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% 04/11/2023 12PM

\begin{example*}
  $\fadj \QQ(\sqrt[3]{2}) \extendover \QQ$ has
  \gls{gal_clos} $\fadj \QQ(\omega, \sqrt[3]{2}) /
  \QQ$ where $\omega = e^{2\pi i / 3}$.
\end{example*}

\newpage
\section{Trace and Norm}

Let $L \extendover K$ be a \gls{finite_ex} \gls{ex}, say
$\exdeg [L : K] = n$. For $\alpha \in L$ the map
\begin{align*}
  L
  &\stackrel{m_\alpha}{\longrightarrow} L \\
  x
  &\longmapsto \alpha x
\end{align*}
is a $K$-linear endomorphism of $L$, hence has a
trace and a determinant.

\begin{flashcard}[trace-and-norm-defn]
\begin{definition*}[Trace and norm]
  \glssymboldefn{trace_alpha}{Tr${}_{L /
  K}(\alpha)$}{Tr${}_{L / K}(\alpha)$}
  \glssymboldefn{norm_alpha}{$N_{L / K}$}{$N_{L / K}(\alpha)$}
  \cloze{
  The \emph{trace} and \emph{norm} of $\alpha$ are
  \[ \Trace_{L \extendover K}(\alpha) =
  \Trace(m_\alpha) \qquad N_{L \extendover
  K}(\alpha) = \det(m_\alpha) \]
  Concretely, if $L$ has $K$-basis $v_1, \ldots,
  v_n$ and $A = (a_{ij})$ is the unique $n \times
  n$ matrix with entries in $K$ such that
  \[ \alpha(v_j) = \sum_{i = 1}^n a_{ij} v_i
  \qquad \text{and} \qquad \forall 1 \le j \le n \]
  then
  \[ \Trace_{L \extendover K} = \Trace A \qquad
  \text{and} \qquad N_{L \extendover K}(\alpha) =
  \det(A) \]
  }
\end{definition*}
\end{flashcard}

\begin{example*}
  $K = \QQ$, $L = \fadj \QQ(\sqrt{d})$, $d \in
  \ZZ$ not a square. If $\alpha = x + y\sqrt{d}$,
  $x, y \in \QQ$, then (since $L$ has $K$-basis
  $1, \sqrt{d}$):
  \begin{align*}
    \aTrace_{\fadj \QQ(\sqrt{d}) \extendover
    \QQ}(\alpha)
    &= \Trace
    \begin{pmatrix}
      x & dy \\
      y & x
    \end{pmatrix}
    = 2x \\
    \aNorm_{\fadj \QQ(\sqrt{d}) \extendover
    \QQ}(\alpha)
    &= \det
    \begin{pmatrix}
      x & dy \\
      y & x
    \end{pmatrix}
    = x^2 - dy^2
  \end{align*}
\end{example*}

\begin{flashcard}[trace-norm-properties-lemma]
\begin{lemma}
  % Lemma 7.1
  \phantom{}
  \begin{enumerate}[(i)]
    \item $\aTrace_{L \extendover K} : L \to K$ is
      a $K$-linear map.
    \item \cloze{$\aNorm_{L \extendover K} : L \to K$ is
      multiplicative, i.e.
      \[ \aNorm_{L \extendover K} (\alpha \beta) =
      \aNorm_{L \extendover K}(\alpha) \aNorm_{L
      \extendover K}(\beta) \qquad \forall \alpha,
      \beta \in L \]}
    \item \cloze{If $\alpha \in K$ tehn
      \begin{align*}
        \aTrace_{L \extendover K}
        &= \exdeg [L : K] \alpha \\
        \aNorm_{L \extendover K}(\alpha)
        &= \alpha^{\exdeg[L : K]}
      \end{align*}}
    \item \cloze{If $\alpha \in L$ then
      \[ \aNorm_{L \extendover K}(\alpha) = 0 \iff
      \alpha = 0 .\]}
  \end{enumerate}
\end{lemma}

\begin{proof}
  \cloze{
  (i) and (ii) follow from the corrsponding
  statements for traces and determinants.

  For (iii), if $\alpha \in K$ then $m_\alpha$ is
  represented by
  \[
  \begin{pmatrix}
    \alpha & 0 & \cdots & 0 \\
    0 & \alpha & \cdots & 0 \\
    \vdots & \vdots & \ddots & \vdots \\
    0 & 0 & \cdots & \alpha
  \end{pmatrix} \]
  which has trace and determinant as indicated.

  For (iv), note
  \[ \aNorm_{L \extendover K}(\alpha) \neq 0 \iff
  \text{$m_\alpha$ is invertible} \iff \alpha \neq
  0 \]
  }
\end{proof}
\end{flashcard}

\begin{flashcard}[triple-field-ex-trace-norm-calculation-lemma]
\begin{lemma}
  % Lemma 7.2
  \label{trace_norm_calculation_lemma}
  Let $M \extendover L \extendover K$ be a
  \gls{field_ex} and $\alpha \in L$. Then
  \begin{align*}
    \aTrace_{M \extendover K}(\alpha)
    &= \cloze{\exdeg [M : L] \Trace_{L \extendover
    K}(\alpha)} \\
    \aNorm_{M \extendover K}(\alpha)
    &= \cloze{\aNorm_{L \extendover K}(\alpha)^{\exdeg [M
    : L]}}
  \end{align*}
\end{lemma}

\begin{proof}
  \cloze{
  If $A$ represents $m_\alpha$ with respect to
  some basis for $L \extendover K$ and $B$
  represents $m_\alpha$ with respect to some basis
  for $M \extendover K$ picked by following the
  proof of the \nameref{tower_law}, then
  \[
  B = \begin{pmatrix}
    A & 0 & \cdots & 0 \\
    0 & A & \cdots & 0 \\
    \vdots & \vdots & \ddots & \vdots \\
    0 & 0 & \cdots & A
  \end{pmatrix} \]
  where $A$ is $\exdeg [L : K] \times \exdeg [L :
  K]$ and $B$ is $\exdeg [M : K] \times \exdeg [M
  : K]$. Then
  \[ \Trace(B) = \exdeg [M : L] \Trace(A) \qquad
  \text{and} \qquad \det(B) = \det(A)^{\exdeg[M :
  L]}. \qedhere \]
  }
\end{proof}
\end{flashcard}

\begin{flashcard}[min-poly-norm-trace-lemma]
\begin{theorem}
  % Theorem 7.3
  \label{min_poly_norm_trace_lemma}
  Let $\exdeg [L : K] < \infty$. Let $\alpha \in
  L$. Let $f$ be the \gls{min_poly} of $\alpha$
  over $K$, say
  \[ f(X) = X^n + a_{n - 1} X^{n - 1} + \cdots +
  a_1 X + a_0 \qquad a_i \in K \]
  Then
  \begin{align*}
    \aTrace_{L \extendover K}(\alpha)
    &= \cloze{-ma_{n - 1}} \\
    \aNorm_{L \extendover K}(\alpha)
    &= \cloze{((-1)^n a_0)^m}
  \end{align*}
  where $m = [L : L(\alpha)]$.
\end{theorem}

\begin{proof}
  \cloze{
  By \cref{trace_norm_calculation_lemma}
  without loss of generality $L = \fadj
  K(\alpha)$, i.e. $m = 1$. If $A$ represents
  $m_\alpha$ with respect to basis $1, \alpha,
  \alpha^2, \ldots, \alpha^{n - 1}$ then
  \[ A =
  \begin{pmatrix}
    0 & 0 & 0 & \cdots & 0 & -a_0 \\
    1 & 0 & 0 & \cdots & 0 & -a_1 \\
    0 & 1 & 0 & \cdots & 0 & -a_2 \\
    \vdots & \vdots & \vdots & \ddots & \vdots &
    \vdots \\
    0 & 0 & 0 & \cdots & 0 & -a_{n - 2} \\
    0 & 0 & 0 & \cdots & 1 & -a_{n - 1} \\
  \end{pmatrix}\]
  Therefore
  \begin{align*}
    \aTrace_{L \extendover K}(\alpha)
    &= \Trace(A) = -a_{n - 1} \\
    \aNorm_{L \extendover K}(\alpha)
    &= \det(A) = (-1)^n a_0 \qedhere
  \end{align*}
  }
\end{proof}
\end{flashcard}

\begin{example*}
  $K = \QQ$, $L = \fadj \QQ(\sqrt{d})$
  \begin{align*}
    \alpha = x + y\sqrt{d}
    &\implies (\alpha - x)^2 = dy^2 \\
    &\implies \text{$\alpha$ is a root of }
    T^2 - \ub{2xT}_{\text{trace}} +
    \ub{x^2 - dy^2}_{\text{norm}} = 0
  \end{align*}
\end{example*}

\begin{flashcard}[transitivity-of-trace-and-norm-thm]
\begin{theorem}[Transitivity of trace and norm]
  % Theorem 7.4
  \label{trace_norm_transitivity}
  Let $M \extendover L \extendover K$ be a
  \gls{finite_ex} \gls{ex} and $\alpha \in M$.
  Then
  \begin{align*}
    \aTrace_{M \extendover K}(\alpha)
    &= \cloze{\aTrace_{L \extendover K}(\aTrace_{M
    \extendover L}(\alpha))} \\
    \aNorm_{M \extendover K}(\alpha)
    &= \cloze{\aNorm_{L \extendover K}(\aNorm_{M
    \extendover L}(\alpha))}
  \end{align*}
\end{theorem}
\end{flashcard}

\begin{proof}[Proof (sketch -- non-examinable)]
  By \cref{trace_norm_calculation_lemma}, without
  loss of generality $M = \fadj L(\alpha)$. Let
  $f$ be the \gls{min_poly} of $\alpha$ over $L$,
  say
  \[ f(X) = X^n + a_{n - 1} X^{n - 1} + \cdots +
  a_1 X + a_0, \qquad a_i \in L \]
  $L \extendover K$ has basis $v_1, \ldots, v_m$
  and $M \extendover L$ has basis $1, \alpha,
  \alpha^2, \ldots, \alpha^{n - 1}$. If $A_i$
  represents $m_{a_i}$ with respect to $v_1,
  \ldots, v_m$ and $B$ represents $m_\alpha$ with
  respect to $(v_i \alpha^{j - 1})_{\substack{1
  \le i \le m \\ 1 \le j \le n}}$ then
  \[ B =
  \begin{pmatrix}
    0 & 0 & 0 & \cdots & 0 & -A_0 \\
    I & 0 & 0 & \cdots & 0 & -A_1 \\
    0 & I & 0 & \cdots & 0 & -A_2 \\
    \vdots & \vdots & \vdots & \ddots & \vdots &
    \vdots \\
    0 & 0 & 0 & \cdots & 0 & -A_{n - 2} \\
    0 & 0 & 0 & \cdots & I & -A_{n - 1} \\
  \end{pmatrix}
  \]
  ($A_i$ is $m \times m$, $B$ is $mn \times mn$).
  We compute
  \begin{align*}
    \aTrace_{M \extendover K}
    &\stackrel{\text{defn}}{=} \Trace(B) \\
    &= -\Trace(A_{n - 1}) \\
    &\stackrel{\text{defn}}{=} \aTrace_{L \extendover
    K}(-a_{n - 1}) \\
    &\stackrel{\text{\cref{min_poly_norm_trace_lemma}}}{=}
    \aTrace_{L \extendover K} (\aTrace_{M
    \extendover L}(\alpha)) \\
    \aNorm_{M \extendover K}(\alpha)
    &\stackrel{\text{defn}}{=} \det(B) \\
    &\stackrel{\text{exercise}}{=}
    (-1)^{mn} \det(A_0) \\
    &\stackrel{\text{defn}}{=} \aNorm_{L
    \extendover K} ((-1)^n a_0) \\
    &\stackrel{\text{\cref{min_poly_norm_trace_lemma}}}{=}
    \aNorm_{L \extendover K} (\aNorm_{M
    \extendover L}(\alpha)) \qedhere
  \end{align*}
\end{proof}

\begin{flashcard}[thm-75]
\begin{theorem}
  % Theorem 7.5
  \label{galois_ex_trace_norm_formulae}
  Let $L \extendover K$ be a \gls{finite_ex}
  \gls{galois} \gls{ex} with $G = \Gal(L
  \extendover K)$. Let $\alpha \in L$. Then
  \[ \aTrace_{L \extendover K}(\alpha) =
  \cloze{\sum_{\sigma \in G} \sigma(\alpha)} \qquad
  \text{and} \qquad \aNorm_{L \extendover
  K}(\alpha) = \cloze{\prod_{\sigma \in G}
  \sigma(\alpha)} \]
\end{theorem}

\begin{proof}
  \cloze{
  By \cref{lec12_first_remark}, the \gls{min_poly}
  of $\alpha$ over $K$ is
  \[ f(X) = \prod_{i = 1}^n (X - \alpha_i) \]
  where $\orb_G(\alpha) = \{\alpha_1, \ldots,
  \alpha_n\}$. Let $m = \exdeg [L : \fadj
  K(\alpha)] = \#\Stab_G(\alpha)$. Now
  \begin{align*}
    \aTrace_{L \extendover K}(\alpha)
    &\stackrel{\text{\cref{min_poly_norm_trace_lemma}}}{=}
    m \sum_{i = 1}^n \alpha_i
    = \sum_{\sigma \in G} \sigma(\alpha) \\
    \aNorm_{L \extendover K}(\alpha)
    &\stackrel{\text{\cref{min_poly_norm_trace_lemma}}}{=}
    \left( \prod_{i = 1}^n \alpha)i \right)^m
    = \prod_{\sigma \in G} \sigma(\alpha)
  \end{align*}
  where the final equality on each line follows by
  the proof of Orbit-Stabiliser Theorem.
}
\end{proof}
\end{flashcard}

\begin{example*}
  $K = \QQ$, $L = \fadj \QQ(\sqrt{d})$. $\Gal(L
  \extendover K) = \{1, \sigma\}$,
  $\sigma(\sqrt{d}) = -\sqrt{d}$. Then for $\alpha
  = x + y\sqrt{d}$, $x, y \in \QQ$,
  \begin{align*}
    \aTrace_{L \extendover K}(\alpha)
    &= (x + y\sqrt{d}) = (x + y\sqrt{d}) + (x -
    y\sqrt{d}) = 2x \\
    \aNorm_{L \extendover K}(\alpha)
    &= (x + y\sqrt{d})(x - y\sqrt{d})
    = x^2 - dy^2
  \end{align*}
  We generalise \cref{galois_ex_trace_norm_formulae}
  to $L \extendover K$ \gls{sep_ex}.
\end{example*}