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Unverified Commit 12e4c86b authored by rrueger's avatar rrueger
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Algebra II: Add Lecture 10. First two theorems and proofs included

parent 38c68a1c
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\input{source-II/lecture-07-2019-04-05}
\input{source-II/lecture-08-2019-04-12}
\input{source-II/lecture-09-2019-05-03}
\input{source-II/lecture-10-2019-05-10}
\input{source-II/lecture-11-2019-05-17}
\end{document}
\chapter{The Cyclotomic Extension}
\section{Cyclotomic Polynomial}
We begin with an observation
%
\begin{align*}
X^n -1 = \prod _{d | n} \Phi _d(X)
\end{align*}
%
\textit{Why should this be true?} Well, $X^n - 1$ contains all the $n$th roots
of unity, which form a group $\Gamma _n$. By Lagrange's theorem, the order $d$
of any root $\gamma _d \in \Gamma _n$ must divide the order of $\Gamma _n$,
namely $n$.
We conclude that every root $\gamma _d \in \Gamma _n$ appears on the R.H.S as a
root of $\Phi _d(X)$, and thus the R.H.S is merely a reordering of all the roots
of $X^n - 1$. We can use this observation to go on to prove
\begin{theorem}[Cyclotomic polynomial live in the world of integers]
\begin{align*}
\Phi _n(X) \in \Z [X]
\end{align*}
\end{theorem}
Before we prove this statement, we remind of Gauss's Lemma from Algebra I.
\begin{siderules}
\begin{remark}[Reminder, Algebra I - Gauss's Lemma]
Let $R$ be a UFD, let $Q = \Frac(R)$, and let $f(x) \in R[x]$. If $f = GH$
in $Q[x]$, then there is a factorization
%
\begin{align*}
f = gh \text{~in~} R[x],
\end{align*}
%
where $\deg(g) = \deg(G)$ and $\deg(h) = \deg(H)$; in fact, $G$ is a
constant multiple of $g$ and $H$ is a constant multiple of $h$. Therefore,
if $f$ does not factor into polynomials of smaller degree in $R[x]$, then
$f$ is irreducible in $Q[x]$.
\end{remark}
\end{siderules}
\begin{proof}[Proof by induction]
For $n=1$, $\Phi _1(X) = X - 1 \in \Z[X]$. Now assume we have proven the
assertion for $k < n$. So we can write
%
\begin{align*}
X^n - 1
= \prod _{d | n} \Phi _d(X)
= \Phi _n \prod _{\substack{d | n \\ d \neq n}} \Phi _d(X)
= \Phi _n \prod _{\substack{d | n \\ d < n}} \Phi _d(X)
\end{align*}
%
and define
%
\begin{align*}
A(X) = \prod _{\substack{d | n \\ d < n}} \Phi _d(X)
\end{align*}
%
as each $\Phi _d(X) \in \Z [X]$ by assumption, we have that $A(X) \in \Z [X]$.
So we have
%
\begin{align*}
X^n - 1 = \Phi _n(X) A(X)
\end{align*}
%
and note that $A(X)$ is monic. Completing the proof is now only an exercise in
applying Gauss's lemma: \textit{A priori} $X^n - 1 = \Phi _n(X) A(X)$ is a
factorisation taking place in $\Q = \Frac (\Z)$. Gauss gives us some other
factorisation $X^n - 1 = F(X)G(X)$ with $F, G \in \Z [X]$ and $F = c \Phi
_n(X)$ and $F = d A(X)$ for $c, d \in \Q$. This amounts to
%
\begin{align*}
X^n - 1 = F(X) G(X) = cd \Phi _n(X) A(X) \implies cd = 1
\end{align*}
%
as $A(X)$ and $\Phi _n(X)$ are monic. However, as $A(X)$ is monic, $d = 1$, so
$c = 1$ and thus $F(X) = \Phi _n(X)
\in \Z [X]$.
\end{proof}
\begin{theorem}
$\Phi _n(x)$ is irreducible over $\Z$.
\end{theorem}
\begin{proof}[Proof by contradiction]
Assume it \textit{is} reducible and remind that $\Phi _n(X)$ is the monic
polynomial with roots exactly the primitive $n$th roots of unity. Then $\Phi
_n(X) = B(X)C(X)$, with $B(X), C(X)$ non-constant and without loss of
generality $B(X)$ is irreducible. If $B(X)$ is not irreducible, keep splitting
the $B(X)$ until there is an irreducible factor. Now, let $\gamma$ be a root
of $B(X)$. By definition, $\gamma$ is a root of $\Phi _n(X)$, hence a
primitive root of unity. The strategy of this proof is to show that, in fact,
$B(X)$ contains \textbf{all} primitive roots of unity, leaving none for $C(X)$
and thus arriving at a contradiction. This strategy will be implemented by
showing two facts for a prime $p$ coprime to n
%
\begin{enumerate}
\item \textit{(Result 1)} $\gamma ^p$ is also a primitive root of unity
\item \textit{(Result 2)} $\gamma ^p$ is also a root of $B(X)$
\end{enumerate}
%
Once these have been shown, it follows that $B(X)$ contains \textbf{all}
primitive roots of unity as the statements hold for all $p$ prime and coprime
to $n$. This results in $C(X)$ merely being a constant. A direct contradiction
to the hypothesis that $\Phi _n(X)$ can be factored into two non-constant
polynomials.
\begin{proof}[Proof Result 1]
The fact that $n$ is coprime to $p$ delivers some $k,m$ such that $kp - mn =
1$. Then
%
\begin{align*}
\gamma ^{pk - nm} & = \gamma \\
\implies
\gamma ^{pk}
& = \gamma \cdot \gamma ^{nm}
= \gamma \cdot {(\gamma ^n)}^m
= \gamma \cdot 1^m
= \gamma
\end{align*}
If $\gamma ^p$ generates $\gamma$ which in turn generates $\Gamma
_n$, then $\gamma ^p$ also generates $\Gamma _n$.
\end{proof}
%
\begin{proof}[Proof Result 2]
To show that this is the case we suppose by contradiction that $\gamma ^p$
is a root of $C(X)$. Now, if $\gamma ^p$ is a root of $C(X)$, then $\gamma$
is a root of $C(X^p)$ and, as before, a root $B(X)$. With $B(X)$ being
irreducible and sharing a root with $C(X)$ we have that $B(X) | C(X^p)$. Now
we perform a reduction of the coefficients by mod $p$.
%
\begin{align*}
\Phi _n(X) & = B(X)C(X) \\
\implies \overline{\Phi _n}(X) & = \overline{B}(X)\overline{C}(X)
\end{align*}
%
Importantly, we still have that $\overline{B}(X) | \overline{C}(X^p)$ and
notably we have that
%
\begin{align*}
\overline{C}(X^p) = {\overline{C}(X)}^p.
\end{align*}
%
so $\overline{B}(X) | {\overline{C}(X)}^p$. This means that they share
non-trivial factors. Specifically, that they share roots. This leads to the
observation, that in $\Z / p\Z [X]$,
%
\begin{align*}
\overline{X^n -1}
= X^n - \overline{1}
= \overline{A}(X) \overline{\Phi _n}(X)
= \overline{A}(X)\overline{B}(X)\overline{C}(X)
\end{align*}
%
has a repeated root. This would imply that
%
\begin{align*}
\gcd (X^n - \overline{1}, \overline{n}X^{n-1}) \neq 1
\end{align*}
%
which, as $\overline{n} \neq 0$ ($p$ and $n$ are coprime), is nonsense.
They clearly don't share a factor.
\end{proof}
\end{proof}
\section{The Return of Gauss-Wantzel}
Previously, we had shown one direction of Theorem~\ref{thm:gauss-wantzel}. We
will now complete the proof. Namely, that the regular n-gon can be constructed
with a straightedge and compass only if $n$ is the product of a power of two and
a Fermat prime.
\begin{proof}
\end{proof}
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