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\begin{document}
\centerline{Math 361 lecture for Friday, Week 4}
\bigskip

\centerline{Cyclotomic fields II}
\bigskip

We continue our discussion from last time.  Let~$\zeta=e^{2\pi i/p}$ where
$p\neq0$ is a prime, and consider the cyclotomic field $K=\Q(\zeta)$.  We
saw that the minimal polynomial for $\zeta$ over~$\Q$ is
\[
  f(x)=x^{p-1}+x^{p-2}+\cdots+x+1= \frac{x^{p}-1}{x-1}=\prod_{i=1}^{p-1}
  (x-\zeta^{i}).
\]
So $[K\mathbin{:}\Q]=p-1$, and the embeddings~$K\to \Q$ are given by~$\sigma_{i}\colon \sigma\mapsto
\zeta^{i}$ for $i=1,\ldots,p-1$.

We will need some of the calculations we did last time:
\begin{align*}
  N(\zeta^{j})&= -1\qquad \text{ for all~$i\in\Z$}\\
  N(1-\zeta)&=p
\end{align*}
and
\[
  T(\zeta^{j})=
  \begin{cases}
    -1& \text{if $j\neq 0\bmod p$}\\
    p-1& \text{if $j=0\bmod p$}.
  \end{cases}
\]

\textbf{Our goal} is to show that
\[
  \mathfrak{O}_{K}=\Z[\zeta]=\Span_{\Z}\{1,\zeta,\ldots,\zeta^{p-2}\}.
\]
\begin{proof}
  Let~$\alpha\in\mathfrak{O}_{K}$.  Since~$\{1,\zeta,\ldots,\zeta^{p-2}\}$ is a
  $\Q$-basis for~$K$, we may write
  \[
    \alpha=a_{0}+a_{1}\zeta++\cdots+a_{p-2}\zeta^{p-2}
  \]
  for some unique $a_{0},\ldots,a_{p-2}\in \Q$.  We must show that~$a_{i}\in\Z$
  for all~$i$.

  Last time, we showed that~$b_{i}:=pa_{i}\in\Z$ for all~$i$.  So it suffices to
  show that $p|b_{i}$ for all~$i$.  Defining $\lambda=1-\zeta$, we have
  \begin{align*}
    p\alpha
    &=pa_{0}+pa_{1}\zeta+\cdots+pa_{p-2}\zeta^{p-2}\\
    &=b_{0}+b_{1}\zeta+\cdots+b_{p-2}\zeta^{p-2}\\
    &=b_{0}+b_{1}(1-\lambda)+\cdots+b_{p-2}(1-\lambda)^{p-2}.
  \end{align*}
  If we expand this last equation as a polynomial in~$\lambda$, what are the
  coefficients?  The constant coefficient is
  \[
    b_{0}+b_{1}+\cdots+b_{p-2}.
  \]
  The coefficient of~$\lambda$ is
  \[
    -b_{1}-b_{2}-\cdots-b_{p-2}.
  \]
  The coefficient of $\lambda^{2}$ is
  \[
    b_{2}\binom{2}{2}+b_{3}\binom{3}{2}+\cdots+b_{p-2}\binom{p-2}{2},
  \]
  and so on.  In general, for $i=0,\ldots,p-2$, the coefficient of~$\lambda^{i}$
  is
  \[
    c_{i}:=(-1)^{i}\sum_{j=i}^{p-2} \binom{j}{i}b_{j}\in\Z.
  \]
  So we have
  \begin{align*}
    p\alpha
    &=b_{0}+b_{1}\zeta+\cdots+b_{p-2}\zeta^{p-2}\\
    &=b_{0}+b_{1}(1-\lambda)+\cdots+b_{p-2}(1-\lambda)^{p-2}\\
    &=c_{0}+c_{1}\lambda+\cdots+c_{p-2}\lambda^{p-2}.
  \end{align*}
  By symmetry, since~$\zeta=1-\lambda$, we have
  \[
    b_{i}=(-1)^{i}\sum_{j=i}^{p-2} \binom{j}{i}c_{j}.
  \]
  Note that~$p$ does not divide any $\binom{j}{i}$ appearing in above.  Hence,
  to achieve our goal of proving $p|b_{i}$ for each~$i$, it suffice to show
  that~$p|c_{i}$ for each~$i$, which we now do by induction.  For the
  case~$i=0$, we have
  \[
    c_{0}:=(-1)^{0}\sum_{j=0}^{p-2}
    \binom{j}{0}b_{j}=b_{0}+\cdots+b_{p-2}=p(a_{0}+\cdots+a_{p-2}).
  \]
  We can not immediately conclude that $p|c_{0}$ since all we know about
  the~$a_{i}$ at this point is that they are rational numbers. However, we can
  use the fact that since~$\alpha\in\mathfrak{O}_{K}$, we know its trace is an
  integer.  Calculate:
  \begin{align*}
    T(\alpha)
    &= T(a_{0}+a_{1}\zeta+\cdots+a_{p-2}\zeta^{p-2})\\
    &= T(a_{0})+T(a_{1}\zeta)+\cdots+T(a_{p-2}\zeta^{p-2}),
  \end{align*}
  and since each $\sigma_{i}$ is the identity on $\Q$,
  \[
    T(a_{k}\zeta^{k})
    =\sum_{i=1}^{p-2} \sigma_{i}(a_{k}\zeta^{k})
    =\sum_{i=1}^{p-2} a_{i}\sigma_{i}(\zeta^{k})
    =\sum_{i=1}^{p-2} a_k T(\zeta^{k}).
  \]
  for each $k=0,\ldots,p-2$. It follows from our earlier trace calculations that
  \begin{align*}
    T(\alpha)&=a_{0}(p-1)-a_{1}-\cdots-a_{p-2}\\
    &=pa_{0}-a_{0}-a_{1}-\cdots-a_{p-2}\\
    &=b_{0}-(a_{0}+\cdots+a_{p-2}).
  \end{align*}
  Since~$T(\alpha)$ and~$b_{0}$ are integers, so is~$a_{0}+\cdots+a_{p-2}$.
  Thus, we may finally conclude that $p|c_{0}$.

  We proceed with the induction step. Let $1\leq k\leq p-2$, and suppose
  that~$p|c_{i}$ for $i< k$.  Thus, we can write $c_{i}=pn_{i}$ for some
  $n_{i}\in \Z$ for $i=0,\ldots,k-1$.  Thus,
  \begin{align*}
    p\alpha &= \sum_{i=1}^{p-2} c_{i}\lambda^{i}\\
    &=pn_{0}+pn_{1}\lambda+\cdots+pn_{k-1}\lambda^{k-1}+\sum_{i=k}^{p-2}
    c_{i}\lambda^{i}.
  \end{align*}

  We now factor~$p$ in~$\mathfrak{O}_{K}$, defining~$\gamma$ as follows:
  \[
    p=N(1-\zeta)=\prod_{i=1}^{p-1}
    (1-\zeta^{i})=(1-\zeta)^{p-1}\underbrace{\prod_{i=1}^{p-1}1+\zeta+\cdots+\zeta^{i-1}}_{\gamma}.
  \]
  Here, we are using the fact
  that~$\frac{1-\zeta^{i}}{1-\zeta}=1+\zeta+\cdots+\zeta^{i-1}$.
  Note that $(1-\zeta)^{p-1}$ and $\gamma$ are in~$\mathfrak{O}_{K}$.  Hence,
  factors in $\mathfrak{O}_{K}$ as
  \[
    p = (1-\zeta)^{p-1}\gamma = \lambda^{p-1}\gamma.
  \]
  Continuing with our above calculation of~$p\alpha$:
  \begin{align*}
   \lambda^{p-1}\gamma \alpha
   &=p\alpha\\
    &=pn_{0}+pn_{1}\lambda+\cdots+pn_{k-1}\lambda^{k-1}+\sum_{i=k}^{p-2} c_{i}\lambda^{i}\\
    &=\lambda^{p-1}+\lambda^{p-1}\gamma
    n_{1}\lambda+\cdots+\lambda^{p-1}\gamma n_{k-1}\lambda^{k-1}+\sum_{i=k}^{p-2} c_{i}\lambda^{i}\\
  \end{align*}
  Solving for the $c_{k}\lambda^{k}$ term, we see
  \[
    c^{k}\lambda^{k}=\lambda^{k+1}\mu
  \]
  for some $\mu\in\mathfrak{O}_{K}$.  It follows that~$c_{k}=\lambda\mu$.  Take
  norms:
  \[
    c_{k}^{p-1}=N(c_{k})=N(\lambda)N(\mu)=p N(\mu)
  \]
  where~$N(\mu)\in\Z$.  Since $p$ divides the integer~$c_{k}^{p-1}$ and~$p$ is
  prime, it follows that~$p|c_{k}$.  That completes the induction and the proof.
\end{proof}
\end{document}
