Cyclotomic polynomial
The
is irreducible in
Proof
By ^P1
can be reduced down to a product of prime cyclotomic polynomials which we know have integer coΓ«fficients. Ξ¦ π Now suppose towards contradiction that
is reducibile. Then its roots Ξ¦ π ( π₯ ) β β€ [ π₯ ] with π π π are divided among the factors, so choose a root β¨ π , π β© = β¨ 1 β© of one irreducible monic factor π π π , so that another root π ( π₯ ) β£ Ξ¦ π ( π₯ ) for a prime π π π π is not a root of π β€ π . Write π ( π₯ ) Ξ¦ π ( π₯ ) = π ( π₯ ) π ( π₯ ) where
are monic. Then π ( π₯ ) , π ( π₯ ) β β€ [ π₯ ] is the minimal polynomial of π ( π₯ ) over π π π and β . π ( π π π π ) = 0 It follows that
is a root of π π π , and hence π ( π₯ π ) , whence π ( π₯ ) β£ π ( π₯ π ) π ( π₯ π ) = π ( π₯ ) β ( π₯ ) for
. Letting underling denote the projection β ( π₯ ) β β€ [ π₯ ] , and invoking the Frobenius automorphism we have β€ [ π₯ ] β π π [ π₯ ] π ββ ( π₯ ) π = π ββ ( π₯ ) β ββ ( π₯ ) so
and π ββ ( π₯ ) have a nontrivial common factor β ββ ( π₯ ) in β ββ ( π₯ ) . It follows π π [ π₯ ] β ββ ( π₯ ) 2 β£ π ββ ( π₯ ) π ββ ( π₯ ) = Ξ¦ π βββ ( π₯ ) so
has a multiple factor. Ξ¦ π βββ ( π₯ ) This implies that
is inseparable. On the other hand, its derivative π₯ π β 1 β π π [ π₯ ] is nonzero (since π π₯ π β 1 β π π [ π₯ ] ), contradicting ^P1. Therefore π β€ π must be irreducible. Ξ¦ π ( π₯ )
Cyclotomic polynomial for a prime power
For the particular case of
Properties
- For all
,π β β
Proof of 1
If
, then every π = π π th root π of unity is also an π th root of 1, which holds in particular for every primitive π th root π of unity. π On the other hand, every
generates a subgroup π β π π , where if π» β€ π π then o r d β‘ π = π . Thus every π» = π π is a primitive π β π π th root of unity for some π . π β£ π Therefore the set of
th roots of unity eauals the union of the sets of primitive π th roots of unity as π ranges over factors of π , whence follows ^P1. π