Cyclotomic polynomials

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updates: 20 November 2011

Cyclotomic polynomials

Let $n$ be a positive integer.

• A $n^{\text{th}}$ root of unity is an element $\omega \in ℂ$ such that ${\omega }^n=1$.
• The cyclic group of $n^{\text{th}}$ roots of unity is $${\mu }_n=\{1,\xi ,{\xi }^2,\dots ,{\xi }^{n-1}\}\text{where}\xi =e^{\frac{2\pi i}{n}}\in ℂ,$$ with the operation of multiplication of complex numbers.

The $n^{\text{th}}$ roots of unity are $$e^{\frac{k2\pi i}{n}}=\cos \left(k2\pi /n\right)+i\cos \left(k2\pi /n\right),\text{for}k=0,1,\dots ,n-1.$$ In the complex plane the elements of ${\mu }_n$ all lie on the circle $S^1=\{z\in ℂ|\left|z\right|=1\}.$

Let $n$ be a positive integer.

• A primitive $n$th root of unity is an element $\omega \in ℂ$ such that ${\omega }^n=1$ and if $m\in {\mathbb{Z}}_{>0}$ and $m<n$ then ${\omega }^m\ne 1$.
• The $n$th cyclotomic polynomial is $${\Phi }_n\left(x\right)=\prod _{\omega }(x-\omega ),$$ where the product is over the primitive $n$th roots of unity in $ℂ$.
• The Euler $\varphi$-function is $\varphi :{\mathbb{Z}}_{>0}\to {\mathbb{Z}}_{>0}$ given by $$\varphi \left(n\right)=\mathrm{deg}{\Phi }_n\left(x\right).$$

Since the $n$th roots of unity are the primitive $d$th roots of unity for the positive integers $d$ dividing $n$, $$x^n-1=\prod _{d|n}{\Phi }_d\left(x\right).$$

Let $n$ be a postitive integer.

1. ${\Phi }_n\left(x\right)\in \mathbb{Z}\left[x\right]$ and ${\Phi }_n\left(x\right)$ is irreducible in $\mathbb{Z}\left[x\right]$.
2. $\varphi \left(n\right)=\mathrm{deg}{\Phi }_n\left(x\right)=\mathrm{Card}\left({\left(\mathbb{Z}/n\mathbb{Z}\right)}^{\times }\right)=\left(\text{the number of primitive}{n}^{\text{th}}\text{roots of unity}\right)$.

Let $\omega$ be a primitive $n^{\text{th}}$ root of unity. Then $\mathbb{Q}\left(\omega \right)$ is the splitting field of $x^n-1\in \mathbb{Q}\left[x\right]$, $$\mathrm{Gal}\left(\mathbb{Q}\right(\omega \left)/\mathbb{Q}\right)\simeq {\left(\mathbb{Z}/n\mathbb{Z}\right)}^{\times },\text{and}\left|\mathbb{Q}\right(\omega ):\mathbb{Q}|=\left|\mathrm{Gal}\right(\mathbb{Q}\left(\omega \right)/\mathbb{Q}\left)\right|=\varphi \left(n\right).$$

Proof. Any element $\sigma \in \mathrm{Gal}\left(\mathbb{Q}\right(\omega \left)/\mathbb{Q}\right)$ is determined by where it sends $\omega$, and it must send $\omega$ to another primitive $n^{\text{th}}$ root of unity. Note that ${\Phi }^n\left(x\right)$ lies in $\mathbb{Q}\left[x\right]$ since it is fixed under $\mathrm{Gal}\left(\mathbb{Q}\right(\omega \left)/\mathbb{Q}\right)$. So ${\Phi }^n\left(x\right)\in \mathbb{Z}\left[x\right]$. THIS PROOF IS INCOMPLETE. $\square$

Notes and References

These notes are a retyped version of notes of Arun Ram from Work2004/BookNewalg/PartV.pdf.

References

[BouTop] N. Bourbaki, General Topology, Chapter VI, Springer-Verlag, Berlin 1989. MR?????

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