\documentclass[12pt]{exam}
\usepackage{amsmath, amsthm, amssymb}
\usepackage{fullpage, paralist}
\usepackage{subfig}
\usepackage{graphicx}
%For diagrams with tikz
\usepackage{tikz-cd}

%For diagrams with xypic
\usepackage[all]{xy}
\CompileMatrices



\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{\cC}{{\mathcal C}}



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

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



\begin{document}

\parskip=\medskipamount \thispagestyle{empty}

\begin{flushright}
  Your Name Goes Here
\end{flushright}
\vspace{.5in}

\centerline{\textbf{Dartmouth College}} \centerline{Mathematics 81/111
  --- Homework 4}

\begin{questions}
  \question We have seen a number of tests to help determine whether a
  polynomial is irreducible, and they will serve you well.  On the
  other hand, neither one size nor one tool fits all problems.

  \begin{parts}
    \part Show that for all $n \ge 1$,
    $f(x) = (x-1)(x-2)\cdots(x-n) -1$ is irreducible in $\Z[x]$.

    \part Show that for all $n \ge 1$ (except $n=4$),
    $f(x) = (x-1)(x-2)\cdots(x-n) +1$ is irreducible in $\Z[x]$.  This
    part requires a bit more persistence than the first part.
  \end{parts}

  
  \question Let $F$ be a field of characteristic 0, and let $m$ and
  $n$ be distinct integers with $\sqrt m \notin F$,
  $\sqrt n \notin F$, and $\sqrt{mn}\notin F$.

  \begin{parts}
    \part Show that $[F(\sqrt m, \sqrt n):F] = 4$.
    
    \part Show by example (with $m/n \notin (\Q^\times)^2$) that the
    above statement can be false if we only assume that
    $\sqrt m \notin F$ and $\sqrt n \notin F$.


    \part Let $m_1, m_2, \dots, m_t$ be square-free integers
    ($m_i \ne 0, \pm1$) which are relatively prime in pairs.  Show
    that $[\Q(\sqrt {m_1}, \sqrt{m_2}, \dots, \sqrt{m_t}):\Q] = 2^t$.
    Hint: A \textit{careful} induction on $t$ may be of use.
  \end{parts}


  \question Show that the class of algebraic field extensions is a
  distinguished class.  Note that `in theory' Lang has a proof of this
  in the text, but at least the second part is incredibly terse. I
  want nice detailed proofs.

  Recall that a class $\cC$ of field extensions is
  \textit{distinguished} if it satisfies three properties:
  \begin{enumerate}
  \item [I.]  Consider a tower of fields $K \subset F \subset E$.  The
    extension $E/K$ is in $\cC$ if and only if $E/F$ and $F/K$ are in
    $\cC$.
  \item [II.] If $E/K$ is in $\cC$, and $F/K$ is any extension of $K$
    (and $E$, $F$ lie in some common field), then $EF/F$ is in $\cC$.

  \item [III.] If $E/K$ and $F/K$ are in $\cC$ (and $E$, $F$ lie in
    some common field), then $EF/K$ is in $\cC$.
  \end{enumerate}

  We have shown that I and II imply III.


  \question Field extensions.

  \begin{parts}
    \part Let $\alpha = \root 11\of5 \in \R$.  Determine (and justify)
    the degree of $\Q(\beta)/\Q$ where
    $\beta = 3 - 2\alpha + 4\alpha^4 - 5\alpha^9$.

    \part Let $n \ge 3$, $\zeta_n = e^{2\pi i /n}$, and consider the
    cyclotomic field $K = \Q(\zeta_n)$, and
    $F = \Q(\zeta_n + \zeta_n^{-1})$.  Show that $F \subset \R$, and
    $[K:F] = 2$.  The field $F$ is called the maximal real subfield of
    $K$.

    \part When $n = 5$ show that $[F:\Q] = 2$, and write
    $F = \Q(\sqrt r)$ for some rational number $r$.  Also write
    $\cos(2\pi/5)$ in terms of radicals of rational numbers.
  \end{parts}

  \question Let $m > 1$ be a square-free integer, and $n \ge 1$ an odd
  integer.  Let $F/\Q$ be any field extension with $[F:\Q] = 2$.  Show
  that $x^n - m$ is irreducible in $F[x]$.

  \question Determine the splitting field over $\Q$ (and its degree)
  of $x^4 + x^2 + 1$.


  \vspace{.5in} The following might also be useful diagrams for you
  \LaTeX ers.  First via xy-pic, then via tikz.

  \[(I) \xymatrix{
      E\ar@{-}[d]\\
      F\ar@{-}[d]\\
      K\\
    }\qquad\qquad(II) \xymatrix{
      &EF&\\
      &&F\ar@{-}[ul]\\
      E\ar@{-}[uur]\ar@{-}[dr]&\\
      &K\ar@{-}[uur]&\\
    }\qquad\qquad(III) \xymatrix{
      &EF&\\
      E\ar@{-}[ur]\ar@{-}[dr]&&F\ar@{-}[ul]\ar@{-}[dl]\\
      &K&\\
    }
  \]

\[
  \xymatrix{
    &EF&\\
    &&G\ar@{-}[ul]\ar@{-}[dr]\\
    E\ar@{-}[uur]&&&F\\
    \\
    &&K\ar@{-}[uur]_{label}\ar@{-}[uull]^{label}&\\
  }
\]

% Now using tikz
\[(I) \begin{tikzcd}
    E\arrow[dash, d]\\
    F\arrow[dash,d]\\
    K\\
  \end{tikzcd}
  \qquad\qquad(II)
  \begin{tikzcd}
    &EF&\\
    &&F\arrow[dash,ul]\\
    E\arrow[dash,uur]\arrow[dash,dr]&\\
    &K\arrow[dash,uur]&\\
  \end{tikzcd}
  \qquad\qquad(III)
  \begin{tikzcd}
    &EF&\\
    E\arrow[dash,ur]\arrow[dash,dr]&&F\arrow[dash,ul]\arrow[dash,dl]\\
    &K&\\
  \end{tikzcd}
\]


% Note the single quote following a label moves it from one side to
% the other
\[
  \begin{tikzcd}
    &EF&\\
    &&G\arrow[dash,ul]\arrow[dash,dr]\\
    E\arrow[dash,uur]&&&F\\
    \\
    &&K\arrow[dash,uur,"label1a", "label1b"']\arrow[dash,uull,"label2a","label2b"']&\\
  \end{tikzcd}
\]

\end{questions}

\end{document}






