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\centerline{Mathematics 352 -- Abstract Algebra II}
\centerline{Problem Set 9}
\centerline{Due: April 11, 2008}
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\noindent
{\it From Hungerford}

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\noindent
{\it Additional Problem.}  In this problem, following on 
the material from Chapter 15 we discussed before Easter,  
you will derive a complete characterization of the regular 
polygons in $\R^2$ that are constructible by 
straightedge and compass.  This question was another famous
and long-standing problem that was solved by Gauss
in the early 19th century.  (He
was so proud of this, in fact, that {\it he had a regular 17-gon, which 
is constructible, carved on his tombstone}!)
\begin{enumerate}
\item  Show that the regular $n$-gon is constructible if and
only if $[\Q(\zeta_n): \Q]$ is
equal to a power of $2$, where $\zeta_n = e^{2\pi i/n}$.  (Hint:
$\zeta_n = \cos\left(\frac{2\pi}{n}\right) + i \sin\left(\frac{2\pi}{n}\right)$.)
\item  Prove that $\Q(\zeta_n)$ is a Galois extension of
$\Q$.
\item  By looking at the Galois group
$Gal_\Q(\Q(\zeta_n))$, deduce that the degree
$[\Q(\zeta_n): \Q]$ is equal to the number of integers $m$,
$1 \le m < n$,
that are {\it relatively prime} to $n$.  (Hint:  You can use without
proof the fact we mentioned in class that if
$\phi_k(x)$ is the minimal polynomial of $\zeta_k$ over
$\Q)$ (the $k$th cyclotomic polynomial),
then 
\[
x^n - 1 = \prod_{d | n} \phi_d(x).
\]
Any element of the Galois group $Gal_\Q(\Q(\zeta_n))$ must
take $\zeta_n$ to an $n$th root of 1 that is {\it not}
a $d$th root of 1 for any $d < n$; do you see why?)
\item  Show that if an integer of the form $2^k + 1$ is 
prime, then $k$ must be a power of 2: $k = 2^r$ for some $r \ge 0$.
\item If the prime factorization of $n$ in $\Z$ is  
\[
n = 2^\ell p_1^{e_1} \cdots p_k^{e_k}\ ,
\]
where the $p_i$ are odd, then the number
of integers $m$, $1 \le m < n$, relatively prime to $n$ is given by
\[
2^{\ell-1} p_1^{e_1-1}(p_1 - 1)\cdots p_k^{e_k-1}(p_k - 1).
\]
Deduce from this statement and parts A) and D) that a regular $n$-gon is 
constructible by ruler and compass if and only if $n$ is equal to $2^\ell$ 
times a product of distinct odd primes of the form $p = 2^{2^r} + 1$.  
(Note:  The odd primes here are called {\it Fermat primes}.  Only five of them
are known to exist: $3 = 2 + 1$ (hence the equilateral triangle is 
constructible -- Euclid's Proposition 1), $5 = 2^2 + 1$, $17 = 2^4 + 1$ (hence
Gauss's 17-gon construction mentioned above!), $257 = 2^8 + 1$,
and $65537 = 2^{16} + 1$.   Fermat apparently thought that {\it all} the numbers 
$2^{2^r} + 1$ would be prime.  But in fact, in all other cases where the 
prime factorization of $2^{2^r} + 1$ has been determined, it is a composite
number!)
\end{enumerate}
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