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\prof{Peter Cholak and Juan Migliore} \class{Math 222}
\quizdate{Friday, March 30, 2001} \pagestyle{empty}

\begin{document}

\noindent Quiz 5

\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}
  Let $F$ be a field. Show that there exist $a,b\in F$ such that
  $x^2+2$ is a divisor of $x^{43}+ax+b$. (Hint: consider the form of
  the remainder $r(x)$ when $x^{43}$ is divided by $x^2+2$. Do not do
  the actual division. The degree of $r(x)$ is ??)
\end{problem}

%We can write $x^{43}=q(x)(x^2+2)+r(x)$ such that $q(x),r(x)\in F[x]$
%and $0\leq \deg r(x)\leq 1$. So $r(x)=cx+d$ for some $c,d\in F$. Take
%$a=-c$ and $b=-d$, we get $x^{43}+ax+b=x^{43}-cx-d=q(x)(x^2+2)$, which
%is divided by $x^2+2$.\\


\begin{problem}
  Factor $x^3+3x+1$ over $\mathbb{Z}_5$ into irreducible factors.
\end{problem}

%By a direct calculation, we see that $x=1$ and $x=2$ are solutions of
%the equation $x^3+3x+1$ mod 5. Dividing $x^3+3x+1$ by $(x-1)(x-2)$, we
%get the quotient $x-2$. So $x^3+3x+1=(x-1)(x-2)^2$.\\

\begin{problem}
  Consider the Galois Field $F = GF(3,x^2 + x + 2)$.  Let $\alpha$ be
  the associated Galois imaginary.

  (a) Show that $\alpha$ is a primitive element in $F$. Work out the
  corresponding cyclic table of $F$.

  (b) Find the inverse of each nonzero element in $F$.  {\em Hint:}
  Use part (a).

  (c) By definition $\alpha$ is one solution to $x^2 + x + 2 = 0$ over
  $\mathbb{Z}_3$.  There should of course be another solution.  It is
  also an element of $F$.  Find this element.  Is it a power of
  $\alpha$?  {\em Hint:} Use long division or try the other
  possibilities.
\end{problem}

%(a) $\alpha^2=2\alpha+1$, $\alpha^3=2\alpha+2$, $\alpha^4=2$,
%$\alpha^5=2\alpha$, $\alpha^6=\alpha+2$, $\alpha^7=\alpha+1$ and
%$\alpha^8=1$. The order of $\alpha$ is 8. Since $\alpha$ is a
%primitive element in $F$, the corresponding cyclic table of $F$ is as
%follows:

%\begin{equation}
%  \begin{split}
%    \alpha^1 &=  \alpha \\
%    \alpha^2 &=  2\alpha + 1\\
%    \alpha^3 &=  2\alpha + 2\\
%    \alpha^4 &= 2 \\
%    \alpha^5 &=  2\alpha \\
%    \alpha^6 &=  \alpha + 2\\
%    \alpha^7 &=  \alpha + 1\\
%    \alpha^8 &=  1 \\
%  \end{split}
%\end{equation}

%(b) We have $(\alpha^h)^{-1}=\alpha^{-h}=\alpha^{8-h}$ for any integer
%$h$. So $\alpha^{-1}=\alpha+1$, $(2\alpha+1)^{-1}=\alpha+2$,
%$(2\alpha+2)^{-1}=2\alpha$, $2^{-1}=2$, $(2\alpha)^{-1}=2\alpha+2$,
%$(\alpha+2)^{-1}+2\alpha+1$, $\alpha+1)^{-1}=\alpha$ and
%$1^{-1}=1$.\par

%(c) Dividing $x^2+x+2$ by $x-\alpha$, we get the quotient
%$x-(2\alpha+2)$. So $2\alpha+2$ is another solution to $x^2 + x + 2 =
%0$ over
%$\Bbb{Z}_3$.\\

\begin{problem}
  Problem 14 from Section 7.2: Let $p$ be a prime number, and let $n$
  be relatively prime to $p^\nu-1$. Prove that there is exactly one
  $n$th root of unity in $GF(p,P(x))$, where $\nu=\deg P(x)$.
\end{problem}

%Suppose $\zeta$ is an $n$th root of unity in $F = GF(p,P(x))$.  Then
%$\zeta^n = 1$ in $F$. Hence the order of $\zeta$ divides both $n$ and
%$p^v-1$ (this follows from Theorem 7.5).  Therefore the order of
%$\zeta$ is $1$.  But $1$ is the only element of $F$ with order 1.
%Thus $\zeta = 1$.

%\begin{problem}
%  Problem 4 from Section 6.3: Find a greatest common divisor of the
%  polynomials $P(x)=x^6+x^5+x^4+2$ and
%  $Q(x)=x^7+x^6+x^5+2x^3+x^2+2x+1$ over $\Bbb{Z}_3$.
%\end{problem}

%We have $Q(x)=x\cdot P(x)+(2x^3+x^2+1)$ and
%$P(x)=(2x^2+x+2)(2x^3+x^2+1)$ over $\Bbb{Z}_3$. Let $r(x)=
%2x^3+x^2+1$. Then a greatest common divisor of $Q(x)$ and $P(x)$ is
%also a greatest common divisor of $P(x)$ and $r(x)$, which is $r(x)$.\\


\end{document}

