%! TEX root = LF.tex
% vim: tw=50
% 12/11/2024 12PM

\begin{remark*}
  $\diff LK \le \O_L$ since $\O_L \subseteq \invdiff LK$.
\end{remark*}

\glssymboldefn{ig}%
Let $I_L$, $I_K$ be the groups of fractional ideals.

\cref{thm_9_7} gives that
\[
  I_L \cong \bigotimes_{\substack{0 \neq P \\
  \text{prime ideals in $\O_L$}}} \Zbb, \qquad
  I_K \cong \bigotimes_{\substack{0 \neq P \\
  \text{prime ideals in $\O_K$}}}
.\]
\glssymboldefn{idN}%
Define $N_{L / K} : I_L \to I_K$ induced by $P
\mapsto \pideal^f$ for $\pideal = P \cap \O_K$ and
$f = f(P / \pideal)$.

\textbf{Fact:}
\begin{picmath}
  \begin{tikzcd}
    L^\times \ar[r] \ar[d, "\idN_{L / K}"]
    & \ig_L \ar[d, "\idN_{L / K}"] \\
    K^\times \ar[r]
    & \ig_K
  \end{tikzcd}
\end{picmath}
(Use \cref{coro_10_10} and
$\ddv_\pideal(N_{L\completep P /
K\completep\pideal}(x)) = f_{P / \pideal}
\ddv_(x)$ for $x \in L\completep P^{\times}$ where
$\ddv_\pideal$ and $\ddv_P$ are the normalised
\glspl{valt} for $L\completep P$, $K\completep
\pideal$).

\begin{theorem}
  \label{thm_12_7}
  % Theorem 12.7
  $\idN_{L / K}(\diff LK) = \discrim_{L / K}$.
\end{theorem}

\begin{proof}
  First assume $\O_K$, $\O_L$ are PIDs. Let $x_1,
  \ldots, x_n$ be an $\O_K$-basis for $\O_L$ and
  $y_1, \ldots, y_n$ be the dual basis with
  respect to trace form. Then $y_1, \ldots, y_n$
  is a basis for $\invdiff LK$. Let $\sigma_1,
  \ldots, \sigma_n : L \to \ol{K}$ be the distinct
  embeddings. Have
  \[
    \sum_{i = 1}^{n} \sigma_i(x_j) \sigma_i(y_k)
    = \Tr(x_j y_k)
    = \delta_{jk}
  .\]
  But
  \[
    \Disc(x_1, \ldots, x_n) =
    \det(\sigma_i(x_j))^2
  .\]
  Thus
  \[
    \Disc(x_1, \ldots, x_n) \Disc(y_1, \ldots,
    y_n) = 1
  .\]
  Write $\invdiff LK = \beta \O_L$ since $\beta
  \in L$. Then
  \begin{align*}
    \discrim_{L / K}^{-1}
    &= (\Disc(x_1, \ldots, x_n)^{-1}) \\
    &= (\Disc(y_1, \ldots, y_n)) \\
    &= (\Disc(\beta x_1, \ldots, \beta x_n))
    &&\text{change of basis matrix is invertible
    in $\O_K$} \\
    &= N_{L / K}(\beta^2)
    \Disc(x_1, \ldots, x_n)
    &&\text{change of basis matrix is
    $[\mult(\beta)]$}
  \end{align*}
  Thus
  \[
    \discrim_{L / K}^{-1} = N_{L / K}(\invdiff
    LK)^2 \discrim_{L / K}
  \]
  so
  \[
    N_{L / K}(\diff LK) = \discrim_{L / K}
  .\]
  In general, \gls{locise} at $S = \O_K \setminus
  \pideal$ and use $S^{-1} \diff LK =
  \diff{S\local \O_L}{S\local \O_K}$. Then
  $S\local \discrim_{L / K} = \discrim_{S\local
  \O_L / S\local \O_K}$. Details omitted.
\end{proof}

\begin{fcthm}[]
  \label{thm_12_8}
  % Theorem 12.8
  Assuming:
  - $\O_L = \O_K[\alpha]$
  - $\alpha$ has monic minimal polynomial $g(X) \in
  \O_K[X]$
  Then:
  $\diff LK = (g'(\alpha))$.
\end{fcthm}

\begin{proof}
  Let $\alpha = \alpha_1, \ldots, \alpha_n$ be the
  roots of $g$. Write
  \[
    \frac{g(X)}{X - \alpha} = \beta_{n - 1} X^{n -
    1} + \cdots + \beta_1 X + \beta_0
  \]
  with $\beta_i \in \O_L$ and $\beta_{n - 1} = 1$.
  We claim
  \[
    \sum_{i = 1}^{n} \frac{g(X)}{X - \alpha_i}
    \frac{\alpha_i^n}{g'(\alpha_i)} = X^r
  \]
  for $0 \le r \le n - 1$.

  Indeed the difference is a palynomial of degree
  $< n$, which vanishes for $X = \alpha_1, \ldots,
  \alpha_n$. Equate coefficients of $X^s$, which
  gives
  \[
    \Tr_{L / K} \left( \frac{\alpha^r
    \beta_s}{g'(\alpha)} \right)
    = \delta_{rs}
  .\]
  Since $1, \alpha, \ldots, \alpha^{n - 1}$ is
  an $\O_K$ basis for $\O_L$, $\invdiff LK$ has an
  $\O_K$ basis
  \[
    \frac{\beta_0}{g'(\alpha)},
    \frac{\beta_1}{g'(\alpha)}, \ldots,
    \ub{\frac{\beta_{n - 1}}{g'(\alpha)}}_
    {\frac{1}{g'(\alpha)}}
  .\]
  Note all of these are $\O_L$ multiples of the
  last term, since the $\beta_i$ are in $\O_L$.
  So $\invdiff LK = \frac{1}{(g'(\alpha))}$, hence
  $\diff LK = (g'(\alpha))$.
\end{proof}

$P$ a prime ideal of $\O_L$, $\pideal = \O_K \cap
P$. $\diff{L\completep P}{K \completep\pideal}$
using $\O_{K\completep \pideal}$, $\O_{L\completep P}$. We identify
$\diff{L\completep P}{K \completep\pideal}$ with a
power $\mathcal{P}$.

\begin{theorem}
  \label{thm_12_9}
  % Theorem 12.9
  $\diff LK = \prod_{\mathcal{P}}
  \diff{L\completep
  P}{K\completep\pideal}$ (finite product, see
  later).
\end{theorem}

\begin{proof}
  Let $x \in L$, $\pideal \subseteq \O_K$. Then
  \[
    \Tr{L / K}(x) = \sum_{\mathcal{P} \mover
    \pideal} \Tr_{L\completep P /
    K\completep\pideal} (x)
    \tag{$*$}
    \label{lec15_eq}
  \]
  (of \cref{coro_10_10}).

  Let $r(\mathcal{P}) = \ddv_{\mathcal{P}}(\diff
  LK)$, $s(\mathcal{P}) = \ddv_{\mathcal{P}}(\diff
  {L\completep P}{K\completep\pideal})$.
  \begin{enumerate}[$\subseteq$]
    \item[$\subseteq$] (i.e. $r(\mathcal{P}) \ge
      s(\mathcal{P})$). Let $x \in L$ with
      $\ddv_{\mathcal{P}}(x) \ge -s(\mathcal{P})$
      for all $\mathcal{P}$. Then
      $\Tr_{L\completep P /
      K\completep\pideal}(xy) \in
      \O_{K\completep\pideal}$, for all $y \in L$
      and for all $\mathcal{P}$.
      Using \eqref{lec15_eq} we get
      \[
        \Tr_{L / K}(xy) \in
        \O_{K\completep\pideal} \qquad \forall y
        \in \O_L, \forall \mathcal{P}
      .\]
      Thus
      \[
        \Tr_{L / K}(xy) \in \O_K \qquad \forall y
        \in \O_L
      \]
      so $\diff LK \subseteq \prod_{\mathcal{P}}
      \diff{L\completep {\mathcal{P}}}{K\completep
      \pideal}$.
    \item[$\supseteq$] (i.e. $r(\mathcal{P}) \le
      s(\mathcal{P})$). Fix $\mathcal{P}$ and let
      $x \in P^{-r(P)} \setminus P^{-r(P)+1}$.
      Then $\ddv_P(x) = -r(P)$, $\ddv_{P'}(x) \ge
      0$ for all $P' \neq P$. By \eqref{lec15_eq},
      we have
      \[
        \Tr_{L\completep P /
        K\completep\pideal}(xy)
        = \Tr_{L / K}(xy)
        - \sum_{\substack{P' \mover \pideal \\ P'
        \neq P}} \Tr_{L\completep P / K\completep
        \pideal}(xy) \qquad \forall y \in \O_L
      \]
      hence
      \[
        \Tr_{L\completep P / K\completep
        \pideal}(xy) \in \O_{K\completep\pideal}
        \qquad \forall y \in \O_{L \completep P}
      .\]
      Hence $x \in \invdiff{L\completep
      P}{K\completep\pideal}$, i.e. $-\ddv_P(x) =
      r(P) \le s(P)$. So $\diff LK \supseteq
      \prod_P \diff{L\completep P}{K\completep
      \pideal}$. \qedhere
  \end{enumerate}
\end{proof}

\begin{corollary}
  \label{coro_12_10}
  % Corollary 12.10
  $\discrim_{L / K} = \prod_{P \mover \pideal}
  \discrim_{L\completep P / K\completep\pideal}$.
\end{corollary}

\begin{proof}
  Apply $\idN_{L / K}$ to $\diff LK = \prod_{P
  \mover \pideal} \diff{L\completep
  P}{K\completep\pideal}$.
\end{proof}

\newpage
\section{Unramified and totally ramified
extensions of local fields}

Let $L / K$ be a finite separable extension of
\gls{nonarch} \glspl{lf}. \cref{coro_11_6} implies
\[
  [L : K] = e_{L / K} f_{L / K}
  \label{lec15_eq2}
  \tag{$*$}
.\]

\begin{fclemma}
  \label{lemma_13_1}
  % Lemma 13.1
  Assuming:
  - $M / L / K$ finite separable extensions of
  \glspl{lf}
  Then:
  \begin{enumerate}[(i)]
    \item $f_{M / K} = f_{L / K} f_{M / L}$
    \item $e_{M / K} = e_{L / K} f_{M / L}$
  \end{enumerate}
\end{fclemma}

\begin{proof}
  \phantom{}
  \begin{enumerate}[(i)]
    \item $f_{M / K} = [k_M : k] = [k_M : k_L][k_L
      : k] = f_{M / L} f_{L / K}$.
    \item (i) and \eqref{lec15_eq2}. \qedhere
  \end{enumerate}
\end{proof}

\begin{fcdefn}[Unramified / ramified / totally
  ramified]
  \glsadjdefn{unramext}{unramified}{extension}%
  \glsadjdefn{ramext}{ramified}{extension}%
  \glsadjdefn{totramext}{totally
  ramified}{extension}%
  \label{defn_13_2}
  % Definition 13.2
  The extension $L / K$ is said to be:
  \begin{itemize}
    \item \emph{unramified} if $e_{L / K} = 1$
      (equivalently $f_{L / K} = [L : K]$).
    \item \emph{ramified} if $e_{L / K} > 1$
      (equivalently $f_{L / K} < [L : K]$).
    \item \emph{totally ramified} if $e_{L / K} =
      [L : K]$
      (equivalently $f_{L / K} = 1$).
  \end{itemize}
\end{fcdefn}