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Let $K$ be a \gls{nf}, $\O_K$ the \gls{algring}.
Let $[K : \QQ] = d$.

\textbf{Aim:} $\exists$ an \gls{intbasis}, that is
$\alpha_1, \ldots, \alpha_d \in \O_K$ such that
\[ \O_K = \alpha_1 \ZZ \oplus \cdots \oplus
\alpha_d \ZZ \]
If $M \subset K$ is a finitely generated
$\ZZ$-module, then
\[ M = \alpha_1 \ZZ \oplus \cdots \oplus \alpha_r
\ZZ \]
Observe $r = \dim_\QQ \Span_\QQ (M)$:
\begin{itemize}
  \item $\alpha_1, \ldots, \alpha_r$ is linearly
    independent over $\QQ$.
  \item $\Span_\QQ(M) = \Span_\QQ(\alpha_1,
    \ldots, \alpha_r)$.
\end{itemize}
Observe $\Span_\QQ \O_K = K$:
\begin{itemize}
  \item If $\alpha \in K$, then $a \alpha \in
    \O_K$ for suitable $a$.
\end{itemize}

\subsubsection*{Discriminant of tuple}

Recall Norm and Trace (from \courseref[Galois
Theory]{Galois}). Let $L / K$ be a finite
extension of fields. For $\alpha \in L$, we can
associate $m_\alpha : x \mapsto \alpha x$ on $L$
considered a vector space over $K$. The norm is
$N_{L / K}(\alpha) = \det(m_\alpha) \in K$. The
trace if $\Trace_{L / K} (\alpha) =
\Trace(m_\alpha) \in K$. Recall the following
properties:
\begin{itemize}
  \item If $\alpha \in K$, $\Trace_{L / K}(\alpha)
    = [L : K]\alpha$, $N_{L / K}(\alpha) =
    \alpha^{[L : K]}$.
  \item $\alpha, \beta \in L$: $\Trace_{L /
    K}(\alpha + \beta) = \Trace_{L / K}(\alpha) +
    \Trace_{L / K}(\beta)$, $N_{L / K}(\alpha
    \beta) = N_{L / K}(\alpha) N_{L / K}(\beta)$.
  \item Let $M / L / K$: $\Trace_{M / K}(\alpha) =
    \Trace_{L / K}(\Trace_{M / L}(\alpha))$,
    similarly with norms.
\end{itemize}
Fix $K$. Let $d = [K : \QQ]$. Then there exists
$d$ distinct embeddings $\sigma_1, \ldots,
\sigma_d : K \to \CC$ (if $K = \QQ(\alpha)$, and
$f$ is the minimal polynomial of $\alpha$, then
$\sigma_1(\alpha), \ldots, \sigma_d(\alpha)$ are
the roots of $f$).

We have:
\[ N_{K / \QQ}(\alpha) = \sigma_1(\alpha) \cdots
\sigma_d(\alpha) \]
\[ \Trace_{K / \QQ}(\alpha) = \sigma_1(\alpha) +
\cdots + \sigma_d(\alpha) \]
If $\alpha \in \O_K$, then $N_{K / \QQ}(\alpha),
\Trace_{K / \QQ}(\alpha) \in \ZZ$.
If $\alpha$ is such that $K = \QQ(\alpha)$, and
\[ f(X) = X^d + a_{d - 1} x^{d - 1} + \cdots + a_0 \]
is its minimal polynomial, then
\[ N_{K / \QQ}(\alpha) = (-1)^d a_0, \qquad
\Trace_{K / \QQ}(\alpha) = -a_{d - 1} .\]

Fix $K$. Write $N = N_{K / \QQ}$, $\Trace = \Trace_{K / \QQ}$.

\begin{flashcard}[disc-of-tuple-defn]
\begin{definition*}[Discriminant]
  \glssymboldefn{disc}{disc}{disc}
  \glspropdefn{tdisc}{discriminant}{tuple}
  \cloze{Let $\sigma_1, \ldots,
  \sigma_d$ be the embeddings $K \to \CC$.
  Let $\alpha_1, \ldots, \alpha_d \in K$. Then we
  write
  \[ \disc(\alpha_1, \ldots, \alpha_d) =
  \det(\sigma_i(\alpha_j)) .\]
  Note that $\det(\sigma_i(\alpha_j))$ denotes the
  determinant of the matrix whose $ij$-th entry is
  $\sigma_i(\alpha_j)$.}
\end{definition*}
\end{flashcard}

\begin{example*}
  \[ \tdisc(1, \alpha, \alpha^2, \ldots, \alpha^{d -
  1}) = \prod_{1 \le i < j \le d}
  (\sigma_i(\alpha) - \sigma_j(\alpha))^2 \]
  If $K = \QQ(\alpha)$ and $f$ is the minimal
  polynomial, then this equals
  \[ (-1)^{\frac{d(d - 1)}{2}} N(f'(\alpha)) .\]
\end{example*}

\begin{note*}
  \[ \ZZ[\alpha] = \ZZ + \alpha \ZZ + \cdots
  + \alpha^{d - 1} \ZZ \]
  for $\alpha \in \O_K$.
\end{note*}

\begin{flashcard}[disc-is-det-tr-lemma]
\begin{lemma*}
  \[ \tdisc(\alpha_1, \ldots, \alpha_d) =
  \det(\Trace(\alpha_i \alpha_j)) \]
\end{lemma*}

\begin{proof}
  \cloze{Write $[x_{ij}]_{ij}$ for the $d \times d$
  matrix with entries $x_{ij}$. Note
  \[ [\sigma_j(\alpha_i)]_{ij}
  [\sigma_j(\alpha_k)]_{jk} = \left[ \sum_{j =
  1}^d \sigma_i(\alpha_i \alpha_j) \right] =
  [\Trace(\alpha_i \alpha_k)]_{ik} \]
  Determinants are multiplicative and invariant
  under transpose.}
\end{proof}
\end{flashcard}

\begin{flashcard}[disc-zero-iff-lindep]
\begin{lemma*}
  \[ \tdisc(\alpha_1, \ldots, \alpha_d) = 0 \iff
  \alpha_1, \ldots, \alpha_d \text{ are linearly
  dependent over $\QQ$} \]
\end{lemma*}

\begin{proof}
  \cloze{If $\alpha_1, \ldots, \alpha_d$ are linearly
  dependent, then the rows of $[\Trace(\alpha_i
  \alpha_j)]$ are also linearly dependent. Then
  $\det = 0$, so $\tdisc = 0$.

  For the converse, suppose for the contrary that
  $\alpha_1, \ldots, \alpha_d$ are linearly
  independent over $\QQ$, and for sake of
  contradiction, assume $\tdisc = 0$, so
  $\tdisc(\Trace(\alpha_i \alpha_j)) = 0$. Then
  there exists some $a_1, \ldots, a_d$ not all $0$ such that
  \[ \sum_{i = 1}^d a_i \Trace(\alpha_i \alpha_j)
  = 0 ~\forall j \]
  This is equivalent to (by additivity of
  $\Trace$):
  \[ \Trace \left( \left( \sum_i a_i \alpha_i
  \right) \alpha_j \right) = 0 ~\forall j \]
  By linear independence of $\alpha_1, \ldots,
  \alpha_d$,
  \begin{itemize}
    \item $\sum_i a_i \alpha_i \neq 0$.
    \item $\exists b_1, \ldots, b_d$ such that
      $\beta^{-1} = \sum_j b_j \alpha_j$.
  \end{itemize}
  Then
  \[ \sum_j b_j \Trace(\beta \cdot \alpha_j) = 0 \]
  hence
  \[ \Trace(\beta \cdot \beta^{-1}) = \Trace(1) =
  0 \]
  which is a contradiction, since $\Trace(1) = d
  \neq 0$.}
\end{proof}
\end{flashcard}

\begin{corollary*}
  $\alpha_1, \ldots, \alpha_d$ are
  linearly independent over $\QQ$ if and only if
  the complex vectors $(\sigma_1(\alpha_j), \ldots,
  \sigma_d(\alpha_j))^\top \in \CC^d$ for $j = 1,
  \ldots, d$ are linearly independent over $\CC$.
\end{corollary*}