Cyclotomic polynomials

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia

Last updates: 20 November 2011

Cyclotomic polynomials

Let n be a positive integer.

The nth 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 μn all lie on the circle S1= { z | |z|=1 } .

1 ξ5 ξ4 ξ3 ξ2 ξ

Let n be a positive integer.

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

Let n be a postitive integer.

  1. Φn(x) [x] and Φn(x) is irreducible in [x].
  2. ϕ(n) =deg Φn(x) = Card( (/n) × ) =(the number of primitive nth roots of unity).

Let ω be a primitive nth root of unity. Then (ω) is the splitting field of xn-1 [x], Gal((ω)/) (/n) ×, and |(ω) :| = | Gal((ω)/) | = ϕ(n) .

Proof. Any element σ Gal((ω)/) is determined by where it sends ω, and it must send ω to another primitive nth root of unity. Note that Φn(x) lies in [x] since it is fixed under Gal((ω)/) . So Φn(x) [x]. THIS PROOF IS INCOMPLETE.

Notes and References

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


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

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