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\prof{Peter Cholak and Juan Migliore}
\class{Math 222}
\quizdate{Wednesday, February 7}
\pagestyle{empty}

\begin{document}


\heading

Be sure to carefully write up your answers.  It is suggested that you
first write out a draft of your proposed questions and then carefully
rewrite that draft to get your final version.  You do \emph{not} have
to write the answers on this sheet of paper.\\

\begin{problem}
  Section 2.3: Problem 18: Explain why the solutions of the
  simultaneous equations $x^8+y^8=a$ and $x^8y^8=b$ have degree 2
  algebraic expressions in $\{a,b\}$.  {\em Hint:} Use Proposition 1.1.
\end{problem}


{\it Solution.} Let $w=x^8$ and $z=y^8$. Then $w$ and $z$ are the
solutions of the quadratic equation $X^2-aX+b=0$ (see Proposition
1.1). So both $w$ and $z$ have degree 2 algebraic expressions in
$\{a,b\}$.  On the other hand, $x \in \sqrt{\sqrt{\sqrt{w}}}$.  So $x$
has a degree 2 algebraic expression in $\{w\}$. Also, $y$ has a degree
2 algebraic expression in $\{z\}$. This shows that the solutions of
the above simultaneous equations have degree 2 algebraic expressions
in $\{a,b\}$.

\begin{problem}
  Show all the solutions of $x^6 + 9x^3 + 8 = 0$ are constructible.
  {\em Hint:} Solve it.
\end{problem}

{\it Solution.} By the factorization, the equation 
becomes\par\vskip-15pt
$$(x^3+1)(x^3+8)=0$$ or further becomes\par\vskip-15pt
$$(x+1)(x^2-x+1)(x+2)(x^2-2x+4)=0$$
The solutions of this equation (hence of the original equation) are $-1,
-2,
\frac{1\pm\root\of{3}i}2, 1\pm\root\of{3}i$. All of them
have degree 2 algebraic expressions in the integers and so they are
constructible.

\begin{problem}
  Show that  a regular 30-gon is constructible.  {\em Hint:} Make use of
the angles in a regular 5-gon and a regular 6-gon, which are both
  constructible.
\end{problem}

{\it Solution.} Note that a regular $n$-gon is constructible if and
only if the complex number \linebreak $\cos\frac{2\pi}n+i\sin\frac{2\pi}n$
is constructible, and if and only if an angle of $\frac{2\pi}n$ is
constructible.  So we need only show the constructibility of the angle
of $\frac{2\pi}{30}$. Since the angles in a regular 5-gon and a
regular 6-gon are both constructible, the complex numbers
$x=\cos\frac{2\pi}5+i\sin \frac{2\pi}5$ and
$y=\cos\frac{2\pi}{6}+i\sin \frac{2\pi}{6}$ are both
constructible. So
$$z=\frac{y}x=\cos(\frac{2\pi}{5}-\frac{2\pi}6)+i\sin
(\frac{2\pi}{5}-\frac{2\pi}6) =\cos\frac{2\pi}{30}+i\sin
\frac{2\pi}{30}$$ is also constructible. This implies that an angle of
$\frac{2\pi}{30}$ is constructible, and hence a regular 30-gon is
constructible.  \par



\begin{problem}
  Section 2.5: Problem 16.  Prove that if $n$ is a positive odd integer
and $\zeta$ is a primitive $n$th root of unity, then so is $\zeta^2$.  Is
this also true for even $n$?  Justify your answer.
\end{problem}

{\it Solution.} If $\zeta$ is a primitive $n$th root of unity then we
have $o(\zeta) = n$, 
$\zeta^n = 1$ and $\zeta^m \neq 1$ for $1 \leq m < n$.  Suppose $\zeta^2$
were not a primitive $n$th root of unity, so $(\zeta^2)^m = \zeta^{2m} =
1$ for some $m < n$.  By Proposition 2.14, it follows that $2m$ is a
multiple of $o(\zeta) = n$.  But since $m < n$, we get $2m < 2n$ so the
only way a multiple of $n$ could equal $2m$ is if $2m=n$.   But
this is impossible since $n$ is odd.

This is not true for even $n$.  For example, $i$ is a primitive fourth
root of unity but $i^2 = -1$ is not.

\end{document}


