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% 09/11/2024 12PM

\begin{fcprop}[]
  \label{prop_11_8}
  % Proposition 11.8
  Assuming:
  - $\O_K$ a \gls{dedd}
  - $L / K$ a finite Galois extension
  - $0 \neq P \subseteq \O_L$ a prime ideal
  - $P \mover \pideal \subseteq \O_K$
  Then:
  \begin{enumerate}[(i)]
    \item $L\completep P / K\completep \pideal$ is
      Galois.
    \item There is a natural map
      \[
        \res : \Gal(L\completep P /
        K\completep\pideal) \to \Gal(L / K)
      \]
      which is injective and has image $\GP$.
  \end{enumerate}
\end{fcprop}

\begin{proof}
  \phantom{}
  \begin{enumerate}[(i)]
    \item $L / K$ Galois implies that $L$ is a
      splitting field of a separable polynomial
      $f(X) \in K[X]$. Hence $L\completep P$ is
      the splitting field of $f(X) \in K\completep
      [X]$, hence $L\completep P /
      K\completep\pideal$ is Galois.
    \item Let $\sigma \in \Gal(L\completep P /
      K\completep\pideal)$, then $\sigma(L) = L$
      since $L / K$ is normal, hence we have a map
      $\res : \Gal(L\completep P / K\completep
      \pideal) \to \Gal(L / K)$, $\sigma \mapsto
      \sigma|_L$. Since $L$ is dense in
      $L\completep P$, $\res$ is injective. By
      \cref{lemma_8_2}, we have
      \[
        |\sigma(x)|_P = |x|_P
      \]
      for all $\sigma \in \Gal(L\completep P /
      K\completep \pideal)$ and $x \in L\completep
      P$. Hence $\sigma(P) = P$ for all $\sigma
      \in \Gal(L\completep P /
      K\completep\pideal)$ and hence $\res(\sigma)
      \in \GP$ for all $\sigma \in
      \Gal(L\completep P / K\completep \pideal)$.

      To show surjectivity, it suffices to show
      that
      \[
        |\GP| = ef = [L\completep P :
        K\completep\pideal]
      .\]
      Write $\pideal\O_L = P_1^{e_1} \cdots
      P_r^{e_r}$, $f = [\O_L / P : \O_K /
      \pideal]$. Then
      \begin{itemize}
        \item $|\GP| = \frac{|\Gal(L / K)|}{r} =
          \frac{efr}{r} = ef$ (using
          \cref{coro_11_5}).
        \item $[L\completep P :
          K\completep\pideal] = ef$. Apply
          \cref{coro_11_6} to $L\completep P /
          K\completep \pideal$, noting that $e, f$
          don't change when we take completions.
          \qedhere
      \end{itemize}
  \end{enumerate}
\end{proof}

\newpage
\part{Ramification Theory}

$p = \pideal_1 \pideal_2$ in $\Zbb[i]$ if and only
if $p = x^2 + y^2$.

We will consider $L / K$ extension of algebraic
number fields with $[L : K] = n$.

\newpage
\section{Different and discriminant}

\begin{notation*}
  \glssymboldefn{Disc}%
  Let $x_1, \ldots, x_n \in L$. Set
  \begin{align*}
    \Delta(x_1, \ldots, x_n)
    &= \det(\Tr_{L / K}(x_i x_j)) \in K \\
    &= \det \left( \sum_{k = 1}^{n} \sigma_k(x_i)
    \sigma_k(x_j)\right) \\
    &= \det(BB^\top)
  \end{align*}
  where $\sigma_k : L \to \ol{K}$ are distinct
  embeddings and $B = (\sigma_i(x_j))$.
\end{notation*}

\textbf{Note:}
\begin{itemize}
  \item If $y_i = \sum_{j = 1}^{n} a_{ij}
    x_j$, $a_{ij} \in K$, then
    \[
      \Disc(y_1, \ldots, y_n) = \det(A)^2 \Disc(x_1,
      \ldots, x_n)
    \]
    where $A = (a_{ij})$.
  \item If $x_1, \ldots, x_n \in \O_L$, then
    $\Disc(x_1, \ldots, x_n) \in \O_K$.
\end{itemize}

\begin{fclemma}[]
  \label{lemma_12_1}
  % Lemma 12.1
  Assuming:
  - $k$ a \gls{perf} field
  - $R$ a $k$-algebra which is finite dimensional
  as a $k$-vector space
  Then:
  the Trace form
  \begin{align*}
    (\bullet, \bullet) : R \times R
    &\to R \\
    (x, y)
    &\mapsto \Tr_{R / k}(xy)
    (\defeq \Tr_k(\mult(xy)))
  \end{align*}
  is non-degenerate if and only if $R = k_1 \times
  \cdots \times k_r$ where $k_i / k$ is a finite
  separable extension of $k$.
\end{fclemma}

\begin{proof}
  Example Sheet 3.
\end{proof}

\begin{fcthm}[]
  \label{thm_12_2}
  % Theorem 12.2
  Assuming:
  - $0 \neq \pideal \subseteq \O_K$ prime ideal
  Then:
  \begin{enumerate}[(i)]
    \item If $\pideal$ \gls{rames} in $L$, then
      for every $x_1, \ldots, x_n \in \O_L$, we
      have $\Disc(x_1, \ldots, x_n) \equiv 0 \bmod
      \pideal$.
    \item If $\pideal$ is un\gls{ramed} in $L$, then
      there exists $x_1, \ldots, x_n$ such that
      $\pideal \nmid (\Disc(x_1, \ldots, x_n))$.
  \end{enumerate}
\end{fcthm}

\begin{proof}
  \phantom{}
  \begin{enumerate}[(i)]
    \item Let $\pideal \O_L = P_1^{e_1} \cdots
      P_r^{e_r}$, $0 \neq P_i \subseteq
      \O_L$ distinct prime ideals, $e_i > 0$.
      Define
      \[
        R \defeq \O_L / \pideal \O_L
        \stackrel{\text{\hyperref[lemma_10_8]{CRT}}}{=}
        \prod_{i = 1}^{r} \O_L / P_i^{e_i}
      .\]
      If $\pideal$ \gls{rames}, then $\O_L /
      \pideal \O_L$ has nilpotents. Hence
      \[
        \Disc(\ol{x}_1, \ldots, \ol{x}_n) = 0
        \qquad \forall \ol{x}_i \in \O_L /
        \pideal \O_L
      .\]
      Then using the fact that
      \begin{picmath}
        \begin{tikzcd}
          \O_L \ar[r]
          \ar[d, "\Tr_{L / K}"]
          & R = \O_L / \pideal \O_L
          \ar[d, "\Tr_{R / k}"] \\
          \O_K
          \ar[r]
          & k = \O_K / \pideal
        \end{tikzcd}
      \end{picmath}
      commutes, we get that
      \[
        \Disc(x_1, \ldots, x_n) \equiv 0 \bmod
        \pideal
        \qquad \forall x_i \in \O_L /
        \pideal \O_L
      .\]
    \item $\pideal$ un\gls{ramed} implies $R =
      \O_L / \pideal \O_L$ is a product of finite
      extensions of $k$. By \cref{lemma_12_1}, we
      get that the Trace form is non-degenerate,
      hence for $\ol{x}_1, \ldots, \ol{x}_n$ a
      basis of $\O_L / \pideal \O_L$ as a $k$
      vector space, we have $\Disc(\ol{x}_1,
      \ldots, \ol{x}_n) \neq 0$. So thee exist
      $x_1, \ldots, x_n \in \O_L$ such that
      \[
        \Disc(x_1, \ldots, x_n) \not\equiv 0 \bmod
        \pideal
        .
        \qedhere
      \]
  \end{enumerate}
\end{proof}

\begin{fcdefn}[Discriminant]
  \glssymboldefn{discrim}%
  \label{defn_12_3}
  % Definition 12.3
  The \emph{discriminant} is the ideal $d_{L / K}
  \subseteq \O_K$ generated by $\Disc(x_1, \ldots,
  x_n)$ for all choices of $x_1, \ldots, x_n \in
  \O_L$.
\end{fcdefn}

\begin{corollary}
  \label{coro_12_4}
  % Corollary 12.4
  $\pideal$ \gls{rames} $L$ if and only if
  $\pideal \mover d_{L / K}$. In particular, only
  finitely many primes \glsref[rames]{ramify} in
  $L$.
\end{corollary}

\begin{fcdefn}[Inverse different]
  \glssymboldefn{invdiff}%
  \label{defn_12_5}
  % Definition 12.5
  The \emph{inverse different} is
  \[
    D_{L / K}^{-1} = \{y \in L : \Tr_{L / K}(xy)
    \in \O_K ~\forall x \in \O_L\}
  ,\]
  an $\O_L$ submodule of $L$.
\end{fcdefn}

\begin{lemma}
  \label{lemma_12_6}
  % Lemma 12.6
  $\invdiff LK$ is a fractional ideal in $L$.
\end{lemma}

\begin{proof}
  Let $x_1, \ldots, x_n \in \O_L$ a $K$-basis for
  $L / K$. Set
  \[
    d \defeq \Disc(x_1, \ldots, x_n) = \det(\Tr_{L
    / K}(x_i x_j))
  ,\]
  which is non-zero since separable.

  For $x \in \invdiff LK$ write $x = \sum_{j =
  1}^{r} \lambda_j x_j$ with $\lambda_j \in K$. We
  show $\lambda_j \in \frac{1}{d} \O_K$. We have
  \[
    \Tr_{L / K}(x x_i) = \sum_{j = 1}^{n} \lambda_j
    \Tr_{L / K}(x_i x_j) \in \O_K
  .\]
  Set $A_{ij} = \Tr_{L / K}(x_i x_j)$. Multiplying
  by $\Adj(A) \in M_n(\O_K)$, we get
  \[
    d \begin{pmatrix}
      \lambda_1 \\
      \vdots \\
      \lambda_n
    \end{pmatrix}
    = \Adj(A) \begin{pmatrix}
      \Tr_{L / K}(x x_1) \\
      \vdots \\
      \Tr_{L / K}(x x_n)
    \end{pmatrix}
  \]
  Since $\lambda_i \in \frac{1}{d} \O_K$, we have
  $x \in \frac{1}{d} \O_L$. Thus $\invdiff LK
  \le \frac{1}{d\O_K}$, so $\invdiff LK$ is a
  fractional ideal.
\end{proof}

\glssymboldefn{diff}%
The inverse $D_{L / K}$ of $\invdiff LK$ is the
different ideal.