%! TEX root = Galois.tex
% vim: tw=50
% 16/11/2023 12PM

Recall that $\mult{(\ZZ / n\ZZ)} = \{a \in \ZZ /
n\ZZ \st \gcd(a, n) = 1\}$ is a group under
multiplication. It has order $\phi(n)$ (Euler
$\phi$-function). Let $K$ be a \gls{field} with
$\characteristic K \nmid n$. Let $\runity_n$ be a
\gls{prim_n_root} (in some
\gls{field_ex} of $K$).

\begin{flashcard}[cyclotoic-ex-hom-thm]
\begin{theorem}
  % Theorem 10.1
  \label{cyclotomic_ex_hom_thm}
  \glssymboldefn{cyclotomic_hom}{$\chi$}{$\chi$}
  There is an injective group homomorphism 
  \[ \cloze{\Gal(\fadj K(\runity_n) \extendover K)
  \stackrel{\chi}{\hookrightarrow} \mult{(\ZZ /
  n\ZZ)}} .\]
  In particular \cloze{$\Gal(\fadj K(\runity_n)
  \extendover K)$ is abelian, and }$\exdeg[\fadj
  K(\runity_n) : K]$ \cloze{divides $\phi(n)$}.
\end{theorem}

\begin{proof}
  \cloze{Let $G = \Gal(\fadj K(\runity_n) \extendover K)$.
  If $\sigma \in G$ then $\runity_n$ and hence
  also $\sigma(\runity_n)$ are roots of $X^n - 1$.
  Therefore $\sigma(\runity_n) = \runity_n^a$ for
  some $a \in \ZZ$. Since $\runity_n$ is a
  \gls{prim_n_root} the value of $a$ is unique
  modulo $n$. We define
  \begin{align*}
    \cychom : G
    &\to \ZZ / n\ZZ \\
    \sigma
    &\mapsto a
  \end{align*}
  Noe let $\sigma, \tau \in G$ with
  $\sigma(\runity_n) = \runity_n^a$,
  $\tau(\runity_n) = \runity_n^b$. Then
  \[ \sigma \tau(\runity_n) = \sigma(\runity_n^b)
  = \runity_n^{ab} ,\]
  so
  \[ \cychom(\sigma \tau) = ab = \cychom(\sigma)
  \cychom(\tau) .\]
  In particular $\cychom(\sigma) \cychom(\sigma^{-1}) =
  \cychom(1) = 1$ so $\cychom(\sigma) \in \mult{(\ZZ /
  n\ZZ)}$ and $\cychom$ is a group homomorphism.
  Since any $\sigma \in G$ is uniquely determined
  by $\sigma(\runity_n)$ it is clear that $\cychom$
  is injective.}
\end{proof}
\end{flashcard}

\begin{remark*}
  If $\cychom(\sigma) = a$ then $\sigma(\runity) =
  \runity^a$ for all $\runity \in \runities_n$. So
  the definition of $\cychom$ does not depend on the
  choice of $\runity_n$.
\end{remark*}

\begin{example*}
  Let $p$ be a prime with $p \equiv 4 \pmod{5}$.
  Let $K = \FF_p$, $L = \qFF_{p^2}$ and $n = 5$.
  Since $5 \mid (p^2 - 1)$, there exists
  $\runity_5 \in L$ a primitive $5$-th root of
  unity. Since $5 \nmid (p - 1)$ we know
  $\runity_5 \notin K$. Therefore $L =
  K(\runity_5)$. By \cref{cyclotomic_ex_hom_thm}
  \[ \ub{\Gal(L \extendover K)}_{\cong C_2}
  \stackrel{\cychom}{\hookrightarrow} \mult{(\ZZ /
  5\ZZ)} .\]
  Therefore $\Im(\cychom) = \{\pm 1\} \subset
  \mult{(\ZZ / 5\ZZ)}$.
\end{example*}

\begin{flashcard}[order-of-Fp-cyclotomic]
\begin{corollary}
  % Corollary 10.2
  Let $K = \FF_p$ and suppose $p \nmid n$. Then
  $\exdeg[\fadj K(\runity_n) : K]$ is \cloze{the order of
  $p$ in $\mult{(\ZZ / n\ZZ)}$.}
\end{corollary}

\begin{proof}
  \cloze{$\Gal(\fadj K(\runity_n) \extendover K)$ is
  generated by \gls{frob_hom} $\phi$ which sends
  $\runity_n \mapsto \runity_n^p$. Therefore
  \begin{align*}
    \exdeg[\fadj K(\runity_n) : K]
    &= \text{order of $\phi$ in $\Gal(\fadj
    K(\runity_n) \extendover K)$} \\
    &= \text{order of $\ub{\cychom(\phi)}_{=p}$ in
    $\mult{(\ZZ / n\ZZ)}$} \qedhere
  \end{align*}
  }
\end{proof}
\end{flashcard}

\begin{flashcard}[cyclotomic-poly-defn]
\begin{definition*}[Cyclotomic polynomial]
  \glsnoundefn{cyc_poly}{cyclotomic
  polynomial}{cyclotomic polynomials}
  \glssymboldefn{cyc_poly}{$\Phi_n$}{$\Phi_n$}
  \cloze{Let $\runity_n = e^{2\pi i / n}$. The $n$-th
  \emph{cyclotomic polynomial} is
  \[ \Phi_n(X) = \prod_{a \in \mult{(\ZZ / n\ZZ)}}
  (X - \runity_n^a) .\]}
\end{definition*}
\end{flashcard}
\vspace{-1em}

Its roots are the \glspl{prim_n_root}. As
$\Gal(\fadj \QQ(\runity_n) \extendover \QQ)$
permutes these, we have $\cycpoly_n \in \radj
\QQ[X]$. Clearly we have $\runity^n = 1$ if and
only if $\runity$ is a \gls{prim_n_root} for some
$d \mid n$. Therefore
\[ X^n - 1 = \prod_{d \mid n} \cycpoly_d(X) .\]
It follows by induction on $n$ that $\cycpoly_n
\in \radj \ZZ[X]$.

\begin{example*}
  \begin{align*}
    \cycpoly_1
    &= X - 1 \\
    \cycpoly_p
    &= \frac{X^p - 1}{X - 1}
    = X^{p - 1} + X^{p - 2} + \cdots + X^2 + X +
    1
    &&\text{($p$ prime)} \\
    \cycpoly_4
    &= X^2 + 1
  \end{align*}
  In general, $\deg \cycpoly_n = \phi(n)$.
\end{example*}

\begin{flashcard}[cyclotomic-irred-over-Q]
\begin{theorem}
  % Theorem 10.3
  \label{cyclotomic_irred_over_Q}
  If $K = \QQ$ then the group homomorphism
  $\cychom$ of \nameref{cyclotomic_ex_hom_thm}
  \cloze{is
  an isomorphism. In particular,
  $\exdeg[\QQ(\runity_n) : \QQ] = \phi(n)$, and
  $\cycpoly_n \in \radj \QQ[X]$ is irreducible.}
\end{theorem}

\begin{proof}
  \cloze{Let $p$ be a prime with $p \nmid n$. We show
  that $\Im \cychom$ contains $p \mod n$. If this
  is true then $\Im \cychom$ contains $a \mod n$
  for every $a$ coprime to $n$ (by considering the
  prime factorisation fo $a$). Therefore $\chi$ is
  surjective as required. Let $f, g \in \radj
  \QQ[X]$ be the \glspl{min_poly} of $\runity_n$
  and $\runity_n^p$ over $\QQ$. If $f = g$ then by
  \cref{f_irred_iff_gal_transitive} there exists
  $\sigma \in \Gal(\QQ(\runity_n) \extendover
  \QQ)$ with $\sigma(\runity_n) = \runity_n^p$ as
  required. If not then $f$, $g$ are distinct
  irreducibles dividing $X^n - 1$. So $f, g \in
  \radj \ZZ[X]$ (using Gauss' lemma) and $fg \mid
  (X^n - 1)$. Now $\runity_n$ is a root of
  $g(X^p)$, so $f(X) \mid g(X^p)$. Reducing modulo
  $p$ gives
  \[ \modred{f}(X) \mid \modred{g}(X^p) =
  \modred{g}(X)^p .\]
  Both $\modred{f}$ and $\modred{g}$ divide the
  \gls{sep_poly} polynomial $X^n - 1 \in \radj
  \FF_p[X]$, so $\modred{f}(X) \mid
  \modred{g}(X)$. Hence
  \[ \modred{f}(X)^2 \mid \modred{f}(X)
  \modred{g}(X) \mid (X^n - 1) \]
  which contradicts the fact that $X^n - 1 \in
  \radj \FF_p[X]$ is \gls{sep_poly}.}
\end{proof}
\end{flashcard}

\begin{flashcard}[gauss-n-gon-ruler-compass-thm]
\begin{theorem}[Gauss]
  % Theorem 10.4
  Let $n \ge 3$. A regular $n$-gon is
  \glsref[constructible_point]{constructible by
  ruler and compass} if and only
  if \cloze{$\phi(n)$ is a power of $2$.}
\end{theorem}

\begin{proof}
  \cloze{
  Let $\runity_n = e^{2\pi i / n}$ and $\alpha =
  \runity_n + \runity_n^{-1} = 2\cos \left(
  \frac{2\pi}{n} \right)$. Since $\alpha \in \RR$,
  $\runity \notin \RR$ and $\runity_n$ is a root
  of $X^2 - \alpha X + 1 \in \radj \fadj
  \QQ(\alpha)[X]$. We have $[\fadj \QQ(\runity_n)
  : \fadj \QQ(\alpha)] = 2$. If a regular $n$-gon
  can be \gls{constructed_point} then $\alpha$ is
  \gls{constructible_real}. Now
  \cref{constructible_requires_power_of_2} implies
  that $\exdeg[\fadj \QQ(\alpha) : \QQ]$ is a
  power of $2$. By \cref{cyclotomic_irred_over_Q},
  $\phi(n) = \exdeg[\fadj \QQ(\runity_n) : \QQ] =
  2\exdeg[\fadj \QQ(\alpha) : \QQ]$ is a power of
  $2$. For the converse we use the converse of
  \cref{compass_construction_is_constructible_field}
  (proof omitted). It remains to show that if
  $\phi(n)$ is a power of $2$ there there exist
  fields
  \[ \QQ = K_m \subset K_{m - 1} \subset \cdots
  \subset K_1 \subset K_0 = \fadj \QQ(\runity_n) \]
  where $\exdeg[K_i : K_{i + 1}] = 2$ for all $i$
  and $K_1 = \fadj \QQ(\alpha)$. By the
  \nameref{ft_of_gal}, it suffices to construct
  subgroups
  \[ \{1\} = H_0 \subset H_1 \subset \cdots
  \subset H_{m - 1} \subset H_m = \mult{(\ZZ /
  n\ZZ)} \]
  where $(H_i : H_{i - 1}) = 2$ for all $i$, and
  $H_1 = \{\pm 1\}$. We must show that if $G =
  \mult{(\ZZ / n\ZZ)}$ is an abelian group with
  order a power of $2$ then there exist subgroups
  $\{1\} = H_0 \subset H_1 \subset H_2 \subset
  \cdots \subset H_m = G$ such that $(H_i : H_{i -
  1}) = 2$ for all $i$. Assuming $H_0, H_1,
  \ldots, H_j$ have been constructed, and $H_j
  \neq G$, we note that $G / H_j$ has order a
  power of $2$ hence contains an element $gH_j$ of
  order $2$. Then set $H_{j + 1} = \langle H_j, g
  \rangle$ and repeat.
  }
\end{proof}
\end{flashcard}
