\setcounter{chapter}{5}
\chapter{Solvability}

In this lecture we will investigate a very specific field extension, namely that
of the rational functions on $n$ variables $\C(X_1, \ldots, X_n)$ over the field
of rational functions in the elementary symmetric functions $\C(e_1, \ldots,
e_n)$. We have seen elementary symmetric functions before in some examples, but
now we will finally define them in general.

\section{Elementary Symmetric Functions}

\begin{definition}[Elementary Symmetric Functions]
  Let $X_1, \ldots, X_n$ be variables. The set of \textbf{elementary symmetric
  functions} $\set{e_0, \ldots, e_n }$ on these $n$ variables is defined as the
  coefficients of the polynomial
  %
  \begin{align*}
    \prod _{k=1} ^n (Z - X_k) = \sum _{k=0} ^n e_{n-k} Z^k
  \end{align*}
  %
  we can also write them in a more concrete closed form:
  %
  \begin{gather*}
    \begin{cases}
      e_0 = 1                                       \\
      e_1 = \sum _{i=1} ^n X_i                      \\
      e_2 = \sum _{1 \leq i<j \leq n} X_i X_j       \\
      e_3 = \sum _{1 \leq i<j<k \leq n} X_i X_j X_k \\
      \cdots                                        \\
      e_n = \prod _{i=1} ^n X_i                     \\
    \end{cases}
  \end{gather*}
  %
  in general we have
  %
  \begin{align*}
    e_k =
    \begin{cases}
    1                                            & k=0 \\
    \sum _{1 \leq i_1 < \cdots < i_k \leq n}
      \left ( \prod _{l=1} ^{k} X_{i_l} \right ) & \text{else} \\
    \end{cases}
  \end{align*}
\end{definition}

\begin{remark}
  Note the strict inequalities between the inner indexes. The $X_i$'s are
  distinct. This gives rise to a definition in words: ``the $k$th symmetric
  elementary function is the sum of all distinct products of $k$ variables''
\end{remark}

\begin{example}[Elementary symmetric Functions on 3 variables]
  \begin{gather*}
    \begin{cases}
      e_0 = 1            \\
      e_1 = X + Y + Z    \\
      e_2 = XY + XZ + YZ \\
      e_3 = XYZ
    \end{cases}
  \end{gather*}
\end{example}

Their name stems from the fact, that owing to the nature of the product form
$\prod _{k=1} ^n (Z-X_k)$, the $e_k$'s are invariant under permutation of the
variables. That is $e_k (X_1, \ldots, X_n) = e_k (X_{\sigma(1)}, \ldots,
X_{\sigma(n)})$ for $\sigma \in S_n$.

\section{The extension over Elementary Symmetric Functions}

We can now go back to our original example, and ask ourselves is $\C(X_1,
\ldots, X_n) / \C(e_1, \ldots, e_n)$ a finite extension? If so, is it Galois? We
will find that the answer to both of these questions is yes. Define $E = \C(X_1,
\ldots, X_n), F = \C(e_1, \ldots, e_n)$ to prove this result.

\begin{proof}(The extension is finite)
  We claim that $E$ is the splitting field over $F$ of the polynomial
  $p(Z)~=~\prod _{j=1} ^n (Z-X_j)$. It surely, by construction, contains all the
  roots of $p$. We must now show that $p$ does not factorise into linear factors
  in any other subfield of $E$ @todo. In particular, as $E$ is the splitting
  field over $F$ of some polynomial with distinct roots, the extension is
  normal, and thus Galois.

  Now that we know that the extension is finite, we can continue to find the
  exact degree. Theorem~\ref{thm:dimension_bound} gives us
  \begin{align*}
    \mathrm{dim}(E/F) \leq n!
  \end{align*}

  \begin{remark}
     The proof for this statement lies in the fact that $E$ is a splitting
     field, meaning that it is no more than the base field with the roots
     adjoined. When we adjoin the first root $r_1$, the maximal degree of the
     extension is $n$, if $p$ is irreducible. Then we may factor $p/(Z-r_1) \in
     F(r_1)$ and continue adjoining the second root $r_2$. Again we may be
     dealing with an irreducible polynomial (in $F(r_1)$), thus the degree of
     the next extension is maximally $n-1$. If we continue this process
     iteratively, we obtain $n!$ as an upper bound of the extension $E/F$.
  \end{remark}

  If we now look at the action of $S_n$ on $\{X_1, \ldots, X_n \}$, with $\sigma
  \in S_n$ acting by $\sigma(X_i) = X_{\sigma (i)}$ we see that $\sigma(e_k) =
  e_k$. So $S_n$ poses a good candidate for the automorphism group as a group
  action. In particular, we know that the $e_k$'s form a basis of $F$, thus we
  know that $S_n$ fixes $F$, and hence must be a subset of the automorphism
  group. Of course however $|S_n| = n!$ and thus $S_n = \Aut(E/F)$.  Finally we
  have
  %
  \begin{gather*}
    n! \leq |\mathrm{Aut}(E/F)| \overset{\text{Gal.}}{=} \dim(E/F) \leq n! \\
    \implies \mathrm{dim}(E/F) = n!
  \end{gather*}
\end{proof}

\section{The subgroups of the extension}

Now that we know that $E/F$ is Galois with dimension $n!$ we can apply theorems
from the last lecture. Specifically, to study the subextensions of $E/F$, we can
look at the subgroups of the Galois group (Part 1 of F.T.G.T).

\begin{figure}[!h]
  \centering
  \begin{tikzpicture}
    \matrix(m)[%
      matrix of math nodes,
      row sep=0.25em,
      column sep=4em,
      minimum width=1em
    ]{
      E =\C(X_1, \ldots, X_n)  & \Gal(E/E) = \set{\id}                  \\
                               & \rotatebox[origin=c]{-90}{$\subseteq$} \\
      K                        & \Gal(E/K) = G                          \\
                               & \rotatebox[origin=c]{-90}{$\subseteq$} \\
      F = \C(e_1, \ldots, e_n) & \Gal(E/F) = S_n                        \\
    };
    \path[-stealth]
      (m-1-1) edge [-] (m-3-1)
      (m-3-1) edge [-] (m-5-1)

      (m-1-1) edge [draw,-,decorate,decoration={snake}]
              node [below] {\tiny corresponds} (m-1-2)
      (m-3-1) edge [draw,-,decorate,decoration={snake}]
              node [below] {\tiny corresponds} (m-3-2)
      (m-5-1) edge [draw,-,decorate,decoration={snake}]
              node [below] {\tiny corresponds} (m-5-2)
      ;
  \end{tikzpicture}
\end{figure}

%\noindent
%Before we do this in general, we can look at following

\begin{example}\label{example:eqn_deg_4}
  We know from group theory that
  %
  \begin{align*}
    \set{\id}
    \triangleleft \Z_2
    \triangleleft A_4
    \triangleleft K_4
    \triangleleft S_4
  \end{align*}
  %
  with
  %
  \begin{align*}
    \Z_2 & = \set{\id, (12)(34)}                     \\
    K_4  & = \set{\id, (12)(34), (13)(24), (14)(23)} \\
    A_4  & = \set{\id, (12)(34), (13)(24), (14)(23),
                       (123), (132), (124), (142),
                       (134),(143), (234), (243)}
  \end{align*}
  %
  and cyclic quotients
  %
  \begin{align*}
    \rquot{S_4}{A_4} \cong \Z_2  \\
    \rquot{A_4}{K_4} \cong \Z_3  \\
    \rquot{K_4}{\Z_2} \cong \Z_3 \\
  \end{align*}
  %
  so we have a tower of extensions that are all Galois each corresponding to a
  subgroup $\Z_2, K_4, A_4, S_4$.
  %
  \begin{remark}
    Of course in general $\Z_2 = \set{\id, (12)} \cong \set{\id, (12)(34)}
    \subseteq S_4$, however to aid notation, we simply write $\Z_2 = \set{\id,
    (12)(34)}$.  Otherwise $\set{\id, (12)} \not\leq K_4$ (it's not even a
    subset).
  \end{remark}
\end{example}

\subsection{Solvability of the General Equation}

\begin{definition}[Solvable]
  A group $G$ is said to be \textbf{solvable} if there exists a sequence of
  subgroups $S_1, \ldots, S_k$ such that
  %
  \begin{align*}
    \set{\id}
    \triangleleft S_1
    \triangleleft \cdots
    \triangleleft S_{k-1}
    \triangleleft G
  \end{align*}
  %
  with $S_i/S_{i-1}$ \textit{abelian} for all subquotients.
\end{definition}

\begin{definition}[Cyclic Galois Extension]
  A Galois field extension $E/F$ is said to be a \textbf{cyclic} if there exist
  subgroups $S_1, \ldots, S_k$ such that
  %
  \begin{align*}
    \set{\id}
    \triangleleft S_1
    \triangleleft \cdots
    \triangleleft S_{k-1}
    \triangleleft G
  \end{align*}
  %
  with $S_i/S_{i-1}$ \textit{cyclic} for all subquotients.
\end{definition}

\begin{theorem}[Lagrange]
  Let $\C \subseteq F$ and $E/F$ a cyclic Galois extension. Then every $E$ can
  be expressed in terms of elements in $F$ using the operations addition,
  subtraction, multiplication, division and extracting the $k$th root.
\end{theorem}

\begin{proof}
\end{proof}

\begin{corollary}
  Together with Example~\ref{example:eqn_deg_4}, we have that every general
  equation of degree 4 can be solved using surds.
\end{corollary}