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\centerline{Mathematics 352 -- Abstract Algebra II}
\centerline{Solutions for Discussion 3}
\centerline{April 28, 2008}
\vskip 10pt
\noindent
A)  The regular $n$-gon is constructible if and only if the point 
$\left(\cos\left(\frac{2\pi}{n}\right),\sin\left(\frac{2\pi}{n}\right)\right)$
is constructible, which is true in turn if and only if 
\begin{equation}\label{const}
\left[\Q\left(\cos\left(\frac{2\pi}{n}\right)\right):\Q\right] = 2^m \text{ and }
\left[\Q\left(\sin\left(\frac{2\pi}{n}\right)\right):\Q\right] = 2^{m'}
\end{equation}
for some $m,m' \ge 0$. 


Assume that (\ref{const}) holds. Note that $x = i \sin\left(\frac{2\pi}{n}\right)$
is a root of the polynomial equation 
\[
x^2 + \left(1 - \cos^2\left(\frac{2\pi}{n}\right)\right) = 0.
\]
For all $n > 2$, this polynomial is irreducible in $\Q\left(\cos\left(\frac{2\pi}{n}\right)\right)[x]$
since it has no real roots but $\Q\left(\cos\left(\frac{2\pi}{n}\right)\right) \subset \R$.  
Hence 
\[
\left[\Q\left(\cos\left(\frac{2\pi}{n}\right), i \sin\left(\frac{2\pi}{n}\right)\right): 
\Q\left(\cos\left(\frac{2\pi}{n}\right)\right)\right] = 2,
\]
so 
\[
\left[\Q\left(\cos\left(\frac{2\pi}{n}\right), i \sin\left(\frac{2\pi}{n}\right)\right): 
\Q\right] = 2^{m+1}.
\]
Now $\Q(\zeta_n)$ is a subfield of 
$\Q\left(\cos\left(\frac{2\pi}{n}\right), i \sin\left(\frac{2\pi}{n}\right)\right)$ 
so $[\Q(\zeta_n):\Q]$ is also a power of 2, since that degree must divide $2^{m+1}$.

Conversely, assume that $[\Q(\zeta_n): \Q] = 2^\ell$ for some $l\ge 0$.  
Note that 
\[
(\zeta_n)^{-1} = \cos\left(\frac{2\pi}{n}\right) - i\sin\left(\frac{2\pi}{n}\right),
\]
so 
\[
\cos\left(\frac{2\pi}{n}\right) = \frac{1}{2}\left(\zeta_n + (\zeta_n)^{-1}\right)
\]
is an element of $\Q(\zeta_n)$.  It follows that
$[\Q\left(\cos\left(\frac{2\pi}{n}\right)\right): \Q]$ divides $2^\ell$, hence 
is also a power of $2$.  Hence 
$\cos\left(\frac{2\pi}{n}\right)$ is constructible.  Hence the point 
$\left(\cos\left(\frac{2\pi}{n}\right),\sin\left(\frac{2\pi}{n}\right)\right)$, and 
the regular $n$-gon are constructible.
\vskip 10pt
\noindent
B) The roots of $x^n - 1 = 0$ in $\C$ are $1,\zeta_n,\zeta_n^2,\ldots,\zeta_n^{n-1}$.
Hence $\Q(\zeta_n) = \Q(1,\zeta_n,\zeta_n^2,\ldots,\zeta_n^{n-1})$ is the splitting
field of $x^n - 1 \in \Q[x]$.  By Theorem 10.15 in Hungerford, $\Q(\zeta_n)$ is
a normal extension of $\Q$.  It is also a separable extension of $\Q$, since
$\Q$ has characteristic zero (Theorem 10.17).  It is finite dimensional since $[\Q(\zeta_n): \Q] < n$.  
Therefore, $\Q(\zeta_n)$ is a Galois extension of $\Q$ by definition.
\vskip 10pt
\noindent
C)  By the observations in part (B), we know that the roots of $x^n - 1 = 0$
in $\C$ are the numbers $\zeta_n^k$ for $k$ with $0 \le k < n$.   Since $\Q(\zeta_n)$
is a Galois extension of $\Q$, it follows by the Fundamental Theorem of Galois Theory 
(Theorem 11.11) that 
\[
[\Q(\zeta_n): \Q] = |{\rm Gal}_\Q \Q(\zeta_n)|.
\]
However, by Theorem 11.3, given $k$ with $0\le k < n$, there exists $\sigma \in {\rm Gal}_\Q \Q(\zeta_n)$
with $\sigma(\zeta_n) = \zeta_n^k$ if and only if $\zeta$ and $\zeta_n^k$ have the same
minimal polynomial, and then by Theorem 11.4 there is exactly one element of the Galois
group for each such $k$.  By the equation
\[
x^n - 1 = \prod_{d|n} \phi_d(x)
\]
we see that $\sigma \in {\rm Gal}_\Q \Q(\zeta_n)$ can only map $\zeta_n$ to other roots
of the $n$th cyclotomic polynomial $\phi_n(x)$, since the roots of the other factors 
are all $d$th roots of unity for $d < n$.  If $c = \gcd(k,n) > 1$, then
$(\zeta_n^k)^c = 1$, so $\zeta_n^k$ is not a primitive $n$th root of unity. 
Hence the roots of $\phi_n(x)$ are precisely the $\zeta_n^k$ where $\gcd(n,k) = 1$, 
which implies what we wanted to show.  
\vskip 10pt
\noindent
D)  We will show the contrapositive:  
If $k > 1$ is not a power of $2$, then $2^k + 1$ is not prime.  
If an integer $k > 1$ is not a power of $2$, then it must be divisible by 
some odd prime $p \ge 3$.  Say $k = pq$ where $q$ is some other integer.  But then
we have, by the factorization of a sum of like odd powers:
\[
2^{pq} + 1 = (2^q)^p + 1^p = (2^q + 1)((2^{q})^{p-1} - (2^{q})^{p-2} + \cdots - 2^q + 1).
\]
(Since $p$ is odd, the last term in the second factor here is always $+1$.)
Since $p$ is odd and $\ge 3$, $2^q+1 < 2^{pq} + 1$.  Hence $2^{pq} + 1$
cannot be a prime because it is divisible by $2^q + 1$.  
\vskip  10pt
\noindent 
E)  By the given formula for the Euler $\phi$-function, if $n = 2^\ell p_1^{e_1}\cdots p_k^{e_k}$, 
where $p_i$ are odd primes, the degree $[\Q(\zeta_n):\Q]$, which equals the number of integers $k$ 
with $1 \le m < n$ that satisfy $\gcd(k,n) = 1$ by part (C), is given by
the equation:
\begin{equation}\label{totient}
[\Q(\zeta_n): \Q] = 2^{\ell-1} p_1^{e_1-1}(p_1 - 1)\cdots p_k^{e_k-1}(p_k - 1).
\end{equation}

If the regular $n$-gon is constructible then $[\Q(\zeta_n):\Q] = 2^m$ for some $m \ge 0$.
However (\ref{totient}) gives a factorization of $[\Q(\zeta_n):\Q]$ and from this and 
the Unique Factorization Theorem in $\Z$ we
see that all $e_i = 1$ (or else $p_i$ divides the degree $[\Q(\zeta_n):\Q]$)
and each $p_i - 1$ must be a power of $2$, so $p_i = 2^{2^{r_i}} + 1$ by part (D).  This 
says that $n$ must be a power of $2$ times a product of distinct primes of the 
form $2^{2^r} + 1$.  

Conversely, if $n$ is a power of $2$ times a product of \emph{distinct} odd primes of the 
form $2^{2^r} + 1$, then (\ref{totient}) shows that $[\Q(\zeta_n):\Q]$ equals
a power of 2 since all the factors on the right hand side of (\ref{totient})
are powers of $2$ (note that our assumptions ensure $e_i = 1$ for all $i$).  Therefore
the regular $n$-gon is constructible by part (A).

\end{document}

