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\prof{Peter Cholak and Juan Migliore}
\class{Math 222}
\quizdate{Monday, February 19, 2001}
\pagestyle{empty}

\begin{document}

\noindent Quiz 3

\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}
  Consider the set of integers modulo 10, $\mathbb{Z}_{10}$.\\ (a)
  Identify the additive inverse of each element in
  $\mathbb{Z}_{10}$.\\
  (b) Identify those elements that have a
  multiplicative inverse and what their inverses are.\\
\end{problem}

{\it Solution.} (a) The additive inverse of $j$ is $10-j$ for
$j\in\Bbb{Z}_{10}$.\par (b) Each of $1,3,7,9$ has a multiplicative
inverse. The inverse of 1 is 1, the inverse of 3 is 7, the inverse of
7 is 3 and
the inverse of 9 is 9.\\

\begin{problem}
  Let $n$ be a positive integer and consider the sum $1 + 2 + 3 +
  \dots + (n-1)$ modulo $n$.  Show that if $n$ is odd, this sum is
  zero in $\mathbb{Z}_n$ and if $n$ is even then the sum is $n/2$ in
  $\mathbb{Z}_n$.  {\em Hint:} Use the formula for this sum which we
  saw when discussing induction.
\end{problem}

{\it Solution.} We know that $S=1 + 2 + 3 + \dots +
(n-1)=\frac{n(n-1)}2$. If $n$ is odd, then $k=\frac{n-1}2\in\Bbb{Z}$
and hence $S=kn\equiv_n 0$, that is, $S$ is zero in $\Bbb{Z}_n$. If
$n$ is even, say $n=2m$, then $S=(n-1)m\equiv_n nm-m\equiv_n
-m\equiv_n m\equiv_n \frac n2$, so $S$ is
$n/2$ in $\Bbb{Z}_n$.\\

\begin{problem}
  In $\mathbb{Z}_{66}$, consider the elements 6, 8, 9, 15, 35 and 55.
  Identify the one that has a multiplicative inverse in
  $\mathbb{Z}_{66}$ and find that inverse.
\end{problem}

\noindent{\it Solution.} The only one whose greatest common divisor 
with 66 is 1 is 35.  (The other greatest common divisors are,
respectively, 6, 2, 3, 3 and 11.)  Using the Euclidean algorithm we
get
\[
\begin{array}{rcl}
66 & = & 1(35) + 31 \\
35 & = & 1(31) + 4 \\
31 & = & 7(4) + 3 \\
4 & = & 1(3) + 1.
\end{array}
\]
Hence $1 = 4 - 1(3) = 4 - [31 - 7(4)] = -31 + 8(4) = -31 + 8[35 - 31]
= -9(31) + 8(35) = -9[66-35] + 8(35) = -9(66) + 17(35) $.  So the
multiplicative inverse of $35$ in $\mathbb{Z}_{66}$ is 17.  \\

\begin{problem}
  Prove that if $p$ is a prime and $\alpha, \beta \in \sqrt[p]{1}$ and
  $\alpha \neq 1$ then there exists an integer $m$ such that $\alpha^m
  = \beta$.  (Hints: First, write $\alpha = \zeta^k$ and $\beta =
  \zeta^r$ where $\zeta$ is the first $p$th roof of unity.  Second, do
  $k$ and $r$ have multiplicative inverses in $\mathbb{Z}_{p}$?)
\end{problem}

\noindent {\it Proof.} Write $\alpha=\zeta^k$ and $\beta=\zeta^r$ with $0\leq
k,r<p$ and $\zeta$ the first $p$th root of unity. By the assumption
$\alpha\ne 1$, we have $0<k<p$. By Proposition 4.3, $k$ has a
multiplicative inverse, say $l$, in $\Bbb{Z}_p$. Then $kl=qp+1$ for
some $q\in\Bbb{Z}$. Let $m=lr$. Then
$$\alpha^m=(\zeta^k)^{lr}=(\zeta^{kl})^r=(\zeta^{qp+1})^r=
((\zeta^p)^q\cdot\zeta)^r=(1\cdot\zeta)^r=\beta.$$


\end{document}

