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Let $R$ be a ring and $G \subset \Aut(R)$ a
subgroup. \emph{Invariant theory} seeks to
describe the cubring $R^G = \{x \in R \st
\sigma(x) = x ~\forall \sigma \in G\}$.

The topic was studied extensively in the 19th
century and was the motivation for Hilbert's Basis
Theorem. It is also important in modern algebraic
geometry for describing the quotient of an
algebraic variety by a group action.

Let $k$ be a field and $L = \fadj k(X_1, \ldots, X_n)$
be the field of rational functions in $n$
variables, i.e. the field of fractions of $R =
\radj k[X_1, \ldots, X_n]$. Let $G = S_n$ act on
$L$ by permutating the $X_i$.

Aim: compute $\fixedf L^G$.

We note that $\fixedf L^G$ contains the
\glspl{elem_symm_func} $\symmpoly_1 = \sum_i
X_i$, $\symmpoly_2
= \sum_{i < j} X_i X_j$, \ldots, $\symmpoly_n = \prod_i
X_i$. By \nameref{symm_func_thm}, $\fixedf R^G = \radj
k[\symmpoly_1, \ldots, \symmpoly_n]$ and there are no polynomial
relations satisfied by the $\symmpoly_i$.

\begin{theorem}
  % Theorem 12.2
  $\fixedf L^G = \fadj k(s_1, \ldots, s_n)$.
\end{theorem}

\begin{proof}[Proof 1]
  Let $\frac{f}{g} \in \fixedf L^G$, $f, g \in
  R$ coprime. Then
  $\frac{\sigma(f)}{\sigma(g)} =
  \frac{f}{g}$ for all $\sigma \in G$. Since
  $R$ is a UFD and the units of $R$ are just
  $\mult{k}$, we have $\sigma(f) = c_\sigma
  f$, $\sigma(g) = c_\sigma g$ for some
  $c_\sigma \in \mult{k}$. But $G$ has
  finite order, say $|G| = N$ (in fact $= n!$).
  Therefore $f = \sigma^N(f) = c_\sigma^N f$
  hence $c_\sigma^N = 1$. Therefore $fg^{N -
  1}, g^N \in \fixedf R^G = \radj k[\symmpoly_1,
  \ldots, \symmpoly_n]$. Therefore
  $\frac{f}{g} = \frac{fg^{N - 1}}{g^N} \in
  \fadj k(\symmpoly_1, \ldots, \symmpoly_n)$.
\end{proof}

\begin{proof}[Proof 2]
  Let 
  \begin{align*}
    f(T)
    &= \prod_{i = 1}^n (T - X_i) \\
    &= T^n - \symmpoly_1 T^{n - 1} + \symmpoly_2
    T^{n - 2} - \cdots + (-1)^n \symmpoly_n
  \end{align*}
  Then $f \in \radj \fadj k(\symmpoly_1, \ldots,
  \symmpoly_n)[T]$ is a polynomial of degree $n$
  and $L$ is a \gls{splitting_field} for $f$
  over $\fadj k(\symmpoly_1, \ldots,
  \symmpoly_n)$. So we have $L \extendover
  \fixedf L^G
  \extendover \fadj k(\symmpoly_1, \ldots,
  \symmpoly_n)$. \es[12]{1} tells us that
  $\exdeg[L : \fadj k(\symmpoly_1, \ldots,
  \symmpoly_n)] \le n!$. But also $\exdeg [L :
  \fixedf L^G] = \#G = n!$ by
  \nameref{artins_thm}. So by
  \nameref{tower_law}, $\fixedf L^G = \fadj
  k(\symmpoly_1, \ldots, \symmpoly_n)$.
\end{proof}

\begin{remark*}
  We have shown that the \gls{galois_group} of a
  ``generic'' monic polynomial of degree $n$ is
  $S_n$.
\end{remark*}

\textbf{Exercise:} Show that for any finite group
$G$ there exists a \gls{finite_ex} \gls{galois}
\gls{ex} $L \extendover K$ with \gls{galois_group}
$G$. This may not be possible if we specify $K$ in
advance.

This may not be possible if we specify $K$ in
advance, for example $K = \CC$ or $K = \FF_p$, and
is a famous open problem when $K = \QQ$ (inverse
\gls{galois_group}).

\begin{corollary}
  % Corollary 12.3
  Let $S_n$ act on $L = \fadj k(X_1, \ldots, X_n)$
  by permuting the $X_i$. If $\characteristic(k) \neq 2$,
  then $\fixedf L^{A_n} = \fadj k(\symmpoly_1,
  \ldots, \symmpoly_n, \delta)$ where $\delta =
  \prod_{i < j} (X_i - X_j)$.
\end{corollary}

\begin{proof}
  $(S_n : A_n) = 2$, hence $\exdeg [L^{A_n} :
  k(\symmpoly_1, .., \symmpoly_n)] = 2$. We have
  $\sigma(\delta) = \sign(\sigma) \delta$ for all
  $\sigma \in S_n$. In particular $\delta \in \fixedf
  L^{A_n}$ and $\delta \notin \fixedf L^{S_n}$.
  Therefore $\fixedf L^{A_n} = \fadj
  k(\symmpoly_1,\ldots, \symmpoly_n, \delta)$.
\end{proof}

\begin{remark*}
  It can be shown that if $R = \radj k[X_1,
  \ldots, X_n]$ then $\fixedf R^{A_n} = \radj
  k[\symmpoly_1, \ldots, \symmpoly_n, \delta]$.

  \textbf{Idea of proof:} Let $f \in \fixedf
  R^{A_n}$. Pick $\sigma \in S_n \setminus A_n$.
  Write $f = \half((f + \sigma f) + (f - \sigma
  f))$. Then show $f - \sigma f$ is divisible by
  $\delta$.
\end{remark*}

\begin{theorem*}[Fundamental Theorem of Algebra]
  $\CC$ is algebraically closed.
\end{theorem*}

\begin{proof}
  We'll use the following facts:
  \begin{enumerate}[(i)]
    \item Every polynomial over $\RR$ of odd
      degree has a root in $\RR$.
    \item Every quadratic over $\CC$ has a root in
      $\CC$ (use quadratic formula,
      $\sqrt{re^{i\theta}} = \sqrt{r}
      e^{i\theta/2}$).
    \item Every group of order $2^n$ has an index
      $2$ subgroup (since every group of order
      $p^k$ has a subgroup of order $p^j$ for all
      $j \le k$, by considering the composition
      series, and using the fact that the centre
      of a $p$-group is always non-trivial).
  \end{enumerate}
  Suppose $L \extendover \CC$ is a \gls{finite_ex}
  with $L \neq \CC$. Replacing $L$ by its
  \gls{gal_clos} over $\RR$, we may assume $L
  \extendover \RR$ is \gls{galois}. Let $G =
  \Gal(L \extendover \RR)$. Let $H \subset G$ be a
  Sylow $2$-subgroup. Then $\exdeg [\fixedf L^H :
  \RR] = (G : H)$ is odd. So if $\alpha \in
  \fixedf L^H$ then $\exdeg[\fadj \RR(\alpha) :
  \RR]$ is odd, hence $\alpha \in \RR$ by (i).
  Therefore $\fixedf L^H = \RR$ and $G = H$ is a
  $2$-group. Let $G_1 = \Gal(L \extendover \CC)
  \subset \Gal(L \extendover \RR) = G$. Since $L
  \neq \CC$ we have $G_1$ non-trivial, so by (iii)
  it has an index $2$ subgroup, say $G_2$. Then
  $\exdeg[\fixedf L^{G_2} : \CC] = (G_1 : G_2) =
  2$ \contradiction by (ii).
\end{proof}