Theorem The nth cyclotomic polynomial is irreducible in Q[x].

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 n.

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 b:Z-->Zp be the canonical ring homomorphism that sends an integer to the corresponding integer modulo p. This extends naturally to a ring homomorphism Z[x]--> Zp[x] if we map anxn+...+a0 to b(an)xn+...+b(a0).

Apply b to the two equations we have obtained so far. Thus we have
xn-b(1)=b(f(x))b(g(x))b(k(x)) (*)
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.

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