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\centerline{ \bigbf Solution, October 6, 1989 }
\bs
\centerline { \bigbf Math 401 -- Second Course in Abstract Algebra, I}
\ms
\centerline { \bigbf Prof. Daniel Grayson }
\ms
\centerline { 2:00 PM MWF, Section F1, 241 Altgeld Hall }
\bs

\problem 6

Consider an automorphism $\sigma$ of a field $K$ whose order $n$ is finite.
Let $F = \{ x \in K \,|\, \sigma(x) = x \}$ be the fixed field of $\sigma$.
Study the structure of the field extension $F \subseteq K$, using the ideas
discovered in treating the roots of the general cubic equation.  Make
assumptions which enable you to prove something.

\bs
\centerline{Solution}

\define\charac{\operatorname{char}}
%\define\dim{\operatorname{dim}}

We may regard $K$ as an $F$-vector space and $\sigma$ as an $F$-linear
transformation of $K$.  Since the equation $\sigma^n=1$  holds, 
the eigenvalues of $\sigma$ are
$n$-th roots of unity.  In order to ensure that the roots of the polynomial
$T^n-1$ which $\sigma$ satisfies are distinct and lie in $F$, we add the following
assumptions:
$\charac K \nmid n$; and $K$ contains a primitive $n$-th root of unity $\zeta$.
By standard linear algebra, we know that $K$ has a basis
consisting of eigenvectors of $\sigma$.  Suppose $v$ and $w$ happen to be
eigenvectors of $\sigma$ belonging to the same eigenvalue $\lambda$.  Then
$v/w$ is again an eigenvector for $\sigma$ with eigenvalue $1$, as the 
computation $ \sigma(v/w) = \sigma v/\sigma
w = \lambda v/\lambda w = v/w$ shows us.  Thus $v/w \in F$, and so $v$ and $w$
are linearly dependent over $F$.  Thus each nonzero eigenspace of $\sigma$ has
dimension~$1$.

Let $G$ denote the group $\{ \omega \in F^\times \mid \omega^n=1\}$ of $n$-th
roots of unity in $F$, and let $H \subseteq G$ be the set of eigenvalues of $\sigma$ over $F$.

We check now that $H$ is a subgroup of $G$.  
The equation $\sigma 1 = 1$ shows that
$1 \in H$.  If $\lambda \in H$ with corresponding eigenvector $v$, then
from $\sigma(1/v) = 1/\sigma(v) = (1/\lambda) (1/v)$ we see
that $1/\lambda$ is an eigenvalue, and thus $1/\lambda \in H$.
Finally, if $\lambda, \mu \in H$, with corresponding eigenvectors $v, w$, then
from $\sigma(v w) = (\lambda v)(\mu w) = (\lambda \mu)( v w)$ we see that $v w$
is an eigenvector with eigenvalue $\lambda \mu$, and thus $\lambda \mu \in H$.

We know that any subgroup of a cyclic group is cyclic, so since $G$ is a cyclic
group (generated by $\zeta$), it follows
that $H$ is a cyclic group, generated, let's say, by $\zeta^d$, where $d \mid
n$ and $d \ge 1$.  Thus all the eigenvalues of $\sigma$ are $n/d$-th roots of
$1$, and $\sigma$ itself satisfies the equation $\sigma^{n/d} = 1$.  Since $n$
was the order of $\sigma$, we see that $d=1$.

We now know that every $n$-th root of $1$ is an eigenvalue of $\sigma$, and
since there are $n$ of these, and each eigenspace has dimension $1$, we see
that $\dim_FK = n$.  Now pick an eigenvector $\alpha \in K$  whose eigenvalue
is $\zeta$.  Then $1, \alpha, \alpha^2, \dots, \alpha^{n-1}$ are eigenvectors
for the eigenvalues $1, \zeta, \zeta^2, \dots, \zeta^{n-1}$.  Since the
eigenvalues are distinct, the eigenvectors $1, \alpha, \alpha^2, \dots,
\alpha^{n-1}$ are linearly independent, and thus they form a $F$-basis for $K$.

Let $a = \alpha^n$.  
The computation $\sigma(a) = \sigma(\alpha^n) = (\zeta \alpha)^n = \zeta^n
\alpha^n = a$ shows that $a \in F$.

There is an $F$-algebra isomorphism $F[x]/(x^n-a) @>\cong>> K$ which
sends the residue class of $x$ to $\alpha$, as both of the algebras in question
have dimension $n$ over $F$.  
This completely identifies $K$ as an $F$-algebra.

We may also write $K=F(a^{1/n})$ or even as $K=F(\alpha)$, with $\alpha^n \in F$.

\end

