\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}