end{itemize}
\end{example*}

\begin{theorem}[]
  \label{thm_9_2}
  % utheorem 9.2
  A ring $R$ is a \gls{dvr} if and only if $R$ is
  a \gls{dedd} with exactly one non-zero prime.
\end{theorem}

\begin{fclemma}[]
  \label{lemma_9_3}
  % Lemma 9.3
  Assuming:
  - $R$ is a Noetherian ring
  - $I \subseteq R$ a non-zero ideal
  Then:
  there exists non-zero
  prime ideals $p_1, \ldots, p_r$ such that $p_1,
  \ldots, p_r \subseteq I$.
\end{fclemma}

\begin{proof}
  Suppose not. Since $R$ is Noetherian, we may
  choose $I$ maximal with this property. Then $I$
  is not prime, so there exists $x, y \in R
  \setminus I$ such that $x, y \in I$.

  Let $I_1 + (x)$, $I_2 = I + (y)$. Then by
  maximality of $I$, there exist $p_1, \ldots,
  p_r$ and $q_1, \ldots, q_s$ such that $p_1
  \cdots p_r \subseteq I_1$ and $q_1 \cdots q_s
  \subseteq I_2$. Then $p_1 \cdots p_r q_1 \cdots
  q_s \subseteq I_1 I_2 \subset I$.
\end{proof}

\begin{fclemma}[]
  \label{lemma_9_4}
  % Lemma 9.4
  Assuming:
  - $R$ is an integral domain
  - $R$ is \gls{intlyc} in $K = \Frac(R)$
  - $0 \neq I \subseteq R$ a finitely generated
  ideal
  - $x \in K$
  Then:
  if $xI \subseteq I$, we have $x \in R$.
\end{fclemma}

\begin{proof}
  Let $I = (c_1, \ldots, c_n)$. We write
  \[
    x c_i = \sum_{j = 1}^{n} a_{ij} c_j
  \]
  for some $a_{ij} \in R$. Let $A$ be the matrix $A
  = (a_{ij})_{1 \le i, j \le n}$ and set $B =
  x\id_n - A \in M_{n \times n}(K)$.

  Then in $K^n$
  \[
    B
    \begin{pmatrix}
      c_1 \\
      \vdots \\
      c_n
    \end{pmatrix}
    = \mathbf{0}
  .\]
  Multiply by $\adj(B)$, the adjugate matrix for $B$. We
  have
  \[
    \det(B) \id_n
    \begin{pmatrix}
      c_1 \\
      \vdots \\
      c_n
    \end{pmatrix}
    = 0
  .\]
  Hence $\det(B) = 0$. But $\det B$ is a monic
  polynomial with coefficients in $R$. Then $x$ is
  \gls{int} over $R$, hence $x \in R$.
\end{proof}

\begin{proof}[Proof of \cref{thm_9_2}]
  \phantom{} \\[-2\baselineskip]
  \begin{enumerate}[$\Rightarrow$]
    \item[$\Rightarrow$] Clear.
    \item[$\Leftarrow$] We need to show $R$ is a
      PID. The assumption implies that $R$ is a
      local ring with unique maximal ideal $m$.

      \textbf{Step 1:} $m$ is principal.

      Let $0 \neq x \in
      m$. By \cref{lemma_9_3}, $(x) \supseteq m^n$
      for some $n \ge 1$. Let $n$ minimal such
      that $(x) \supset m^n$, then we may choose
      $y \in m^{n - 1} \setminus (x)$.

      Set $\pi =
      \frac{x}{y}$. Then we have $ym \subseteq m^n
      \subseteq (x)$ and hence $\pi^{-1}m
      \subseteq R$. If $\pi^{-1} m \subseteq m$,
      then $\pi^{-1} \in R$ by \cref{lemma_9_4}
      and $y \in (x)$, contradiction. Hence
      $\pi^{-1} m = R$, so $m = \pi R$ is
      principal.

      \textbf{Step 2:} $R$ is a PID.

      Let $I \subseteq R$ be a non-zero ideal.
      Consider a sequence of fractional ideals $I
      \subseteq \pi^{-1} I \subseteq \pi^{-2} I
      \subseteq \cdots$ in $K$. Then since
      $\pi^{-1} \notin R$, we have $\pi^{-k} I
      \neq \pi^{-(k + 1)} I$ for all $k$ by
      \cref{lemma_9_4}. Therefore since $R$ is
      Noetherian, we may choose $n$ maximal such
      that $\pi^{-n} I \subseteq R$. If $\pi^{-n}
      I \subseteq m = (\pi)$, then $\pi^{-(n + 1)}
      \subseteq R$. So we must have $\pi^{-n} I =
      R$, and hence $I = (\pi^n)$. \qedhere
  \end{enumerate}
\end{proof}

\glssymboldefn{local}%
\glsnoundefn{locon}{localisation}{NA}%
\glsnoundefn{locise}{localise}{NA}%
Let $R$ be an integral domain and $S \subseteq R$
a multiplicatively closed subset ($x, y \in S$
implies $xy \in S$, and also have $1 \in S$). The
localisation $S^{-1} R$ of $R$ with respect to
$S$ is the ring
\[
  S^{-1} R = \left\{ \frac{r}{s} \bigg| r \in R, s
  \in S\right\} \subseteq \Frac(R)
.\]
If $p$ is a prime ideal in $R$, we write $R_{(p)}$
for the localisation with respect to $S = R
\setminus p$.

\begin{example*}
  \phantom{}
  \begin{itemize}
    \item $p = (0)$, then $R\localp{(p)} = \Frac(R)$.
    \item $R = \Zbb$, $\Zbb\localp{(p)} = \left\{
      \frac{a}{b} \big| a \in \Zbb, (b, p) = 1
      \right\}$, where $p$ is a rational prime.
  \end{itemize}
\end{example*}

\textbf{Facts:} (not proved in this course, but
can be found in a typical course / textbook on
commutative algebra)
\begin{itemize}
  \item $R$ Noetherian implies $S\local R$ is
    Noetherian.
  \item \phantom{} \begin{center}
      \includegraphics[width=0.6\linewidth]{images/c87300de5b8a4e6b.png}
    \end{center}
\end{itemize}

\begin{corollary}
  \label{coro_9_5}
  % Corollary 9.5
  Let $R$ be a \gls{dedd} and $p \subseteq R$ a
  non-zero prime ideal. Then $R\localp{(p)}$ is a
  \gls{dvr}.
\end{corollary}

\begin{proof}
  By properties of \gls{locon}, $R\localp{(p)}$ is
  a Noetherian integral domain with a unique
  non-zero prime ideal $pR\localp{(p)}$.

  It suffices to show $R\localp{(p)}$ is
  \gls{intlyc} in $\Frac(R\localp{(p)}) =
  \Frac(R)$ (since then $R\localp{(p)}$ is a
  \gls{dedd} hence by \cref{thm_9_2},
  $R\localp{(p)}$ is a \gls{dvr}).

  Let $x \in \Frac(R)$ be \gls{int} over
  $R\localp{(p)}$. Multiplying by denominators of
  a monic polynomial satisfied by $x$, we obtain
  \[
    sx^n + a_{n - 1} x^{n - 1} + \cdots + a_0 = 0
  ,\]
  with $a_i \in R$, $s \in S = R \setminus p$.
  Multiply by $s^{n - 1}$. Then $xs$ is integral
  over $R$, so $xs \in R$. Hence $x \in
  R\localp{(p)}$.
\end{proof}