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\bec {\Large\bf The Cyclotomic Polynomials} \ece
\begin{define} $\gam$ is a {\em primitive} $n$-th root of unity if
$\gam^n=1$ but $\gam^j\neq 1$ for all $1\leq j\leq n-1$.
\end{define}
If $\gam$ is a {\em primitive} $n$-th root of unity then $\gam^n=1$
so $\gam=\kappa^k$ where we set $\kappa=e^{2\pi i/n}$ and $0\leq k< n$.
If $\gcd(k,n)=d>1$ then $\gam^{n/d}=1$ so $\gam$ is not a
{\em primitive} $n$-th root of unity. If $k,n$ {\em are} relatively
prime then $\gam^{kj}=1$ iff $n|(kj)$ which occus iff $n|j$ so
$\gam^k$ {\em is} a primitive $n$-th root of unity.
\noi {\tt Example:} There are four primitive $12$-th roots of unity,
$\kappa,\kappa^5,\kappa^7,\kappa^{11}$ with $\kappa = e^{2\pi i/12}$.
\begin{define} The $n$-the Cyclotomic Polynomial, denoted $\Phi_n(x)$
is $\prod (x-\kappa)$ where the product ranges over all primitive
$n$-th roots of unity.
\end{define}
\noi {\tt Examples:}
\\ $\Phi_1(x)=x-1$
\\ $\Phi_2(x)= x-(-1) = x+1$
\\ $\Phi_3(x) = (x^3-1)/(x-1) = x^2+x+1$
\begin{claim} \[ x^n-1 = \prod_{d|n} \Phi_d(n) \]
\end{claim}
\noi {\tt Proof:} Both are monic polynomials and have exactly the
same roots, since any $\kappa$ with $\kappa^n=1$ is a primitive $d$-th
root of unity for some $d|n$ and, conversely, if $\kappa$ is a primitive
$d$-th root of unity for some $d|n$ then $\kappa^n = (\kappa^d)^{n/d} = 1$.
From the Claim we see (formally, by induction) that all $\Phi_n(x)$ are monic
polynomials with integer coefficients. Now we state our goal:
\begin{theorem} \label{one} $\Phi_n(x)$ is irreducible in $Q[x]$. \end{theorem}
This will take a while. First a technical result that will play a key role.
\begin{lemma}\label{three}
Let $p$ be prime, $n\geq 2$, with $p$ not dividing $n$. Set $F(x)=x^n-1$
but considered in $Z_p[x]$. Then $F(x)$ has no multiple irreducible factors. That is,
there is no factorization $F(x)=a^2(x)b(x)$ with $a(x)$ irreducible.
\end{lemma}
\noi If we did have $F(x)=a^2(x)b(x)$ then
we could then extend $Z_p$
to $F=Z_p(\ah)$ with $\ah$ a root of $a(x)$ and then $\ah$ would be a multiple
root of $F(x)$. But we proved in class that there is an extension of the
ground field (here $Z_p$) in which $F(x)$ has a multiple root if and only if
$\gcd(F(x),F'(x))\neq 1$. Here $F'(x)=nx^{n-1}$ and as (critically) $p$ does not
divide $n$, $n\neq 0$ in $Z_p$. In $Z_p[x]$ we have $F'(x)$ has a constant times
the irreducible (as degree one) polynomial $x$ to the power $n-1$. And $x$ is
not a factor of $x^n-1$. So $F(x),F'(x)$ have no common factors. Such the
extension by $\ah$ cannot occur, and $F(x)$ can have no multiple factors.
(Alternatively, $(n^{-1}x)(nx^{n-1}-(x^n-1)=1$ so $\gcd(nx^{n-1},x^n-1)=1$ in
$Z_p[x]$.)
\par Now we come to the guts of the argument.
\begin{lemma}\label{two}
If $\eps$ is a primitive $n$-th root of unity and
$p$ is a prime not dividing $n$ then $\eps,\eps^p$ have the same
minimal polynomial over $Q[x]$.
\end{lemma}
\noi First we show that Lemma \ref{two} implies the Theorem \ref{one}. Set $\kappa=e^{2\pi i/n}$,
as before. Any primitive $n$-th root of unity can be written $\kappa^k$ with
$k,n$ relatively prime so we can write $k=p_1\cdots p_l$ with the $p_j$ primes
(possibly repeating) that do not divide $n$. Set $I_0=1$, $I_j=p_1\cdots p_j$,
the partial product, for $1\leq j\leq l$. For each $0\leq j < l$ we have
that $\kappa^{I_j}$ is a primitive $n$-the root of unity so from the Lemma
$\kappa^{I_j}$ and $\kappa^{I_{j+1}}$ have the same minimal polynomial over
$Q[x]$ and so $\kappa=\kappa^{I_0}$ and $\kappa^k = \kappa^{I_l}$ would have
the same minimal polynomial. This means that the minimal polynomial for
$\kappa$ must have as roots all of the primitive $n$-th roots of unity $\kappa^k$.
Thus the minimal polynomial must be divisible by $\Phi_n(x)$ and as $\Phi_n(x)\in Q[x]$
it must {\em be} $\Phi_n(x)$.
\noi {\tt Proof of Lemma \ref{two}:} Let $f(x),g(x)$ denote the minimal polynomials of
$\eps,\eps^p$ respectively. As these are factors of $F(x)=x^n-1$ they are monic
with integer coefficients. Assume they are not equal. Then they are both in the factorization
of $x^n-1$ into irreducible polynomials. That is, we can write $F(x)=f(x)g(x)h(x)$ with
$f,g,h$ monic polynomials.
Critically $g(x^p)$ has root $\eps$ so $f(x)|g(x^p)$. Now {\em reduce modulo
$p$}. Looking at $Z_p[x]$ we have Sophomore's Delight and $g(x^p)=g(x)^p$!!
Thus
\[ f(x)|g(x^p)=g(x)^p \mbox{ in } Z_p[x] \]
We don't know how $f(x),g(x)$ reduce in $Z_p[x]$ but the irreducible factors of
$g(x)$ are the same (with higher multiplicity) as the irreducible factors of
$g(x)^p$. Therefore $f(x)$ and $g(x)$ must have a common irreducible factor $a(x)$ in $Z_p(x)$.
Therefore in the factorization of $F(x)$ in $Z_p[x]$ there is multiple factor --
we can write $F(x)=a^2(x)b(x)$. This contracts Lemma \ref{three}.
{\tt The Galois Group:} Let $n\geq 3$. Set $\eps=e^{2\pi i/n}$ and set $K=Q(\eps)$.
The $\eps$ is a primitive $n$-th root of unity. As $\Phi_n(x)$ is irreducible over
$Q$, $\Phi_n(x)$ is the minimal polynomial for $\eps$ over $Q$. As all $\eps^j\in K$,
$K$ is the splitting field of $\Phi_n(x)$. Now we can describe the Galois Group
$\Gamma(K:Q)$. For each $1\leq j < n$ with $j,n$ relatively prime, $\eps^j$ is a
root of $\Phi_n(x)$ and so there is a $\sig_j\in \Gamma(K:Q)$ defined by
\[ \sig_j(\eps) = \eps^j \]
The Galois Group consists of precisely these $\sig_j$. We multiply by noting
\[ (\sig_j(\sig_k(\eps))= \sig_j(\eps^k) = \eps^{jk} = \sig_{jk}(\eps) \]
Note $\eps^{jk}$ is determined by the value of $jk$ modulo $n$.
Thus
\[ \sig_j\sig_k = \sig_{jk} \]
where $jk$ is defined modulo $n$.
\par We define the structure $Z_n^*$. Its elements are the $1\leq j