th cyclotomic polynomial
It follows from the theorem that a primitive complex
nth root of unity has the nth cyclotomic polynomial
as its minimal polynomial over Q.
Proof of the theorem:
By Gauss's Lemma,
it is enough to show that cycn(x)
is irreducible in Z[x].
Note that cycn is monic so is certainly
primitive. Thus, if it fails to be irreducible
then there exist polynomials f(x),g(x) in Z[x]
with smaller degree than cycn(x) such that
cycn(x)=f(x)g(x). Since cycn(x) is monic
note that f(x) and g(x) are too (at least WLOG).
Let e be root of f(x) and hence a primitive nth
root of unity. Choose a prime number p that does not divide
Lemma ep is also a root of f(x).
Proof of the lemma:
Suppose not. Then g(ep)=0.
so it follows that e must be a root of g(xp).
Since f(x) is the minimal polynomial of e over Q
it follows that f(x)|g(xp) in Q[x].
Thus we have g(xp)=f(x)h(x), for some
h(x) with rational coeffcients. Thinking about contents
for a moment, we see that h(x) has integer coefficients
and is monic.
Now let k(x) be the product, over all
divisors d of n such that 1 <= d <n,
of cycd(x). Since xn-1=cycn(x)k(x)
we deduce that xn-1=f(x)g(x)k(x). Now let
be the canonical ring homomorphism
that sends an integer to the corresponding integer
modulo p. This extends naturally to a ring homomorphism
if we map anxn+...+a0 to
Apply b to the two equations we have obtained so far. Thus
But the properties of the Frobenius endomorphism tell us that
b(g(xp))=b(g(x))p and so we deduce
that b(g(x))p=b(f(x))b(h(x)). By unique factorization of polynomials
we can deduce that b(f(x)) and b(g(x)) have a common
irreducible factor in Zp[x].
Thus (*) tells us that xn-b(1) has a repeated zero
in a splitting field. Since d/d(xn-b(1))=b(n)xn-1
which is nonzero (as (p,n)=1) there can
be no such repeated root. This contradiction completes the proof of the lemma.
We now finish the proof of the theorem. Let w be
some primitive nth root of 1, say w=er,
for some r with (r,n)=1. Now, choose a prime factor
p of r. Thus r=ps, for some s and
(s,n)=1. Now w=(es)p.
So an induction based on the lemma shows that
w is a root of f(x). Thus, cycn(x)=f(x)
and we are done.