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\prof{Peter Cholak and Juan Migliore} \class{Math 222}
\quizdate{Wednesday, April 11, 2001} \pagestyle{empty}

\begin{document}

\noindent Quiz 6

\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 7.3, Number 10 (slightly modified) (5 points): 
\begin{itemize}
\item[a.] For any element $a\in GF(p,P(x))$, let $b \in GF(p,P(x))$
be an element such that $b^{p-1} = a$.  Show that then $(kb)^{p-1} =
a$ for $1 \leq k \leq p-1$.

\item[b.] For any element $a\in GF(p,P(x))$, let  $r(a)$ denote the
number of distinct elements $b$ in $GF(p,P(x))$  such that
$b^{p-1}=a$. Prove that if $a\ne 0$, then $r(a)=0$ or  $p-1$.  (You
have to show that these are the {\em only} two possibilities.)
\end{itemize}
\end{problem}


{\it Proof.} (a) We know that if $1 \leq k \leq p-1$ then $k^{(p-1)}
\equiv_p 1$, so $(kb)^{p-1} = b^{p-1} = a$.
For (b) we must show that {\em if}  there exists some $b\in
GF(p,P(x))$ with $b^{p-1}=a$, then $r(a)=p-1$.  (If there doesn't
exist any such element then $r(a) = 0$.) Let $b$ be such an element.
Then by (a), $b,2b,\cdots,(p-1)b$ are distinct solutions of the
polynomial equation $x^{p-1}-a=0$ in $GF(p,P(x))$.  But
$\deg(x^{p-1}-a)=p-1$; this equation has at most
$p-1$ distinct solutions by Proposition 6.7. So
we get $r(a)= p-1$.\\

%\begin{problem}
%  Let $p$ be any prime and $n$ any positive integer.  Prove the
%  polynomial $x^n-1$ has repeated zeros in $\mathbb{Z}_p$ iff $p$ is a
%  factor of $n$.
%\end{problem}

\begin{problem} 
  Section 8.2, Problems 35 and 36:
\begin{itemize}  
\item[a.] (5 points) Prove that if the set $S$ is finite and $\sigma$
is a function of $S$ into itself, then the following are equivalent:

\begin{itemize}
\item[(i)] If $x_1$ and $x_2$ are distinct elements of $S$ then
$\sigma(x_1) \neq \sigma(x_2)$ (i.e.\ $\sigma$ is {\em one-to-one}).

\item[(ii)] If $y$ is any element of $S$ then there is an element $x$
in $S$ such that $y = \sigma(x)$ (i.e.\ $\sigma$ is {\em onto}).

\end{itemize}

\noindent (Hint: consider the set $\sigma (S) = \{ \sigma(x) \ | \ x
\in S \}$.)

\item[b.] (5 points)  Show, by means of examples, that when $S$ is an
infinite set then neither of the conditions (i) or (ii)
necessarily implies the other.  Specifically, give an example where 

\begin{itemize}

\item[$\bullet$] (i) is true but (ii) is false;

\item[$\bullet$] (ii) is true but (i) is false;

\item[$\bullet$] both (i) and (ii) are true.

\end{itemize}
\end{itemize}
\end{problem}

{\em Solution.} 
First we'll do (a).  If (i) holds then the number of elements of
$\sigma(S)$ is the same as the number of elements of $S$.  Since
$\sigma(S)
\subset S$, this means $\sigma(S) = S$ (since $S$ is finite), i.e.\
$\sigma$ is onto.  Conversely, if (ii) holds then $\sigma(S) = S$, so
the number of elements is the same.  So $\sigma$ can't collapse any
elements $x_1$ and $x_2$  to the same $y$, i.e.\ $\sigma$ is
one-to-one.

For (b), let $S = {\mathbb R}$.  The funtion $\sigma(x) = e^x$ is
one-to-one but not onto.  The function $\sigma(x) = x^3-x$ is onto
(use calculus) but not one-to-one (since $\sigma(0) = \sigma(1) =
0$).  The function $\sigma(x) = x$ is both one-to-one and onto. \\

\medskip

\begin{problem} (5 points)
  Let $\sigma = \left(
  \begin{array}{ccccccccc}
    1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
    2 & 3 & 4 & 1 & 6 & 7 & 5 & 9 & 8
  \end{array} \right)$ and  $\tau = \left(
  \begin{array}{ccccccccc}
    1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
    9 & 3 & 4 & 2 & 7 & 5 & 1 & 6 & 8
  \end{array} \right)$. 
\begin{itemize}
\item[a.]  Express $\sigma$, $\tau$, $\sigma^{-1}$ and $\sigma\tau$
in disjoint cycle notation.

\item[b.]  
  Find the orders of $\sigma$, $\tau$, and $\sigma\tau$.
 
\item[c.] Write $\sigma$ and $\tau$ as a product of transpositions.
\item[d.]  Find the parity of $\sigma$, $\tau$, $\sigma\tau$ and
  $\sigma\tau^2$.
\end{itemize}
\end{problem}

{\it Solution.} (a) $\sigma=(1234)(567)(89)$, $\tau=(198657)(234)$. \\
$\sigma^{-1}=(4321)(765)(98)$, $\sigma\tau=\pmatrix
1&2&3&4&5&6&7&8&9\\ 8&4&1&3&5&6&2&7&9\endpmatrix=(187243)$.  \\ 
(b) $o(\sigma)=12$,
$o(\tau)=6$, and $o(\sigma\tau)=6$. \\  (c) $\sigma=(1\ 2\ 3\ 4)(5\
6\  7)(8 \ 9)= (1\ 2)(2\ 3)(3\ 4)(5\ 6)(6\ 7)(8\ 9)$. $\tau=(1\ 9\ 8\
6\  5\ 7)(2\ 3\ 4)=(1\ 9)(9\ 8)(8\ 6)(6\ 5)(5\ 7) (2\ 3)(3\ 4)$. \\
(d)
$\sigma$ is even and $\tau$ is odd.  Then $\sigma\tau$ is odd and
$\sigma\tau^2$ is even.


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