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\begin{flushright}
  Your Name Here
\end{flushright}


\begin{center}
  \textbf{Math 105}\\ Homework 1
\end{center}


\begin{enumerate}
\item Let $K$ be a number field of degree $n$ over $\Q$, $q$ a prime
  in $\Z$, and suppose that \newline
  $q\O_K = Q_1^{e_1}\cdots Q_r^{e_r}$ is the factorization of $q\O_K$
  as a product of prime ideals in $\O_K$.  Show that
  \[[\O_K:q\O_K] = q^n = \prod_{i=1}^r [\O_K:Q_i]^{e_i}.
  \]
  Note that this proves that $n = [K:\Q] = \sum_{i=1}^r e_i f_i$ where
  $[\O_K:Q_i]=q^{f_i}$.

\item Let $K$ be a number field, $[K:\Q]=n$ and let $\alpha \in \O_K$
  with $K = \Q(\alpha)$.  It is not always the case that the ring of
  integers of a number field has the form $\Z[\alpha]$, but we can
  often show that it is by gaining information about how large
  $\Z[\alpha]$ is as a subring of $\O_K$.

  Suppose that $[\O_K:\Z[\alpha]] = m$.  Show that
  $\Delta(1, \alpha, \dots, \alpha^{n-1}) = m^2 d_K$ where $d_K$ is
  the discriminant of $K$.

\item Let $L/K$ be an extension of number fields with $n = [L:K]$.
  Suppose that $L = K(\alpha)$, and put $f = m_{\alpha,K}$ be its
  minimal polynomial.  Show that
  \[\Delta(1, \alpha, \dots, \alpha^{n-1}) =
    (-1)^{\frac{n(n-1)}{2}}N_{L/K}(f'(\alpha)),
  \]
  where $N_{L/K}(\alpha) = \prod_{\sigma\in E} \sigma(\alpha)$ is the
  norm from $L$ to $K$ and $f'$ is the formal derivative of $f$; here
  $E$ is the set of all embedding of $L$ into an algebraic closure of
  $K$ which fix $K$ pointwise.

\item Let $K$ be a number field with $K = \Q(\alpha)$ for some
  $\alpha\in \O_K$, and suppose that\newline $[\O_K:\Z[\alpha]] = m$.
  Let
  $f = m_{\alpha,\Q} = x^n + a_{n-1}x^{n-1} + \cdots + a_0 \in \Z[x]$
  be the minimal polynomial of $\alpha$.  Show that if $f$ is
  Eisenstein with respect to some rational prime $p$ ($p \mid a_i$ for
  all~$i$, but $p^2 \nmid a_0$), then $p \nmid m$.

  Some hints.
  \begin{itemize}
  \item First show that the Eisenstein condition implies that
    $\alpha^n/p \in \O_K$ and that\newline
    $p^2 \nmid N_{K/\Q}(\alpha)$.
  \item Next assume that $p\mid m = [\O_K:\Z[\alpha]]$. We shall try
    to obtain a contradiction.  To begin, show there exists
    $\xi =\O_K\setminus \Z[\alpha]$ with
    $p\xi = b_0 + b_1\alpha + \cdots + b_{n-1}\alpha^{n-1}\in
    \Z[\alpha]$, and not all $b_i$ divisible by $p$.
  \item Let $j$ be the smallest index with $p \nmid b_j$ and put
    $\eta= (b_j\alpha^j + \cdots + b_{n-1}\alpha^{n-1})p^{-1}$.  Show
    that $\eta$ and $\zeta= b_j\alpha^{n-1} p^{-1}$ are elements of
    $\O_K$.
  \item Explore $N_{K/\Q}(p\zeta)$ to produce the contradiction.
  \end{itemize}

  \vfill \rightline{(continued on reverse)}
  \newpage
            
\item Let $K = \Q(\root3\of 5)$.  The goal is to prove that
  $\O_K = \Z[\root3\of 5]$ using some of the tools you have developed
  in the problems above.
  \begin{enumerate}
  \item Show that the discriminant
    $\Delta(1, \root3\of 5, \root 3\of {25}) = -3^3\cdot 5^2$. Note
    that this implies an upper bound on the index
    $[\O_K:\Z[\root3\of5]]$.
  \item Note that $f = x^3-5$ is Eisenstein with respect to the prime
    5, and $g = (x+5)^3 -5$ is Eisenstein with respect to the prime 3.
    How can you use this to reach the desired conclusion?
  \end{enumerate}

\item Let $p$ be a prime in $\Z$, $\zeta$ a primitive $p^n$th root of
  unity in $\C$, and put $K = \Q(\zeta)$.  Show that $\O_K=\Z[\zeta]$.

  Some hints.

  \begin{itemize}
  \item Show that the discriminant $d_K = (-1)^{p(p-1)/2}p^N$ where
    $N= n\phi(p^n) - p^{n-1}$ where $\phi$ is the Euler totient
    function.  Recall that the minimal polynomial of $\zeta$ is the
    $p^n$th cyclotomic polynomial which for prime powers is easy to
    compute.

  \item Then show that the element $(1-\zeta)$ has minimal polynomial
    which is Eisenstein with respect to the prime $p$.
  \end{itemize}
\end{enumerate}

\end{document}


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