## Cyclotomic polynomials

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 $μn= {1,ξ, ξ2,…, ξn-1 } where ξ=e 2πin ∈ℂ,$ with the operation of multiplication of complex numbers.

The ${n}^{\text{th}}$ roots of unity are $e k2πin =cos(k2π/n) + icos(k2π/n) ,for k=0,1,…, n-1.$ In the complex plane the elements of ${\mu }_{n}$ all lie on the circle ${S}^{1}=\left\{z\in ℂ\phantom{\rule{0.2em}{0ex}}|\phantom{\rule{0.2em}{0ex}}|z|=1\right\}.$

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 $m\in {ℤ}_{>0}$ and $m then ${\omega }^{m}\ne 1$.
• The $n$th cyclotomic polynomial is $Φn(x)= ∏ω (x-ω) ,$ where the product is over the primitive $n$th roots of unity in $ℂ$.
• The Euler $\varphi$-function is $\varphi :{ℤ}_{>0}\to {ℤ}_{>0}$ given by $ϕ(n)= degΦn(x) .$

Since the $n$th roots of unity are the primitive $d$th roots of unity for the positive integers $d$ dividing $n$, $xn-1 = ∏d|n Φd(x) .$

Let $n$ be a postitive integer.

1. ${\Phi }_{n}\left(x\right)\in ℤ\left[x\right]$ and ${\Phi }_{n}\left(x\right)$ is irreducible in $ℤ\left[x\right]$.
2. $\varphi \left(n\right)=\mathrm{deg}\phantom{\rule{0.2em}{0ex}}{\Phi }_{n}\left(x\right)=\mathrm{Card}\left({\left(ℤ/nℤ\right)}^{×}\right)=\left(\text{the number of primitive}\phantom{\rule{0.5em}{0ex}}{n}^{\text{th}}\phantom{\rule{0.5em}{0ex}}\text{roots of unity}\right)$.

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

Proof. Any element $\sigma \in \mathrm{Gal}\left(ℚ\left(\omega \right)/ℚ\right)$ 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 $ℚ\left[x\right]$ since it is fixed under $\mathrm{Gal}\left(ℚ\left(\omega \right)/ℚ\right)$. So ${\Phi }^{n}\left(x\right)\in ℤ\left[x\right]$. 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?????