\documentclass[12pt]{exam}
\usepackage{amsmath, amsthm, amssymb}
\usepackage{fullpage, paralist}
\usepackage{subfig}
\usepackage{graphicx}


\newcommand{\Z}{{\mathbb Z}}
\newcommand{\Q}{{\mathbb Q}}
\newcommand{\R}{{\mathbb R}}
\newcommand{\C}{{\mathbb C}}

\newcommand{\A}{{\mathcal A}}
\newcommand{\I}{{\mathcal I}}


\newcommand{\ds}{\displaystyle}
\newcommand{\la}{\langle}
\newcommand{\ra}{\rangle}
\newtheorem{thm}{Theorem}[section]

\theoremstyle{remark}
\newtheorem*{nnremark}{Remark}

\printanswers
%\noprintanswers
\unframedsolutions


\begin{document}

\parskip=\medskipamount \thispagestyle{empty}


\centerline{\textbf{Dartmouth College}} \centerline{Mathematics 81/111
  --- Homework 1}

\begin{questions}

\question In class (cf. [S] Corollary 7.3.13), we showed that if $A$
  is an integral domain, then any polynomial $f\in A[x]$ of degree $n$
  has at most $n$ distinct roots in $A$.  The crux of the argument is
  if $a$ is a root of $f$, then $f = (x-a) g(x)$ for some $g \in A[x]$
  having degree $n-1$.  So if $b\ne a$ is another root of $f$, then
  $0=f(b) = (b-a)g(b)$, and since $A$ is has no zero divisors,
  $g(b) =0$ and the result follows by induction.

  So now let $A = \mathbb H$ be Hamilton's quaternions.  $\mathbb H$
  is a four-dimensional vector space over $\R$ with basis
  $\{1, i, j, k\}$.  It is made into a non-commutative division ring
  via the relations $i^2 = j^2=k^2 = -1$ and $ij = k = -ji$.  More
  information is in the exercises to \S7.1 of [S].

  So let $f(x) = x^2 + 1$.  Clearly $f(\pm i)=f(\pm j)=f(\pm k) = 0$.
  Mimicking the argument above, we see that $i$ is a root of $f$, so
  we write $f = x^2 + 1 = (x-i) g(x) = (x-i)(x+i)$.  Now $j$ is
  another root of $f$ and $j\ne i$ (why?) and if we write
  $0 = f(j) = (j-i)g(j) = (j-i)(j+i)$, we are supposed to conclude
  that $g(j) = j+i = 0$, but that is clearly not true (again, why?).
  So the main question is where has the argument failed and why?


\question Let $\varphi: A\to B$ be a ring homomorphism.
  \begin{parts}
  \part Suppose that $J \subseteq B$ is an ideal. Show
    \begin{itemize}
    \item $\varphi^{-1}(J)$ is an ideal of $A$.
    \item $\varphi(\varphi^{-1}(J)) \subseteq J$, and give an example
      to show equality need not hold.
    \item Show that equality does hold if $\varphi$ is surjective.
    \end{itemize}
  \part Next let $I \subset A$ be an ideal.
    \begin{itemize}
    \item Show by example that $\varphi(I)$ need not be an ideal of
      $B$, but is an ideal if $\varphi$ is surjective.
    \item Assuming that $\varphi$ is surjective, determine
      $\varphi^{-1}(\varphi(I))$.
    \end{itemize}
  \part Use the above to show that there is a one-to-one
    correspondence between the ideals of $A/I$ and the ideals of $A$
    which contain $I$.
  \end{parts}


\question Let $A$ be a ring and $I$ a proper ideal in $A$.  Let
  $\pi :A\to A/I$ be the canonical map.  For $f \in A[x]$, let $f^\pi$
  (Lang's notation) denote the image of $f$ in $(A/I)[x]$.
  \begin{parts}
  \part Show that $A[x]/(I, f(x)) \cong (A/I)[x]/(f^\pi)$.
  \part Use the previous part to that if $p\in \Z$ is a prime, then
    $(x,p)$ is a maximal ideal in $\Z[x]$.
  \part Characterize all ideals $I$ with
    $(x) \subseteq I \subseteq \Z[x]$.  There is certainly more than
    one way to proceed.
  \end{parts}

\question Let $D$ be an integral domain, and $a, b \in D$ not both
  zero.  Recall that we say that $d \in D$ is \textbf{a gcd} of $a$
  and $b$ if $d \mid a$, $d\mid b$ and if $c$ is any other common
  divisor of $a$ and $b$ then $c \mid d$.  We know (or will show) that
  $\Q[x]$ is a UFD, so in particular gcds exist.  Let
  $A = \Q[x^2, x^3]$, the subring of $\Q[x]$ generated by $x^2$ and
  $x^3$.
  \begin{parts}
  \part Show that $A$ is not a UFD by showing the $\gcd(x^5, x^6)$
    does not exist in $A$.

  \part Determine the structure of $\Q[x^2,x^3]/(x^3)$ as a
    $\Q$-algebra. You might start by determining its structure as a
    vector space over $\Q$.
  \end{parts}


\question Let $\zeta_n = e^{2\pi i/n} \in \C$ be a primitive $n$th
  root of unity, that it a generator for the cyclic group, $\mu_n$, of
  $nth$ roots of unity (all the complex roots of $x^n -1$).
  \begin{parts}
  \part Suppose that $k, m$ are integers relatively prime to $n$.
    Show that $\ds \frac{\zeta_n^m-1}{\zeta_n^k - 1}$ is a unit in the
    ring $\Z[\zeta_n]$.
  \part Show that if $n >3$ is odd, then $(\zeta_n^2-1)/(\zeta_n -1)$
    has infinite order in $\Z[\zeta_n]^\times$.
  \end{parts}
\end{questions}

\end{document}







%%% Local Variables:
%%% mode: latex
%%% TeX-master: t
%%% End:
