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\begin{fcdefn}[Valuation on a Dedekind domain]
  \label{defn_9_6}
  \glssymboldefn{ddv}%
  % Definition 9.6
  If $R$ is a \gls{dedd}, and $p \subseteq R$ a
  non-zero prime ideal, we write $v_p$ for the
  normalised \gls{valt} on $\Frac(R) =
  \Frac(R\localp{(p)})$ corresponding to the
  \gls{dvr} $R\localp{(p)}$.
\end{fcdefn}

\begin{example*}
  $R = \Zbb$, $p = (p)$, then $v_p$ is the
  $p$-adiv \gls{valt}.
\end{example*}

\begin{fcthm}[]
  \label{thm_9_7}
  % Theorem 9.7
  Assuming:
  - $R$ is a \gls{dedd}
  - $I \subseteq R$ a non-zero ideal
  Then:
  $I$can be written uniquely as  aproduct of
  prime ideals:
  \[
    I = p_1^{e_1} \cdots p_r^{e_r}
  \]
  (with $p_i$ distinct).
\end{fcthm}

\begin{remark*}
  Clear for PIDs (PID implies UFD).
\end{remark*}

\begin{proof}[Proof (Sketch)]
  We quote the following properties of
  \gls{locon}:
  \begin{enumerate}[(i)]
    \item $I = J \iff I R\localp{(p)} =
      JR\localp{(p)}$ for all prime ideals $p$.
    \item If $R$ a \gls{dedd}, $p_1, p_2$ non-zero
      ideals, then
      \[
        p_1 R\localp{(p_2)} = \begin{cases}
          p_2 R\localp{(p_2)} & p_1 = p_2 \\
          R\localp{(p_2)} & p_1 \neq p_2
        \end{cases}
      \]
  \end{enumerate}
  Let $I \subseteq R$ be a non-zero ideal. By
  \cref{lemma_9_3}, there are distinct prime
  ideals $p_1, \ldots, p_r$ such that
  $p_1^{\beta_1} \cdots p_r^{\beta_r} \subseteq
  I$, where $\beta_i > 0$.

  Let $0 \neq p$ be a prime ideal, $p \notin
  \{p_1, \ldots, p_r\}$. Then property (ii) gives
  that $p_i R\localp{(p)} = R\localp{(p)}$, and
  hence $I R\localp{(p)} = R\localp{(p)}$.

  \cref{coro_9_5} gives $IR\localp{(p_)} = (p_i
  R\localp{(p_i)})^{\alpha_i} = p_i^{\alpha_i}
  R\localp{(p_i)}$ for some $0 \le \alpha_i \le
  \beta_i$. Thus $I = p_1^{\alpha_1} \cdots
  p_r^{\alpha_r}$ by property (i).

  For uniqueness, if $I = p_1^{\alpha_1} \cdots
  p_r^{\alpha_r} = p_1^{\gamma_1} \cdots
  p_r^{\gamma_r}$ then $p_i^{\alpha_i}
  R\localp{(p_i)} = p_i^{\gamma_i}
  R\localp{(p_i)}$ hence $a_i = \gamma_i$ by
  unique factorisation in \glspl{dvr}.
\end{proof}

\newpage
\section{Dedekind domains and extensions}

Let $L / K$ be a finite extension. For $x \in L$,
we write $\Tr_{L / K}(x) \in K$ for the trace of
the $K$-linear map $L \to L$, $y \mapsto xy$.

If $L / K$ is separable of degree $n$ and
$\sigma_1, \ldots, \sigma_n : L \to \ol{K}$denotes
the set of embeddings of $L$ into an algebraic
closure $\ol{K}$, then $\Tr_{L / K}(x) = \sum_{i =
1}^{n} \sigma_i(x) \in K$.

\begin{fclemma}[]
  % Lemma 10.1
  Assuming:
  - $L / K$ a finite separable extension of fields
  Then:
  the symmetric bilinear pairing
  \begin{align*}
    (\bullet, \bullet)
    &\to K \\
    (x, y)
    &\mapsto \Tr_{L / K}(xy)
  \end{align*}
  is non-degenerate.
\end{fclemma}

\begin{proof}
  $L / K$ separable tells us that $L = K(\alpha)$
  for some $\alpha \in L$. Consider the matrix $A$
  for $(\bullet, \bullet)$ in the $K$-basis for
  $L$ given by $1, \alpha, \ldots \alpha^{n - 1}$.

  Then $A_{ij} = \Tr_{L / K}(\alpha^{i + j}) =
  [BB^\top]_{ij}$ where
  \[
    B = \begin{pmatrix}
      1 & 1 & \cdots & 1 \\
      \sigma_1(\alpha) & \sigma_2(\alpha) & \cdots
      \sigma_n(\alpha) \\
      \vdots & \vdots & \ddots & \vdots \\
      \sigma_1(\alpha^{n - 1}) &
      \sigma_2(\alpha^{n - 1}) & \cdots &
      \sigma_n(\alpha^{n - 1})
    \end{pmatrix}
  \]
  So
  \[
    \det A = \det(B)^2 = \left[ \prod_{1 \le i < j
    \le n} (\sigma_i(\alpha) - \sigma_j(\alpha))
    \right]^2
  \]
  (Vandermonde determinant), which is non-zero
  since $\sigma_i(\alpha) \neq \sigma_j(\alpha)$
  for $i \neq j$ by separability.
\end{proof}

\textbf{Exercise:} On Example Sheet 3 we will show
that a finite extension $L / K$ is separable if
and only if the trace form is non-degenerate.

\begin{fcthm}[]
  \label{thm_10_2}
  % Theorem 10.2
  Assuming:
  - $\O_K$ a \gls{dedd}
  - $L$ a finite separable extension of $K =
  \Frac(\O_K)$
  Then:
  the \gls{intc} $\O_L$ of $\O_K$ in $L$ is a
  \gls{dedd}.
\end{fcthm}

\begin{proof}
  $\O_L$ a subring of $L$, hence $\O_L$ is an
  integral domain.

  Need to show:
  \begin{enumerate}[(i)]
    \item $\O_L$ is Noetherian.
    \item $\O_L$ is \gls{intlyc} in $L$.
    \item Every $\neq 0$ prime ideal $P$ in $\O_L$
      is maximal.
  \end{enumerate}
  Proofs:
  \begin{enumerate}[(i)]
    \item Let $e_1, \ldots, e_n \in L$ be a
      $K$-basis for $L$. Upon scaling by $K$, we
      may assume $e_i \in \O_L$ for all $i$.

      Let $f_i \in L$ be the dual basis with
      respect to the trace form $(\bullet,
      \bullet)$. Let $x \in \O_L$, and write $x =
      \sum_{i = 1}^n \lambda_i f_i, \lambda_i \in
      K$. Then $\lambda_i = \Tr_{L / K}(x e_i) \in
      \O_K$.

      (For any $z \in \O_L$, $\Tr_{L /
      K}(z)$ is a sum of elements in $\ol{K}$
      which are \gls{int} over $\O_K$. Hence
      $\Tr_{L / K}(z) \in K$ is \gls{int} over
      $\O_K$, hence $\Tr_{L / K}(z) \in \O_K$.)

      Thus $\O_L \subseteq \O_K f_1 + \cdots +
      \O_K f_n \subseteq L$. Since $\O_K$ is
      Noetherian, $\O_L$ is finitely generated as
      an $\O_K$-module, hence $\O_L$ is
      Noetherian.
    \item Example Sheet 2.
    \item Let $P$ be a non-zero prime ideal of
      $\O_L$, and $p \defeq P \cap \O_K$ be a
      prime ideal of $\O_K$. Let $0 \neq x \in P$.
      Then $x$ satisfies an equation
      \[
        x^n + a_{n - 1} x^{n - 1} + \cdots + a_0 =
        0, \qquad a_i \in \O_K
      ,\]
      with $a_0 \neq 0$. Then $a_0 \in P \cap
      \O_K$ is a non-zero element of $p$, hence
      $p$ is non-zero, hence $p$ is maximal.

      We have $\O_K / p \injto \O_L / P$, and
      $\O_L / P$ is a finite dimensional vector
      space over $\O_K / p$. Since $\O_L / P$ is
      an integral domain and finite, it is a field. \qedhere
  \end{enumerate}
\end{proof}

\begin{remark*}
  \cref{thm_10_2} holds without the assumption
  that $L / K$ is separable.
\end{remark*}

\begin{corollary}
  % Corollary 10.3
  The ring of integers of a number field is a
  \gls{dedd}.
\end{corollary}

\textbf{Convention:} $\O_K$ is the ring of
integers of a number field -- $p \le \O_K$ a
non-zero prime ideal.
We normalise $|\bullet|_p$ (absolute value
associated to $\ddv_p$, as defined in
\cref{defn_9_6}) by $|x|_p = (N p)^{-\ddv_p(x)}$,
where $N_p = |\O_K / p|$.
