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% 07/11/2024 12PM

\begin{example*}
  $K = \Qbb$, $L = \Qbb(i)$, $f(X) = X^2 + 1$.
  \nameref{hensel} gives us that $\sqrt{-1} \in
  \Qbb_5$. Hence $(5)$ splies in $\Qbb(i)$, i.e.
  $5\O_L = \pideal_1\pideal_2$.
\end{example*}

\begin{corollary}
  \label{coro_10_10}
  % Corollary 10.10
  Let $0 \neq \pideal \subseteq \O_K$ a prime
  ideal. For $x \in L$ we have
  \[
    N_{L / K}(x) = \prod_{P \mover \pideal}
    N_{L\completep P / L\completep\pideal}(x)
  .\]
\end{corollary}

\begin{proof}
  Let $B_1, \ldots, B_r$ be bases for
  $L\completep{P_1}, \ldots, L\completep{P_r}$ as
  $K\completep\pideal$-vector spaces. Then $B =
  \bigcup_i B_i$ is a basis for $L \otimes_K
  K\completep\pideal$ over $K\completep\pideal$.
  Let $[\mult(x)]_B$ (respectively
  $[\mult(x)]_{B_i}$) denote the matrix for
  $\mult(x) : L\otimes_K K\completep\pideal \to
  L\otimes_K K\completep\pideal$ (respectively
  $L\completep{P_i} \to L\completep{P_i}$) with
  respect to the basis $B$ (respectively $B_i$).
  Then
  \[
    [\mult(x)]_B = \begin{pmatrix}
      [\mult(x)]_{B_1}
      &
      & \\
      & \ddots
      & \\
      &
      & [\mult(x)]_{B_r}
    \end{pmatrix}
  \]
  hence
  \begin{align*}
    N_{L / K}(x)
    &= \det([\mult(x)]_B) \\
    &= \prod_{i = 1}^{r} \det [\mult(x)]_{B_i} \\
    &= \prod_{i = 1}^{r} N_{L\completep{P_i} /
    K\completep\pideal}(x)
    \qedhere
  \end{align*}
\end{proof}

\newpage
\section{Decomposition groups}

\begin{fcdefn}[Ramification]
  \glsadjdefn{rames}{ramifies}{prime ideal}%
  \glsnoundefn{ramind}{ramification
  index}{ramification indexes}%
  \glsadjdefn{ramed}{ramified}{prime ideal}%
  \label{defn_11_1}
  % Definition 11.1
  Let $0 \neq \pideal$ be a prime ideal of $\O_K$,
  and
  \[
    \pideal \O_L = P_1^{e_1} \cdots P_r^{e_r}
  \]
  with $P_i$ distinct prime ideals in $\O_L$, and
  $e_i > 0$.

  \begin{enumerate}[(i)]
    \item $e_i$ is the \emph{ramification index} of $P_i$
      over $\pideal$.
    \item We say $\pideal$ \emph{ramifies} in $L$ if some $e_i >
      1$.
  \end{enumerate}
\end{fcdefn}

\begin{example*}
  $\O_K = \Cbb[t]$, $\O_L = \Cbb[T]$. $\O_K \to
  \O_L$ sends $t \mapsto T^n$. Then $t\O_L = T^n
  \O_L$, so the \gls{ramind} of $(T)$ over $(t)$
  is $n$.

  Corresponds geometrically to the degree $n$ of
  covering of Riemann surfaces $\Cbb \to \Cbb$, $x
  \mapsto x^n$.
\end{example*}

\begin{fcdefn}[Residue class degree]
  \label{defn_11_2}
  % Definition 11.2
  $f_i \defeq [\O_L / P_i : \O_K / \pideal]$ is
  the \emph{residue class degree} of $P_i$ over
  $\pideal$.
\end{fcdefn}

\begin{theorem}
  \label{thm_13_3}
  % Theorem 13.3
  $\sum_{i = 1}^{r} e_i f_i = [L : K]$.
\end{theorem}

\begin{proof}
  Let $S = \O_K \setminus (\pideal)$. Exercise (properties
  of \gls{locon}):
  \begin{enumerate}[(1)]
    \item $S\local \O_L$ is the \gls{intc} of
      $S\local \O_K$ in $L$.
    \item $S\local\completep\pideal S\local \O_L
      \cong S\local P_1^{e_1} \cdots S\local
      P_r^{e_r}$.
    \item $S\local \O_L / S\local P_i \cong \O_L /
      P_i$ and $S\local \O_K / S\local \pideal
      \cong \O_K / \pideal$.
  \end{enumerate}
  In particular, (2) and (3) imply $e_i$ and
  $f_i$ don't change when we replace $\O_K$ and
  $\O_L$ by $S\local \O_K$ and $S\local \O_L$.

  Thus we may assume that $\O_K$ is a \gls{dvr}
  (hence a PID). By \nameref{lemma_10_8}, we have
  \[
    \O_L / \pideal \O_L \cong \prod_{i = 1}^{r}
    \O_L / P_i^{e_i}
  .\]
  We count dimension as $k \defeq \O_K / \pideal$
  vector spaces.

  RHS: for each $i$, there exists a decreasing
  sequence of $k$-suibspaces
  \[
    0 \subseteq P_i^{e_i - 1} / P_i^{e_i}
    \subseteq \cdots \subseteq P_i / P_i^{e_i}
    \subseteq \O_L / P_i^{e_i}
  .\]
  Thus $\dim_k \O_L / P_i^{e_i} = \sum_{j = 0}^{e_i
  - 1} \dim_k(P_i^j / P_i^{j + 1})$. Note that
  $P_i^j / P_i^{j + 1}$ is an $\O_L / P_i$-module
  and $x \in P_i^j \setminus P_i^{j + 1}$ is a
  generator (for example can prove this after
  \gls{locon} at $P_i$).

  Then $\dim_k P_i^j / P_i^{j + 1} = f_i$ and we
  have
  \[
    \dim_k \O_L / P_i^{e_i} = e_i f_i
  ,\]
  and hence
  \[
    \dim_k \prod_{i = 1}^{r} \O_L / P_i^{e_i} =
    \sum_{i = 1}^{r} e_i f_i
  .\]
  
  LHS: Structure theorem for finitely generated
  modules over PIDs tells us that $\O_L$ is a free
  module over $\O_K$ of rank $n$.

  Thus $\O_L / \pideal \O_L \cong (\O_K /
  \pideal)^n$ as $k$-vector spaces, hence $\dim_k
  \O_L / \pideal\O_L = n$.
\end{proof}

Geometric analogue:

$f : X \to Y$ a degree $n$ cover of compact
Riemann surfaces. For $y \in Y$:
\[
  n = \sum_{x \in f^{-1}(y)} e-x
\]
where $e_x$ is the \gls{ramind} of $x$. Now assume
$L / K$ is Galois. Then for any $\sigma \in \Gal(L
/ K)$, $\sigma(P_i) \cap \O_K = \pideal$ and hence
$\sigma(P_i) \in \{P_1, \ldots, P_r\}$.

\begin{proposition}
  \label{prop_11_4}
  % Proposition 11.4
  The action of $\Gal(L / K)$ on $\{P_1, \ldots,
  P_r\}$ is transitive.
\end{proposition}

\begin{proof}
  Suppose not, so that there exists $i \neq j$
  such that $\sigma(P_i) \neq P_j$ for all $\sigma
  \in \Gal(L / K)$.

  By \nameref{lemma_10_8}, we may choose $x \in
  \O_L$ such that $x \equiv 0 \mod P_i$, $x \equiv
  1 \mod \sigma(P_i)$ for all $\sigma \in \Gal(L /
  K)$. Then 
  \[
    N_{L / K}(x) = \prod_{\sigma \in \Gal(L / K)}
    \sigma(x) \in \O_K \cap P_i = \pideal
    \subseteq P_j
  .\]
  Since $P_j$ prime, there exists $\tau \in \Gal(L
  / K)$ such that $\tau(x) \in P_j$. Hence $x \in
  \tau^{-1}(P_j)$, i.e. $x \equiv 0 \mod
  \tau^{-1}(P_j)$, contradiction.
\end{proof}

\begin{corollary}
  \label{coro_11_5}
  % Corollary 11.5
  Suppose $L / K$ is Galois. Then $e_1 = \cdots =
  e_r = e$, $f_1 = \cdots = f_r = f$, and we have
  $n = efr$.
\end{corollary}

\begin{proof}
  For any $\sigma \in \Gal(L / K)$ we have
  \begin{enumerate}[(i)]
    \item $\pideal \O_L = \sigma(\pideal) \O_L =
      \sigma(P_1)^{e_1} \cdots \sigma(P_r)^{e_r}$,
      hence $e_1 = \cdots = e_r$.
    \item $\O_L / P_i \cong \O_L / \sigma(P_i)$
      via $\sigma$. Hence $f_1 = \cdots = f_r$.
      \qedhere
  \end{enumerate}
\end{proof}

\glsnoundefn{cramind}{ramification
index}{ramification indexes}%
If $L / K$is an extension of complete \glspl{dvf}
with normalised \glspl{valt} $v_L$, $v_K$ and
\glspl{unif} $\pi_L, \pi_K$, then the ramification
index is $e = e_{L / K} = v_L(\pi_K)$. The residue
class degree is $f \defeq f_{L / K} = [k_L : k]$.

\begin{corollary}
  \label{coro_11_6}
  % Corollary 11.6
  Let $L / K$ be a finite separable extension.
  Then $[L : K] = ef$.
\end{corollary}

$\O_K$ a \gls{dedd}:

\begin{fcdefn}[Decomposition]
  \glsnoundefn{decom}{decomposition}{decompositions}%
  \glssymboldefn{GP}%
  \label{defn_11_7}
  % Definition 11.7
  Let $L / K$ be a finite Galois extension. The
  decomposition at a prime $P$ of $\O_L$ is the
  subgroup of $\Gal(L / K)$ defined by
  \[
    G_P = \{\sigma \in \Gal(L / K) \st \sigma(P) =
    P\}
  .\]
\end{fcdefn}