**Theorem** The

*n*th

cyclotomic polynomial is

irreducible in

**Q***[x]*.

It follows from the theorem that a primitive complex
*n*th root of unity has the *n*th cyclotomic polynomial
as its minimal polynomial over **Q**.

**Proof of the theorem:**
By Gauss's Lemma,
it is enough to show that *cyc*_{n}(x)
is irreducible in **Z***[x]*.
Note that *cyc*_{n} 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 *cyc*_{n}(x) such that
*cyc*_{n}(x)=f(x)g(x). Since *cyc*_{n}(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 *n*th
root of unity. Choose a prime number *p* that does not divide
*n*.

**Lemma** *e*^{p} is also a root of *f(x)*.

**Proof of the lemma:**
Suppose not. Then *g(e*^{p})=0.
so it follows that *e* must be a root of *g(x*^{p}).
Since *f(x)* is the minimal polynomial of *e* over **Q**
it follows that *f(x)|g(x*^{p}) in **Q***[x]*.
Thus we have *g(x*^{p})=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 *cyc*_{d}(x). Since *x*^{n}-1=cyc_{n}(x)k(x)
we deduce that *x*^{n}-1=f(x)g(x)k(x). Now let
*b:***Z***-->***Z**_{p}
be the canonical ring homomorphism
that sends an integer to the corresponding integer
modulo *p*. This extends naturally to a ring homomorphism
**Z***[x]-->
***Z**_{p}*[x]*
if we map *a*_{n}x^{n}+...+a_{0} to
*b(a*_{n})x^{n}+...+b(a_{0}).

Apply *b* to the two equations we have obtained so far. Thus
we have

*b(g(x*^{p}))=b(f(x))b(h(x))

and

*x*^{n}-b(1)=b(f(x))b(g(x))b(k(x)) (*)

But the properties of the Frobenius endomorphism tell us that
*b(g(x*^{p}))=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 **Z**_{p}*[x]*.
Thus (*) tells us that *x*^{n}-b(1) has a repeated zero
in a splitting field. Since *d/d(x*^{n}-b(1))=b(n)x^{n-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 *n*th root of 1, say *w=e*^{r},
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=(e*^{s})^{p}.
So an induction based on the lemma shows that
*w* is a root of *f(x)*. Thus, *cyc*_{n}(x)=f(x)
and we are done.