\centerline{Exercises XII}

\bigskip\noindent
In the following exercises feel free to use the following facts.

\medskip\noindent
{\bf Fact 1.} Suppose that $K$ is a field of characteristic zero and $p$ a prime
number. Let $a$ be an element of $K$ which is not a $p$th power in $K$. Then
the polynomial $X^p-a$ is irreducible over $K$.

\medskip\noindent
{\bf Fact 2.} Let $L/K$ be a finite extension of degree $d$. Consider the map
$\det : L \to K$ which maps $\alpha \in L$ to the determinant of the
$K$-linear map $L \to L$, $x \mapsto \alpha x$. You check easily that
$\det$ (sometimes called the norm map) satisfies the following two
properties:
\item{(a)} $\det(\alpha\beta)=\det(\alpha)\det(\beta)$, and
\item{(b)} $\det(\alpha)=\alpha^d$ if $\alpha \in K$.

\medskip\noindent
{\bf 1.} Let $K$, $a\in K$, and $p$ be as in Fact 1. Let $L=K(a^{1/p})$.
Compute $\det(a^{1/p})$, where $\det$ is as in Fact 2. (Careful!)

\medskip\noindent
{\bf 2.} Let $K$ be a field of characteristic $0$, and let $p$ be an odd
prime. Suppose that $a$ is an element of $K$ which is not a $p$th power in
$K$. Show that $X^{p^2}-a$ is irreducible over $K$. Hint: show that
$a^{1/p}$ is not a $p$th power in $K(a^{1/p})$.

\medskip\noindent
{\bf 3.} Show by an example that the statement of exercise 2 is wrong if
$p=2$. 

\medskip\noindent
{\bf 4.} In this exercise $K$ is a field with the following properties:
\itemitem{(a)} The characteristic of $K$ is $0$,
\itemitem{(b)} the algebraic closure $\bar K$ of $K$ is finite over $K$, and
\itemitem{(c)} $K \not = \bar K$.

\noindent
Show that the finite group $G=Gal(\bar K/K)$ is a $2$-group. Hint: Suppose
$G$ contains an element $\sigma$ of odd prime order $p$ in $G$. Show that
the fixed field $\bar K^{<\sigma>}$ contains a primitive $p$th root of
unity. Then use Kummer theory and Exercise 2.

\medskip\noindent
{\bf Remark.} This is the beginning of the proof that shows that fields as
in Exercise 4 are special. Namely, it can only happen if $[\bar K:K]=2$ and
$K$ is a ``real closed field''.

\medskip\noindent
{\bf 5.} Let $p_1=2, p_2=3, p_3=5, p_4=7$, etc be the prime numbers.
Consider the group 
$$
\prod_{n\in {\bf N}} {\bf Z}/p_n{\bf Z} .
$$
Show that any finite index subgroup is given by a sub-product
$$
\prod_{n\in S} {\bf Z}/p_n{\bf Z} .
$$
for some subset $S \subset {\bf N}$ containing all but finitely many $n\in
{\bf N}$. (Feel free to use that the result holds for finite products of cyclic
groups of pairwise distinct prime orders. But only if you actually
understand what I mean by that...)

\medskip\noindent
{\bf Warning.} It isn't true that any subgroup of the product is a
``sub-product''.

\bye

