%% need to revise problems 3,4,5 with answers
%% revise solution to problem 8.
%% check 1,2,6-8.


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\prof{Peter Cholak and Juan Migliore} \class{Math 222 - Exam 1}
\quizdate{Wednesday, February 28} \pagestyle{empty}

\begin{document}

\heading

\noindent {\bf Except where noted, be sure to show all your work.}

\bigskip

\begin{problem}(4 points each for a total of 20 points)
   {\bf Define} the following terms:

   a) $\sqrt[n]{z}$ over the complex numbers.\\

\vspace{1.3in}
%$\sqrt[n]{z}$ over the complex numbers is the set of all the
%complex numbers $x$ with $x^n=z$.
%\\


b) A \emph{primitive} complex $n$th root of unity.   \\

\vspace{1.3in}
%A complex root of unity whose order is $n$.
%\\

c)  $m$ and $n$ are \emph{relatively prime}. \\

\vspace{1.3in}
%The integers $m$ and $n$ are \emph{relatively prime} if their
%greatest common divisor is 1.
%\\

d) The {\em multiplicative inverse} of a nonzero number $a$ in
$\mathbb{Z}_p$ ($p$ a prime).\\
\vspace{1.3in}
%The {\em multiplicative inverse} of a nonzero number $a$ in
%$\mathbb{Z}_p$ is the number $b$ in $\mathbb{Z}_p$ with $ab\equiv
%ba\equiv 1$ modulo $p$.\\

e) An equation $a_0x^n + a_1 x ^{n-1} + \ldots a_{n-1}x + a_n = 0$ is
\emph{solvable by radicals} (or \emph{algebraically
   resolvable}) if $\ldots$ \\

%each of its root has an algebraic expression in the coefficients $\{
%a_0, a_1, \ldots, a_n\}$.

\end{problem}
\newpage

\begin{problem}(2 points each  for a total of 10 points)
   Answer {\bf True} or {\bf False} - no work required:

   a) the possible orders of elements in $\mathbb{Z}_{19}$ are
   $1$ and $19$.\\

\vspace{1.3in}
  % \emph{Answer:} False. The possible orders of elements in
%   $\mathbb{Z}_{17}$ are all the positive divisors of
%   $18=19-1$. They are just $1,2,3,6,9,18$.\\

   b) $4$ is the multuplicative inverse of $3$ in
   $\mathbb{Z}_{11}$. \\

\vspace{1.3in}
   %\emph{Answer:} True. $4 \times 3 = 12 \equiv_{11} 1$.\\



   c) The coefficient of $a^6b^8$ in $(a^2 - 2 b)^{11}$ is
divisible
   by $11$. \\
\vspace{1.3in}
   %\emph{Answer:} True. The coefficient of $a^6b^8$ in $(a^2 - 2
%   b)^{11}$ is $\binom {11}{3}(-2)^8= \frac{11!} {(3!)(8!)}(-2)^8$,
%   which is divisible
%   by 11.\\

   d) $3$ has a multuplicative inverse in $\mathbb{Z}_{18}$.\\
\vspace{1.3in}
  % \emph{Answer:} False.  The gcd of 3 and 18 is 3.  Since 3 and 18 are
%   not relatively prime, 3 does not have a multuplicative inverse in
%   $\mathbb{Z}_{18}$.\\

   e)  $3$ is an primitive element of $\mathbb{Z}_{13}$. \\

   %\emph{Answer:} False. The order of $3$ is 3 (since $3^3 = 27
%\equiv_{13} 1$),  not 12.

   \end{problem}
\newpage

\begin{problem}(10 points)
   Find all complex solutions to $x^6 - 4x^3 + 3=0$. Which of these
   solutions are constructible? Why or why not? (You may use the first
   third root of unity, $\omega$, in your answer.)
\end{problem}\\
\newpage
%\emph{Solution.}  $x^6 - 4x^3 + 3 = (x^3)^2 - 4(x^3) + 3 = (x^3
%-1)(x^3 -3)$ so the roots are $1, \omega, \omega^2, \sqrt[3]{3},
%\sqrt[3]{3} \omega$ and  $\sqrt[3]{3} \omega^2$.  The first three are
%constructible but not the second three because of the cube root.
%\\


\begin{problem} (10 points)
   Find the multiplicative inverse of 23 in $\mathbb{Z}_{31}$.
\end{problem}
\newpage
%\emph{Solution.} Using the Euclidean algorithm we have
%\[
%\begin{array}{rcl}
%31 & = & 1(23) + 8 \\
%23 & = & 2(8) + 7 \\
%8 & = & 1(7) + 1
%\end{array}
%\]
%so
%\[
%\begin{array}{rcl}
%1 & = & 8 - 1(7) \\
%& = & 8 - 1[23 - 2(8)] \\
%& = & 3(8) - 1(23) \\
%& = & 3[31 - 1(23)] - 1(23) \\
%& = & 3(31) - 4(23)
%\end{array}
%\]
%and hence 4 is the multiplicative inverse of 23 in
%$\mathbb{Z}_{31}$.\\



\begin{problem} (10 points)
   Show that for all odd $k$, $n^k - n$ is divisible by 3.
\end{problem}
\newpage
%{\it Proof.}  It is enough to show that $n^k - n \equiv_3 0$.  If $n
%\equiv_3 0$ then clearly $n^k - n \equiv_3 0$. So we can assume $n
%\not\equiv_3 0$.  Now we will use the fact that if $n \not\equiv_3 0$
%then $n^2 \equiv_3 1$.  Let $k= 2m+1$.  Then $n^k - n = n^{2m+1}-n =
%n(n^2)^m-n \equiv_3 n(1)^m - n = n - n = 0$. So in either case, $n^k -
%n \equiv_3 0$.\\

\begin{problem}(10 points)
  It happens to be true that 1997 and 1999 are both prime numbers (you
  don't have to check this).  Explain why the polynomial $x^{1997} -
  1$ has no roots in $\mathbb{Z}_{1999}$ other than $x=1$.  (Hint
  think about the order of the root.)
\end{problem}
\newpage

%\emph{Solution.} If there were such a root, its order would be
%exactly 1997 since 1997 has no divisors other than itself and 1.
%But any element of
%$\mathbb{Z}_{1999}$ satisfies the equation $x^{1998} - 1
%\equiv_{1999} 0$, so its order divides 1998.  Clearly 1997 does not
%divide 1998.
%\\

\begin{problem} (15 points)
   Let $a$ and $b$ be nonzero integers such that $g = (a,b)$.  Prove
   that $\displaystyle\left(\frac{a}{g},\frac{b}{g}\right)=1$.
\end{problem}
\newpage
%{\it Proof.} Let $d = \displaystyle\left(\frac ag,\frac bg\right)$.
%Then $\displaystyle \frac ag=dh$ and $\displaystyle \frac bg=dk$ for
%some $h,k\in\Bbb{Z}$. Then $a=gdh$ and $b=gdk$. Hence $gd$ is a common
%divisor of $a$ and $b$.  Since $g$ is the greatest common divisor of
%$a$ and $b$, we have $gd\leq g$. Both $d$ and $g$ are positive
%integers, forcing $d=1$.\\

\begin{problem} (15 points)
   For any prime $p$, if $a^p \equiv_p b^p$ then $a^p \equiv_{p^2}
   b^p$. (Hint: use Proposition 5.3.  )
\end{problem}

%{\it Proof.} Suppose $a^p \equiv_p b^p$. By Fermat's theorem, $a
%\equiv_p a^p \equiv_p b^p \equiv_p b$.  Hence as integers we have $a
%= b+kp$ for some integer $k$.  This means that
%\[
%a^p = (b+kp)^p = b^p + \binom{p}{1} b^{p-1}(kp) + \binom{p}{2}
%b^{p-2}(kp)^2 + \cdots.
%\]
%Since $\binom{p}{1} = p$, the second term is $kb^{p-1}p^2$, which is
%divisible by $p^2$.  All terms after that are of the form
%$\binom{p}{i} b^{p-i}(kp)^i$ for $i \geq 2$, so they are all
%divisible by $p^2$ as well.  Hence $a^p - b^p$ is divisible by $p^2$,
%so $a^p \equiv_{p^2} b^p$.


\end{document}



