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\begin{flashcard}[ramification-index-defn]
\begin{definition*}[Ramification index]
  \glssymboldefn{ramind}{$e(Q \mid P)$}{$e(Q \mid
  P)$}
  \glsnoundefn{ramind}{ramification
  index}{ramification indices}
  \cloze{Given an extension of \glspl{nf} $L / K$, and
  primes $P \subset \O_K$, $Q \subset \O_L$ such
  that $P$ \gls{lo} $Q$, we define $e(Q \mid P)$ to
  be the largest $e \in \ZZ$ such that $Q^e \mid
  P\O_L$.}
\end{definition*}
\end{flashcard}

\vspace{-1em}
\textbf{Observe:} $\O_L \to \O_L / Q$ sends $\O_K$
to $\O_K / P$ because $Q \cap \O_K = P$, so $\O_L
/ Q \mid \O_K / P$.

\begin{flashcard}[inertial-degree-defn]
\begin{definition*}[Inertial degree]
  \glssymboldefn{inertdeg}{$f(Q \mid P)$}{$f(Q
  \mid P)$}
  \glsnoundefn{inertdeg}{inertial degree}{inertial
  degrees}
  \cloze{If $P$ \gls{lo} $Q$, we define the
  \emph{inertial degree}
  \[ f(Q \mid P) = [\O_L / Q : \O_K / P] .\]}
\end{definition*}
\end{flashcard}

\vspace{-1em}
Let $M / L$, let $R \subset \O_M$ be a prime that
\gls{lo} $Q$. Then $R$ \gls{lo} $P$, and
\begin{align*}
  \ramind(R \mid P)
  &= \ramind(R \mid Q) \ramind (Q \mid P) \\
  \inertdeg(R \mid P)
  &= \inertdeg(R \mid Q) \inertdeg(Q \mid P)
\end{align*}

\begin{lemma*}
  For all ideals $I$, $\exists k \in \ZZ_{> 0}$
  such that $I^k$ is a principal ideal.
\end{lemma*}

\begin{proof}
  Later. This Lemma is only stated now so that we
  can use it in the following proofs.
\end{proof}

\begin{flashcard}[ideal-lifting-norm-lemma]
\begin{proposition*}
  Let $L / K$. Let $I \subset \O_K$. Then
  $\N(I\O_L) = \N(I)^{[L : K]}$.
\end{proposition*}

\begin{proof}
  \cloze{True for principal ideals. Indeed, if $I =
  \alpha\O_K$ for some $\alpha \in \O_K$, then
  \begin{align*}
    I\O_L
    &= \alpha \O_L \\
    \N(\alpha\O_K)
    &= N_{K / \QQ}(\alpha) \\
    \N(\alpha\O_L)
    &= N_{L / \QQ}(\alpha) = N_{K /
    \QQ}(\alpha)^{[L : K]}
  \end{align*}
  Now to prove for a general ideal $I$, pick $k >
  0$ such that $I^k$ is principal. Then by the
  above, the equality holds for $I^k$. Hence it
  holds for $I$, by multiplicativity of $\N(I)$
  (since $\N(I)$ is a positive integer).}
\end{proof}
\end{flashcard}

\begin{flashcard}[primes-deg-sum-thm]
\begin{theorem*}
  Let $Q_1, \ldots, Q_r$ be the primes in $\O_L$
  that \gls{la} $P \subset \O_K$. Then:
  \[ [L : K] = \sum_{j = 1}^r \ramind(Q_j \mid P)
  \inertdeg(Q_j \mid P) .\]
\end{theorem*}

\begin{proof}
  \cloze{$P \O_L = Q_1^{\ramind(Q_1 \mid P)} \cdots
  Q_r^{\ramind(Q_r \mid P)}$ (by the definition
  of \gls{ramind}). Then
  \[ \N(P\O_L) = \N(Q_1)^{\ramind(Q_1 \mid P)}
  \cdots \N(Q_r)^{\ramind(Q_r \mid P)} =
  \N(P)^{\sum_{i = 1}^r \ramind(Q_i \mid P)
  \inertdeg(Q_j \mid P)} .\]
  By the above Proposition,
  \[ \N(P\O_L) = \N(P)^{[L : K]} .\]
  So the desired equality follows, since $\N(P) >
  1$.}
\end{proof}
\end{flashcard}

\begin{flashcard}[dedekind-thm-factorisation]
\begin{theorem*}[Dedekind]
  \label{dedekind}
  \cloze{Let $K$ be a \gls{nf}. Let $P \subset \O_K$ a
  prime. Let $p$ be the rational prime below $P$.
  Let $g \in \O_K[X]$ be monic and irreducible.
  Let $\alpha$ be a root of $g$, and let $L =
  K(\alpha)$. Assume $p \nmid [\O_L :
  \O_K[\alpha]]$. Let $\ol{g}$ be the image of $g$
  in $(\O_K / P)[X]$. Let
  \[ \ol{g} = \ol{g}_1^{e_1} \cdots \ol{g}_r^{e_r} \]
  be the factorisation of $\ol{g}$ into
  irreducibles in the $(\O_K / P)[X]$. Let $g_j
  \in \O_K[X]$ monic such that $g_j \equiv
  \ol{g}_j \pmod{P}$ for all $j$. Then $Q_j =
  P\O_L + g_j(\alpha)\O_L$ is a prime in $\O_L$
  with $\inertdeg(Q_j \mid P) = \deg g_j$, and
  \[ P\O_L = Q_1^{e_1} \cdots Q_r^{e_r} .\]}
\end{theorem*}
\end{flashcard}

\begin{flashcard}[monogenic-nf-defn]
\begin{definition*}[Monogenic]
  \glsnoundefn{monogenic}{monogenic}{\gls{nf}}
  \cloze{A \gls{nf} $K$ is \emph{monogenic} if there is
  $\alpha$ such that $\O_K = \ZZ[\alpha]$.}
\end{definition*}
\end{flashcard}

\begin{flashcard}[dedekind-pf-1]
\begin{proposition*}
  \prompt{First step of \nameref{dedekind}:}
  \cloze{\[ Q_1^{e_1} \cdots Q_r^{e_r} \subset
  P\O_L \]}
\end{proposition*}

\begin{proof}
  \cloze{Pick $e_j$ elements (not necessarily distrinct)
  from each $P\O_L \cup \{g_j(\alpha)\}$,
  and multiply them together.
  Collect all such products in a set $A$. By
  definition, $Q_1^{e_1} \cdots Q_r^{e_r}$ is
  generated by $A$. So it is enough to show that
  $A \subset P\O_L$. All but one element in $A$
  has a factor in $P\O_L$. The exception is
  $g_1(\alpha)^{e_1} \cdots g_r(\alpha)^{e_r}
  \equiv g(\alpha) = 0 \pmod{P\O_L}$. Hence
  $g_1(\alpha)^{e_1} \cdots g_r(\alpha)^{e_r} \in
  P\O_L$.}
\end{proof}
\end{flashcard}