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% 18/11/2023 12PM

\begin{corollary*}
  % Corollary 10.5 -- dodgy numbering so removed
  % this
  A regular $n$-hon is
  \glsref[constructible_point]{constructible
  by ruler and compass} if and only if $n$ is a
  power of $2$ and distinct primes of the form
  $F_k = 2^{2^k} + 1$.
\end{corollary*}

\begin{proof}
  If $n = \prod_i p_i^{\alpha_i}$ then
  \[ \phi(n) = \prod_i p_i^{\alpha_i - 1} (p_i - 1) \]
  so $\phi(n)$ is a power of $2$ if and only if
  $n$ is a product of a power of $2$ and distinct
  odd primes of the form $2^m + 1$. If $2^m + 1$
  is prime then $m$ must be a power of $2$. Indeed
  if $m = ab$ with $b > 1$ odd, then putting $x =
  2^a$ in
  \[ x^b + 1 = (x + 1)(x^{b - 1} - x^{b - 2} +
  \cdots - x + 1) \]
  gives a non-trivial factorisation.
\end{proof}

\begin{center}
  \begin{tabular}{c|ccccc}
    $k$ & $0$ & $1$ & $2$ & $3$ & $4$ \\
    \hline
    $F_k = 2^{2^k} + 1$ & $3$ & $5$ & $17$ & $257$
    & $65537$
  \end{tabular}
\end{center}
$F_0, \ldots, F_4$ are all prime. This prompted
Fermat to guess that all the $F_k$ might be prime.
However in 1732
Euler showed that
\[ F_5 = 641 \times 6700417 .\]
Since then many other Fermat numbers have been
proved composite and no more have been shown to be
prime.

\begin{flashcard}
\begin{theorem}[Linear independence of field
  embeddings]
  % Theorem 10.5
  \label{lindep_of_fembeddings}
  \cloze{
  Let $K, L$ be \glspl{field} and $\sigma_1,
  \ldots, \sigma_n : K \hookrightarrow L$ distinct
  \gls{field} \glspl{embedding} ($n \ge 1$). If
  $\lambda_1, \ldots, \lambda_n \in L$ satisfy
  \[ \lambda_1 \sigma_1(x) + \cdots + \lambda_n
  \sigma_n(x) = 0 \qquad \forall x \in K \]
  then $\lambda_1 = \lambda_2 = \cdots = \lambda_n
  = 0$.
  }
\end{theorem}

\begin{proof}
  \cloze{
  Induction on $n$. Trivially true for $n = 1$.
  Now suppose $n \ge 2$ and
  \[ \lambda_1 \sigma_1(x) + \cdots + \lambda_n
  \sigma_n(x) = 0 \qquad \forall x \in K
  \label{lec20_l67_eq} \tag{1} .\]
  Pick $y \in K$ such that $\sigma_1(y) \neq
  \sigma_2(y)$. Replacing $x$ by $xy$ in
  \eqref{lec20_l67_eq} gives
  \[ \lambda_1 \sigma_1(x) \sigma_1(y) + \cdots +
  \lambda_n \sigma_n(x) \sigma_n(y) = 0 \qquad
  \forall x \in K \label{lec20_l73_eq} \tag{2} \]
  Taking $\sigma_1(y) \times \eqref{lec20_l67_eq}
  - \eqref{lec20_l73_eq}$ gives a new relation
  with only $n - 1$ terms. It must be trivial by
  the induction hypothesis. Therefore $\sigma_1(y)
  \lambda_i = \sigma_i(y) \lambda_i$ for all $2
  \le i \le n$. Since $\sigma_1(y) \neq
  \sigma_2(y)$ we have $\lambda_2 = 0$. Therefore
  \eqref{lec20_l67_eq} has only $n - 1$ terms, so
  the induction hypothesis tells us that all
  $\lambda_i = 0$.
  }
\end{proof}
\end{flashcard}

\subsection{Kummer Theory}

We continue to assume $\characteristic K \nmid n$,
but now further assume that $\runities_n \subset
K$, i.e. $K$ contains a \gls{prim_n_root}
$\runity_n$.

Let $\alpha \in \mult{K}$. Let $L \extendover K$
be a \gls{splitting_field} of $f(X) = X^n - a$.
Since $f\fd(X) = nX^{n - 1}$ and $n 1_K \neq 0$ we
have $\gcd(f, f\fd) = 1$, so $f$ is \gls{sep_poly}
by
\cref{classification_of_finite_galois_extensions}
$L \extendover K$ is \gls{galois}.

Let $\alpha \in L$ be a root of $f$. Then
\[ f(X) = \prod_{j = 0}^{n - 1} (X - \runity_n^j
\alpha) \]
Therefore $L = \fadj K(\alpha, \runity_n \alpha,
\ldots, \runity_n^{n - 1} \alpha) = \fadj
K(\alpha)$. We sometimes write $\sqrt[n]{\alpha}$
for $\alpha$.

\begin{flashcard}[kummer-extension-defn]
\begin{definition*}[Kummer extension]
  \glsnoundefn{kumm_ex}{Kummer extension}{Kummer
  extensions}
  \cloze{$\fadj K(\sqrt[n]{\alpha}) \extendover K$ is
  called a \emph{Kummer extension}.}
\end{definition*}
\end{flashcard}

\begin{flashcard}[kummer-gal-is-cyclic]
\begin{theorem}
  % Theorem 10.6
  \label{kummer_gal_is_cyclic}
  Assume $\runities_n \subset K$ and $a \in \mult{K}$.
  There is an injective group homomorphism
  \[ \Gal(\fadj (\sqrt[n]{a}) \extendover K)
  \stackrel[\theta]{}{\hookrightarrow} \runities_n \]
  In particular \cloze{$\Gal(\fadj K(\sqrt[n]{a})
  \extendover K)$ is a cyclic group and
  $\exdeg[\fadj K(\sqrt[n](a)) : K]$ divides $n$.}
\end{theorem}

\begin{proof}
  \cloze{Let $G = \Gal(\fadj K(\sqrt[n]{a}) \extendover
  K)$. If $\sigma \in G$ then $\sqrt[n]{a}$ and
  hence also $\sigma(\sqrt[n]{a})$ are roots of
  $X^n - a$, so $\sigma(\sqrt[n]{a}) = \runity_n^r
  \sqrt[n]{a}$ for some $0 \le r < n$. We define
  \begin{align*}
    \theta : G
    &\to \runities_n \\
    \sigma
    &\mapsto \runity_n^r =
    \frac{\sigma(\sqrt[n]{a})}{\sqrt[n]{a}}
  \end{align*}
  Now let $\sigma, \tau \in G$. Then
  \begin{align*}
    \sigma(\sqrt[n]{a})
    &= \runity_n^r \sqrt[n]{a} \\
    \tau(\sqrt[n]{a})
    &= \runity_n^s \sqrt[n]{a}
  \end{align*}
  Then
  \[ \sigma \tau(\sqrt[n]{\alpha}) =
  \sigma(\runity_n^s \sqrt[n]{a}) = \runity_n^{r +
  s} \sqrt[n]{a} \]
  So $\theta(\sigma \tau) = \runity^{r + s} =
  \runity_n^r \runity_n^s = \theta(\sigma)
  \theta(\tau)$. Therefore $\theta$ is a group
  homomorphism. Since any $\sigma \in G$ is
  uniquely determined by
  $\sigma(\sqrt[n]{a})$, it is clear that
  $\theta$ is injective.}
\end{proof}
\end{flashcard}

\begin{remark*}
  The definition of $\theta$ does not depend on
  the choice of $\sqrt[n]{a}$. Indeed if $\alpha^n
  = \beta^n = a$, then
  \begin{align*}
    \left( \frac{\alpha}{\beta} \right)^n = 1
    &\implies \frac{\alpha}{\beta} \in \runities_n
    \subset K \\
    &\implies \sigma \left( \frac{\alpha}{\beta}
    \right) = \frac{\alpha}{\beta} \qquad \forall
    \sigma \in G \\
    &\implies \frac{\sigma(\alpha)}{\alpha} =
    \frac{\sigma(\beta)}{\beta} \qquad \forall
    \sigma \in G
  \end{align*}
\end{remark*}

\begin{flashcard}[mult-n-notation]
\begin{notation*}
  \glssymboldefn{multn}{$(K^*)^n$}{$(K^*)^n$}
  \[ (\mult{K})^n = \{x^n : x \in \mult{K}\}
  \subset \mult{K} .\]
  This is a subgroup since $\mult{K}$ is abelian.
\end{notation*}
\end{flashcard}

\begin{flashcard}[Xn-a-irred-iff-coro]
\begin{corollary}
  % Corollary 10.7
  Assume $\runities_n \subset K$ and $a \in \mult{K}$.
  Then
  \[ \exdeg[\fadj K(\sqrt[n]{a}) : K] =
  \text{order of $a$ in
  $\frac{\mult{K}}{\multn{K}}$} .\]
  In particular \cloze{$X^n - a$ is irreducible in $\radj
  K[X]$ if and only if $a$ is not an $a$-th power
  in $K$ for any $1 < d \mid n$.}
\end{corollary}

\begin{proof}
  \cloze{Let $\alpha = \sqrt[n]{a}$ and $G = \Gal(\fadj
  K(\alpha) \extendover K)$.
  \begin{align*}
    a^m \in \multn{K}
    &\iff a^m \in \mult{K}
    &&\text{(using $\runities_n \subset K$)} \\
    &\iff \sigma(\alpha^m) = \alpha^m
    &&\forall \sigma \in G \\
    &\iff \theta(\sigma)^m = 1
    &&\forall \sigma \in G \\
    &\iff \Im \theta \subset \runities_m \\
    &\iff \exdeg[\fadj K(\alpha) : K] =
    \#\Im(\theta) \text{ divides $m$}
  \end{align*}
  Therefore $\exdeg[\fadj K(\alpha) : K]$ is the
  least $m$ such that $a^m \in \multn{K}$. Now:
  \begin{align*}
    \text{$X^n - a$ is irreducible in $\radj K[X]$}
    &\iff \exdeg[\fadj K(\alpha) : K] = n \\
    &\iff \text{$a$ has order $n$ in
    $\frac{\mult{K}}{\multn{K}}$} \\
    &\iff \nexists m \mid n, m < n \text{ such
    that $a^m \in \multn{K}$} \\
    &\iff \nexists 1 < d \mid n \text{ with $a \in
    \multn[d]{K}$}
  \end{align*}
  (the last $\iff$ is by putting $n = md$ and use
  that $\mu_n \subset K$).}
\end{proof}
\end{flashcard}

Special case: $n = 2$, $\characteristic K \neq 2$.
Then $\exdeg[\fadj K(\sqrt{a}) : K] = 2$ provided
$a \notin \multn[2]{K}$.