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

\centerline{Cyclotomic fields I}
\bigskip

Let 
\[
  \zeta_{m}=e^{2\pi i/m}=\cos(2\pi/m)+i\sin(2\pi/m),
\]
for $m\in \Z_{\geq 2}$, and consider the
field~$K=\Q(\zeta_{m})$.  The powers of~$\zeta_{m}$ are the \emph{$m$-th roots
of unity}:
\[
  x^{m}-1 =\prod_{k=1}^{m} (x-\zeta^{k}). 
\]
Since~$\zeta_{m}$ satisfies a monic polynomial with integer coefficients, it is
an algebraic integer.
\begin{eg}\ 
  \begin{enumerate} 
    \item \textbf{Case $m=2$.}  We have $\zeta_{2}=-1$ and $K=\Q$.  
    \item \textbf{Case $m=3$.} We have
      \begin{align*}
        \zeta_{3} 
	&= \cos(2\pi/3)+i\sin(2\pi/3)=\frac{1}{2}+i
	\frac{\sqrt{3}}{2}=\frac{1+i\sqrt{3}}{2}\\
	x^3-1
	&= (x-1)(x^2+x+1),
      \end{align*}
      and the minimal polynomial for~$\zeta_{3}$ is $x^2+x+1$. So
      $[\Q(\zeta_{3})\mathbin{:}\Q]=2$.
    \item \textbf{Case $m=4$.} We have
      \begin{align*}
        \zeta_{4} 
	&= \cos(2\pi/4)+i\sin(2\pi/4)=i\\
	x^4-1
	&= (x^2-1)(x^2+1)=(x-1)(x+1)(x^2+1)
      \end{align*}
      and the minimal polynomial for~$\zeta_{4}$ is $x^2+1$. So
      $[\Q(\zeta_{3})\mathbin{:}\Q]=2$, also.
  \end{enumerate}
\end{eg}

\begin{thm}\label{thm:04wed}  We have
  \[
    [\Q(\zeta_{m})\mathbin{:}\Q]=\phi(m)
  \]
  where~$\phi$ is the Euler totient function:
  \[
    \phi(m)=|\{a: 1\leq a<m \text{ and } \gcd(a,m)=1\}| =
    m\prod_{\substack{p|n\\\text{$p$ prime}}}\left(1-\frac{1}{p}\right).
  \]
  Further, $\mathfrak{O}_{\Q(\zeta_{m})}$ has integral basis
  $1,\zeta_{m},\zeta_{m}^{2},\ldots,\zeta_{m}^{\phi(m)-1}$, i.e.,
  $\mathfrak{O}_{\Q(\zeta_{m})}=\Q[\zeta_{m}]$. 
\end{thm}

\begin{eg}
  Let $m=4$. Then numbers $1$ and~$3$ are relatively prime
  to $m$.  Therefore,
  \[
    \phi(4)=2= 4\left(1-\frac{1}{2}\right).
  \]
  The ring of integers in~$\Q(i)$ is $\Z[i]=\Span_{\Z}\{1,i\}$.
\end{eg}

Our next goal is to prove the above theorem in the case where $m$ is
an odd prime.  \textbf{For the rest of this lecture}, let
\[
  \zeta=\zeta_{p} =e^{2\pi i/p}
\]
where~$p$ is an prime.  (In the case~$p=2$, we have $\zeta=1$ and
$\Q(\zeta)=\Q$.)
\bigskip

\textbf{Minimal polynomial of $\zeta$.}  We use the following criterion for
irreducibility to find the minimal polynomial for~$\zeta$.

\begin{thm}
  (Eisenstein's criterion.) Let
  \[
    f = a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0}\in\Z[x].
  \]
  Suppose there is a prime~$q\in\Z$ such that
  \begin{itemize}
    \item $q|a_{i}$ for $i=0,1,\ldots,n-1$,
    \item $q\nmid a_{n}$, and
    \item $q^2\nmid a_{0}$.
  \end{itemize}
  Then up to a constant factor,~$f$ is irreducible in~$\Z[x]$ and, hence, $f$ is
  irreducible in $\Q[x]$.
\end{thm}
\begin{proof}
 See Theorem 1.8 in our text for the (easy) proof.
\end{proof}

\begin{prop}
  The minimal polynomial for~$\zeta$ is $f(x)=x^{p-1}+x^{p-2}+\cdots+x+1$.
\end{prop}
\begin{proof}
  First note that
  \[
    x^{p-1}+x^{p-2}+\cdots+x+1 = \frac{x^{p}-1}{x-1},
  \]
  and, thus, all of the $p$-th roots of unity except~$1$ are zeros of~$f$.
  So it remains to show that $f$ is irreducible over $\Q$.
  For that, it suffices to show that~$f(x+1)$ is irreducible since $f(x)=g(x)h(x)$ if and
  only if $f(x+1)=g(x+1)h(x+1)$.  We have
  \begin{align*}
    f(x+1)
    &= \frac{(x+1)^{p}-1}{(x+1)-1}\\[8pt]
    &= \frac{x^{p}+\binom{p}{p-1}x^{p-2}+\cdots+\binom{p}{1}x+1 -1}{x}\\[8pt]
    &=x^{p-1}+\binom{p}{p-1}x^{p-2}+\binom{p}{p-2}x^{p-3}+\cdots+\binom{p}{1}.
  \end{align*}
  Eisenstein's criterion now applies. Note that
  $\binom{p}{k}=\frac{p!}{k!(p-k)!}$ is divisible by $p$ for $1\leq k\leq p-1$.
\end{proof}

\begin{cor}
  We have $[K\mathbin{:}\Q]=p-1$ and $\left\{1,\zeta,\ldots,\zeta^{p-2}\right\}$
  is a $\Q$-basis for~$K$.
\end{cor}

Up to this point we have proved almost all of Theorem~\ref{thm:04wed} for the case
where $m$ is prime.  The only thing that is left is to show
$1,\zeta,\ldots,\zeta^{p-2}$ is a $\Z$-basis $\mathfrak{O}_{\Q(\zeta)}$. 
For that, we will need to calculate some norms and traces and to prove a useful
lemma.  The complete proof of this special case of Theorem~\ref{thm:04wed} will
then be finished in the next lecture.
\bigskip

\textbf{Some norms and traces.}  The minimal polynomial for~$\zeta$ factors as
\[
  f(x)=\prod_{i=1}^{n-1} (x-\zeta^{i}). 
\]
Therefore, the field embeddings are given by
\begin{align*}
  \sigma_{i}\colon K=\Q(\zeta)&\to K\subset \C\\
  \zeta&\mapsto \zeta^{i}.
\end{align*}

The field polynomial for $\alpha\in \Q(\zeta)$ is 
\begin{align*}
  f_{\alpha}(x) 
  &= \prod_{i=1}^{p-1} (x-\sigma_{i}(\alpha))\\
  &= x^{p-1} -
  (\sigma_{1}(\alpha)++\cdots+\sigma_{p-1}(\alpha))x^{p-2}+\cdots+(-1)^{p-1}\sigma_{1}(\alpha)\cdots\sigma_{p-1}(\alpha),
\end{align*}
and the norm and trace of~$\alpha$ are given by certain coefficients of
$f_{\alpha}$:
\[
  N(\alpha) = \prod_{i=1}^{p-1} \sigma_{i}(\alpha),\quad T(\alpha)
  =\sum_{i=1}^{p-1} \sigma_{i}(\alpha).
\]
In particular, the field polynomial for $\zeta$ is its minimal polynomial:
\[
  f_{\zeta}(x) = f(x) = \prod_{i=1}^{p-1}
  (x-\zeta^{i})=x^{p-1}+x^{p-2}+\cdots+x+1, 
\]
from which we see (recalling that $p$ is odd),
\[
  N(\zeta)=\zeta\cdot \zeta^{2}\cdots \zeta^{p-1} = (-1)^{p-1} = 1
\]
and
\[
  T(\zeta) = \zeta+\zeta^{2}+\cdots\zeta^{p-1}	= -1.
\]
By multiplicativity of the norm,
\[
  N(\zeta^{j})=1
\]
for all $j\in\Z$.  What about the trace of powers of~$\zeta$?   First note if
$1\leq i<k\leq p-1$,
\[
  \sigma_{i}(\zeta^{j})=\sigma_{k}(\zeta^{j})
  \Longleftrightarrow
  \zeta^{ij}=\zeta^{kj}
  \Longleftrightarrow
  \zeta^{(i-k)j}=1
  \Longleftrightarrow
  (i-k)j = 0 \bmod p
  \Longleftrightarrow
  j = 0 \bmod p.
\]
Therefore, for $j \neq 0\bmod p$,
\[
 \{\sigma_{1}(\zeta^{j}),\ldots,\sigma_{p-1}(\zeta^{j})\}
 =
 \{\zeta,\zeta^{2},\ldots,\zeta^{p-1}\},
\]
from which it follows that
\[
  T(\zeta^{j})= \sum_{i=1}^{p-1}
  \sigma_{i}(\zeta^{j})=\zeta+\zeta^{2}+\cdots+\zeta^{p-1}=-1.
\]
If $j=0\bmod p$, then $\zeta^{j}=1$, and $T(1)=\sum_{i=1}^{p-1}
\sigma_{i}(1)=\sum_{i=1}^{p-1} 1=p-1$.

One last useful calculation:  
\[
  N(1-\zeta) = \prod_{i=1}^{p-1} \sigma_{i}(1-\zeta)=\prod_{i=1}^{p-1}
  (1-\zeta^{i})=f(1)=\underbrace{1+\cdots+1}_{p \text{ times}}=p. 
\]
To summarize:

\begin{prop}
  Let $\zeta=e^{2\pi i/p}$.  Then
  \begin{align*}
    N(\zeta^{j})&= 1\qquad \text{ for all~$i\in\Z$}\\
    N(1-\zeta)&=p
  \end{align*}
  and
  \[
    T(\zeta^{j})=
    \begin{cases}
      -1& \text{if $j\neq 0\bmod p$}\\
      p-1& \text{if $j=0\bmod p$}.
    \end{cases}
  \]
\end{prop}
\medskip

\textbf{A useful lemma.} Let $K=\Q(\zeta)$ where $\zeta=\zeta_{p}$ for an odd
prime~$p$.  Let~$\alpha\in \mathfrak{O}_{K}$.  Then since
$\{1,\zeta,\ldots,\zeta^{p-2}\}$ is a $\Q$-basis for~$K$, we can write
\[
  \alpha = a_{0}+a_{1}\zeta+\cdots+a_{p-2}\zeta^{p-2}
\]
for some unique $a_{i}\in\Q$.  To complete the proof of Theorem~\ref{thm:04wed},
we need to show the $a_{i}$ are integers.  We will prove that in the next
lecture with the help of the following lemma:

\begin{lem}
  For $0\leq k\leq p-2$,
  \[
    T(\alpha \zeta^{-k}-\alpha \zeta)=pa_{k}\in\Z.
  \]
\end{lem}

\begin{proof} First note that since $\alpha\in\mathfrak{O}_{K}$, as is
  $\zeta^{-k}=\zeta^{p-k}$ and $\zeta$, it follows that 
  $\alpha \zeta^{-k}-\alpha \zeta\in \mathfrak{O}_{K}$ (recall that
  $\mathfrak{O}_{K}$ is a ring).  Therefore, 
  $T(\alpha \zeta^{-k}-\alpha \zeta)$ is an integer.

  We then calculate:
  \begin{align*}
    T(\alpha \zeta^{-k}-\alpha \zeta)
    &= T(\alpha \zeta^{-k})-T(\alpha \zeta)\\
    &=T(a_{0}\zeta^{-k}+a_{1}\zeta^{-k+1}+\cdots+a_{k}+\cdots+a_{p-2}\zeta^{-k+p-2})\\
    &\qquad
    -T(a_{0}\zeta+a_{1}\zeta^{2}+\cdots+a_{p-2}\zeta^{p-1})\\
    &=-a_{0}-a_{1}-\cdots-a_{k-1}+(p-1)a_{k}-a_{k+1}-\cdots-a_{p-2}\\
    &\qquad
    -(-a_{0}-a_{1}-\cdots-a_{p-2})\\
    &=pa_{k}.
  \end{align*}

  Note, for instance, that
  \[
    T(a_{0}\zeta^{-k})
    =\sum_{i=1}^{p-1} \sigma_{i}(a_{0}\zeta^{-k})
    =\sum_{i=1}^{p-1} a_{0}\sigma_{i}(\zeta^{-k})
    =a_{0}\sum_{i=1}^{p-1} \sigma_{i}(\zeta^{-k})
    =a_{0}T(\zeta^{-k})
    =-a_{0}
  \]
  since~$a_{0}\in\Q$ and $-k\neq0\bmod p$.
\end{proof}

\end{document}
