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\begin{flashcard}[Sigma-vector-defn]
\begin{definition*}
  Let $K$ be a \gls{nf}. Recall that we have $d$
  embeddings $\sigma_1, \ldots, \sigma_d : K \to
  \CC$, where $d = [K : \QQ]$. We write $r$ for the
  number of $\sigma_j$ such that $\sigma_j(K)
  \subset \RR$.
  Furthermore, we order the $\sigma_i$ such that
  $\sigma_1, \ldots, \sigma_r$ are
  precisely the real embeddings.

  Write $s = \frac{d - r}{2}$. There are $s$ pairs
  of complex conjugate embeddings. Denote them by
  $\tau_1, \ol{\tau_1}, \ldots, \tau_s, \ol{\tau_s}$
  (relabelling of $\sigma_{r + 1}, \ldots,
  \sigma_d$).

  \glssymboldefn{vSigma}{$\Sigma$}{$\Sigma$}
  Define $\Sigma : K \to \RR^d$ by
  \[ \Sigma(\alpha) = \cloze{\begin{pmatrix}
    \sigma_1(\alpha) \\
    \vdots \\
    \sigma_r(\alpha) \\
    \Re(\tau_1(\alpha)) \\
    \Im(\tau_1(\alpha)) \\
    \vdots \\
    \Re(\tau_s(\alpha)) \\
    \Im(\tau_s(\alpha))
  \end{pmatrix}} \]
  This is $\QQ$-linear.
\end{definition*}
\end{flashcard}

\begin{flashcard}[Sigma-det-lemma]
\begin{lemma*}
  Let $\alpha_1, \ldots, \alpha_d \in K$. Then
  \[ \tdisc(\alpha_1, \ldots, \alpha_d) = \cloze{(-4)^s
  \det (}\vSigma(\alpha_1), \cloze{\ldots,
  \vSigma(\alpha_d))^2} \]
\end{lemma*}

\begin{proof}
  \cloze{The matrix $[\sigma_i(\alpha_j)]_{ij}$ has the
  following rows somewhere:
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/917b08413bc549a1.png}
  \end{center}
  $\det(\sigma_i(\alpha_j)) = \pm (-2i)^s
  \det(\Sigma(\alpha_1), \ldots,
  \Sigma(\alpha_d)))$. Squaring this we get the
  claim.}
\end{proof}
\end{flashcard}

\begin{flashcard}[lattice-in-Rd]
\begin{definition*}[Lattice]
  \glsnoundefn{latt}{lattice}{lattices}
  \cloze{A \emph{lattice} in $\RR^d$ is an additive
  subgroup of the form
  \[ \Lambda = v_1 \ZZ \oplus \cdots \oplus v_d
  \ZZ \]
  where $v_1, \ldots, v_d \in \RR^d$.}
\end{definition*}
\end{flashcard}

\begin{flashcard}[fundamental-domain-defn]
\begin{definition*}[Fundamental domain]
  \glsnoundefn{fundom}{fundamental
  domain}{fundamental domains}
  \cloze{A \emph{fundamental domain} is a
  \emph{Borel} set which
  contains exactly one point from each coset of
  some \gls{latt} $\Lambda$.}\fcscrap{

  See \courseref[Probability \& Measure]{PM} for a
  definition of Borel sets. The rough idea is that Borel sets
  are the sets for which we have a well-defined
  notion of volume.}
\end{definition*}
\end{flashcard}

\begin{example*}
  Fundamental parallelepiped:
  \[ [0, 1) \cdot v_1 + \cdots + [0, 1) \cdot v_d \]
\end{example*}

\begin{lemma*}
  All \gls{fundom} have the same volume.
\end{lemma*}

\begin{proof}
  Out of the scope of this course (but should be
  fairly simple if you have studied
  \courseref[Probability \& Measure]{PM}).
\end{proof}

\begin{flashcard}[coVol-notation]
\begin{notation*}
  \glssymboldefn{coVol}{coVol}{coVol}
  We use $\coVol(\cloze{\Lambda})$ to denote
  \cloze{the volume of any \gls{fundom} of
  $\Lambda$ (this is well-defined by the above
  lemma).}
\end{notation*}
\end{flashcard}

\textbf{Observe:}
\[ \Vol([0, 1) v_1 + \cdots + [0, 1)v_d) =
|\det(v_1, \ldots, v_d)| \]
\[ \tdisc(\alpha_1, \ldots, \alpha_d) = (-4)^s
\coVol(\vSigma(\alpha_1 \ZZ + \cdots +
\vSigma(\alpha_d)\ZZ)^2 .\]

\begin{flashcard}[disc-of-module-defn]
\begin{definition*}[Discriminant of a module]
  \glssymboldefn{mdisc}{disc}{disc}
  \glspropdefn{mdisc}{discriminant}{module}
  \cloze{The \emph{discriminant} of a module of rank $d$
  is the discriminant of any basis of it (this is
  well-defined by part (3) of the following
  proposition).}
\end{definition*}
\end{flashcard}

\begin{flashcard}[disc-volume-le-prop]
\begin{proposition*}
  Let $\alpha_1, \ldots, \alpha_d, \beta_1,
  \ldots, \beta_d \in K$ which are linearly
  independent over $\QQ$. Let $A \in \QQ^{d \times
  d}$ such that
  \[ (\beta_1, \ldots, \beta_d)^\top = A(\alpha_1,
  \ldots, \alpha_d)^\top .\]
  \begin{enumerate}[(1)]
    \item Then
      \[ \disc(\beta_1, \ldots, \beta_d) = \det(A)^2
      \disc(\alpha_1, \ldots, \alpha_d) .\]
    \item If $\beta_1, \ldots, \beta_d \in \ZZ\alpha_1 +
      \cdots + \ZZ\alpha_d$, then
      \[ |\disc(\beta_1, \ldots, \beta_d)| \ge
      |\disc(\alpha_1, \ldots, \alpha_d)| .\]
    \item If the $\alpha$'s and $\beta$'s generate the
      same module, then the discriminants are the
      same.
  \end{enumerate}
\end{proposition*}

\begin{proof}
  \cloze{\[ [\sigma_j(\beta_i)]_{ij} =
  A[\sigma_j(\alpha_i)] \]
  First claim (1) follows by the definition of
  \gls{tdisc} and the properties of $\det$.

  For (2), there exists $A \in \ZZ^{d \times d}$
  such that $(\beta_1, \ldots, \beta_d)^\top =
  A(\alpha_1, \ldots, \alpha_d)^\top$, and
  $|\det(A)| \ge 1$ since $\det(A) \neq 0$.

  For (3), we already have $\ge$ by (2). For
  $\le$, we can exchange the $\alpha$'s and
  $\beta$'s.}
\end{proof}
\end{flashcard}

\begin{flashcard}[modules-disc-formula]
\begin{proposition*}
  Let $M_1 \subset M_2$ be two modules of rank $d$
  in $K$. Then
  \[ \mdisc(M_1) = \cloze{|M_2/M_1|^2} \mdisc(M_2) \]
\end{proposition*}
\end{flashcard}

\vspace{-1em}
Recall from \courseref{GRM}:

\begin{theorem*}
  Let $M_1 \subset M_2$ be two free $\ZZ$-modules
  of rank $d$. Then $M_2$ has a basis $\alpha_1,
  \ldots, \alpha_d$ and there are $\alpha_1,
  \ldots, \alpha_d \in \ZZ$ such that $\alpha_1
  \mid \alpha_2 \mid \cdots \mid \alpha_d$ and
  $a_1\alpha_1, \ldots, a_d\alpha_d$ is a basis for
  $M$.
\end{theorem*}