CYCLOTOMIC POLYNOMIALS
1. The Derivative and Repeated Factors
The usual definition of derivative in calculus involves the nonalgebraic notion of
limit that requires a field such as R or C (or others) where limits are sensible for
analytic reasons.
However, polynomial differentiation is an algebraic notion over any field. One
can simply define the polynomial derivative to be as it is from calculus,
! n−1
n
X
X
i
D
ai X =
(i + 1)ai+1 X i .
i=0
i=0
Alternatively (and working a bit casually here), one can introduce a second indeterminate T and observe that the polynomial
f (X + T ) − f (X) ∈ k[X][T ]
is satisfied by T = 0, and so it takes the form
f (X + T ) − f (X) = T g(X)(T ),
g ∈ k[X][T ].
The derivative of f is then defined as the T -constant term of g(X)(T ),
f 0 (X) = g(X)(0) ∈ k[X].
The familiar facts that the derivative is linear, i.e., (f + cg)0 = f 0 + cg 0 , and that
the derivative satisfies the product rule, i.e., (f g)0 = f g 0 + f 0 g, can be rederived
from this purely algebraic definition of the derivative. In fact, yet a third approach
is to define X 0 = 1 and then stipulate that differentiation is the unique extension
of this rule that is linear and satisfies the product rule. (For example, 1 = X 0 =
(X · 1)0 = X · 10 + X 0 · 1 = X · 10 + 1, so 10 = 0.)
Proposition 1.1. Let k be a field, and let f ∈ k[X] be a polynomial. Then:
If gcd(f, f 0 ) = 1 then f has no repeated factors.
Proof. To establish the contrapositive, suppose that f = g 2 h in k[X]. The product
rule gives
f 0 = 2gg 0 h + g 2 h0 = g(2g 0 h + gh0 ).
Thus g | f 0 , and so g | gcd(f, f 0 ), and so gcd(f, f 0 ) 6= 1.
2. Definition of the Cyclotomic Polynomials
Working in Z[X] we want to define cyclotomic polynomials
Φn ,
n ∈ Z≥1 .
Φn (X) = Q
Xn − 1
,
d|n Φd (X)
Provisionally, define
d<n
1
n ≥ 1.
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CYCLOTOMIC POLYNOMIALS
Here it is understood that for n = 1 the denominator is the empty product, i.e.,
Φ1 (X) = X − 1.
Certainly each Φn lies in the quotient field Z(X) of Z[X] but we will show considerably more. In fact, always Φn liesQ
in Z[X] and is irreducible. For example,
repeatedly using the identity X n − 1 = d|n Φd (X) in various ways,
X2 − 1
= X + 1 = −Φ1 (−X),
X −1
p
X −1
= X p−1 + · · · + X + 1,
p prime,
Φp (X) =
X −1
e
e−1
Xp − 1
Φpe (X) = pe−1
= Φp (X p ),
p prime,
−1
X
d
d−1
d−1
X2 p + 1
X2 p − 1
=
= Φp (−X 2 ), p > 2 prime,
Φ2d p (X) =
d−1 p
d−1
2
2
(X
− 1)Φ2d (X)
X
+1
Φ2 (X) =
d e
Φ2d pe (X) =
(X 2d−1 pe
X2 p − 1
− 1)Φ2d (X)Φ2d p (X) · · · Φ2d pe−1 (X)
d−1 e
=
X2 p + 1
d−1
−Φ1 (−X 2 )Φp (−X 2d−1 ) · · · Φpe−1 (−X 2d−1 )
(induction on e)
d−1 e
d−1 e−1
X2 p + 1
= Φp (−X 2 p ),
d−1 pe−1
2
X
+1
X 10 + X 5 + 1
(X 15 − 1)
=
Φ15 (X) =
(X 5 − 1)Φ3 (X)
X2 + X + 1
=
p > 2 prime,
= X 8 − X 7 + X 5 − X 4 + X 3 − X + 1,
Φ21 (X) =
X 21 − 1
X 14 + X 7 + 1
=
− 1)Φ3 (X)
X2 + X + 1
(X 7
= X 12 − X 11 + X 9 − X 8 + X 6 − X 4 + X 3 − X + 1.
These identities give Φn for all n < 30.
We begin with two observations about the polynomial X n − 1 where n ∈ Z≥1 .
• X n − 1 has no repeated factors in Z[X] for any n ∈ Z≥1 . Indeed, working
in the larger ring Q[X] we have
(X n − 1) − (1/n)X · (X n − 1)0 = (X n − 1) − (1/n)X · nX n−1 = −1 ∈ Q× ,
so that gcd(X n − 1, (X n − 1)0 ) = 1. As explained above, X n − 1 therefore
has no repeated factors in Q[X], much less in Z[X]. (For future reference,
we note that this argument also works in (Z/pZ)[X] where p is prime, so
long as p - n.)
• For any n, m ∈ Z≥1 ,
gcd(X n − 1, X m − 1) = X gcd(n,m) − 1.
Indeed, taking n > m, the calculation
X n − 1 − X n−m (X m − 1) = X n−m − 1
shows that
hX n − 1, X m − 1i = hX n−m − 1, X m − 1i,
CYCLOTOMIC POLYNOMIALS
3
and, letting n = qm + r where 0 ≤ r < n and repeating the calculation q
times,
hX n − 1, X m − 1i = hX r − 1, X m − 1i.
That is, carrying out the Euclidean algorithm on X n − 1 and X m − 1
in Q(X) slowly carries out the Euclidean algorithm on n and m in Z. (If
the Euclidean algorithm on n and m takes k division-steps, with quotients
q1 through qk , then the Euclidean algorithm on X n − 1 and X m − 1 takes
q1 + · · · + qk steps.) The result follows.
Recall the definition
Xn − 1
Φn (X) = Q
, n ≥ 1.
d|n Φd (X)
d<n
Note that
• Φ1 ∈ Z[X], and
• (vacuously) no Φk and Φm for distinct k, m < 1 share a factor.
The two bullets are the base case of an induction argument. Fix some n > 1 and
assume that
• all Φm for m ≤ n lie in Z[X], and
• no Φk and Φm for distinct k, m < n share a factor.
Let m < n, and let k = gcd(m, n) ≤ m < n. Any common factor of Φm and Φn is
a common factor of X m − 1 and X n − 1, hence a factor of X k − 1. However, the
calculation
Xn − 1
n
X −1
Xn − 1
Q
Φn (X) = Q
= k
X −1
d|n Φd (X) d|n Φd (X)
d≤k
d<n
shows that no factor of Φn is a factor of X k − 1. In sum, the strict inequality in
the second bullet weakens: no Φk and Φm for distinct k, m ≤ n share a factor.
Equivalently, no Φk and Φm for distinct k, m < n + 1 share a factor.
Now, for each proper divisor d of n + 1, each factor of Φd is a factor of X d − 1,
hence a factor of X n+1 − 1. And the product
Q
d|n+1 Φd (X)
d<n+1
n+1
contains no repeat factors of X
− 1. Thus Φn+1 ∈ Z[X]. In sum,
• all Φm for m ≤ n + 1 lie in Z[X], and
• no Φk and Φm for distinct k, m < n + 1 share a factor.
By induction, Φn ∈ Z[X] for all n ∈ Z≥1 , and no two Φn share a factor.
P
The totient function identity d|n ϕ(d) = n quickly combines with the formula
Q
n
d|n Φd (X) = X − 1 and a little induction to show that the degree of the cyclotomic polynomial is the totient function of its index,
n ≥ 1.
deg(Φn ) = ϕ(n),
Substitute X = 0 in the identity X n − 1 =
Q
d|n
d≤n
Φd (X) to see that the constant
coefficient of Φd for all divisors d of n is ±1. In particular, the constant coefficient
of Φn is ±1.
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CYCLOTOMIC POLYNOMIALS
If we view the integers Z as a subring of the complex number field C then the
nth cyclotomic polynomial factors as
Y
(X − ζne ), where ζn = e2πi/n .
Φn (X) =
0≤e<n
gcd(e,n)=1
Indeed, the formula produces the monic polynomial in C[X] whose roots are the
complex numbers z such that z n = 1 but z m 6= 1 for all 1 ≤ m < n.
However, we do not want to think of the cyclotomic polynomials innately in
complex terms (even though the word cyclotomic literally refers to dividing the
circle), because the integers are also compatible with other algebraic structures
that are incompatible with the complex numbers in turn. Rather, we want to think
of the roots of Φn as the elements having order exactly n in whatever environment
is suitable. In the next example, the environment is Z/pZ where p is a prime that
does not divide n. Earlier we observed that our arguments apply here as well.
3. Application: An Infinite Congruence Class of Primes
Theorem 3.1. Let n > 1 be a positive integer. There exist infinitely many primes p
such that p = 1 mod n.
Proof. Suppose that the only primes equal to 1 mod n are p1 , · · · , pt . The idea is
to find some other prime p and some integer a such that
Φn (a) = 0 mod p
and p - n.
Granting the display, it says that the multiplicative order of a modulo p is exactly n.
Since (Z/pZ)× is cyclic of order p − 1, we thus have n | p − 1, i.e., p = 1 mod n. So
the original list of such primes was not exhaustive after all, and hence there is no
such finite list.
To carry out the idea, define for any positive integer `,
a = a` = ` · n · p1 · · · pt .
Then
Φn (a` ) > 1 for all large enough `.
Take such `, and with ` chosen, omit it as a subscript of a. Consider any prime
divisor p of Φn (a). Since Φn (a) = ±1 mod a, we have p - a. That is, since n | a,
Φn (a) = 0 mod p
and p - n.
Thus the proof is complete, as explained in the previous paragraph,
4. Application: Wedderburn’s Theorem
The cyclotomic polynomials combine with the counting formulas of group actions
to provide a lovely proof of the following result.
Theorem 4.1. Let D be a finite division ring, i.e., a field except that multiplication
might not commute. Then D is a field.
Proof. The center of D is a finite field k; let q denote its order. Since D is a vector
space over k, the order of D is therefore q n for some n. We want to show that
n = 1.
For any x ∈ D, the centralizer
Dx = {y ∈ D : yx = xy}
CYCLOTOMIC POLYNOMIALS
5
is again a division ring containing k, and its multiplicative group Dx× is a subgroup
of D× . Hence |Dx | = q nx where q nx − 1 divides q n − 1, so that nx divides n. And
if x ∈
/ k then the divisibilities are proper. Let D× act on D by conjugation. The
class formula gives
X
|D| = |k| +
|D× |/|Dx× |
x
where the sum adds the sizes of the nontrivial orbits, or
X qn − 1
qn = q +
q nx − 1
x
summing over representatives of nontrivial orbits.
Recall that each nx arising from the sum is a proper divisor of n. The formulas
Y
Φn (q) Φd (q) = q n − 1,
d|n
Y
qn − 1
Φd (q) = nx
Φn (q) ,
q −1
d|n
d-nx
(in which all Φ∗ (q)-values are integers) show that Φn (q) divides the left side and
each term of the sum in the formula
X qn − 1
.
qn − 1 = q − 1 +
q nx − 1
x
Consequently Φn (q) divides q − 1.
But viewing Z as a subring of the complex number system C lets us show that
Φn (q) can not divide q − 1 unless n = 1,
Y
|Φn (q)|2 =
(q − cos(2πk/n))2 + (sin(2πk/n))2
0≤k<n
gcd(k,n)=1
If n > 1 then each term of the product is greater than (q−1)2 , and so |Φn (q)| > q−1.
This completes the proof.
5. Irreducibility of Prime-Power Cyclotomic Polynomials
The irreducibility of prime-power cyclotomic polynomials Φpe (X) in Z[X] (and
hence in Q[X]) can be established by an argument set entirely in Z[X] and its
quotient-structures.
To avoid interrupting the pending argument, we establish a small preliminary
result.
ε
ε
Lemma 5.1. Let p be prime. For all ε ∈ Z≥0 , (X − 1)p = X p − 1 in (Z/pZ)[X].
Proof. The result is immediate if ε = 0. And for the induction step, working
in (Z/pZ)[X],
p X
ε
ε+1
p
pε+1
pε p
pε
p
X ip (−1)p−i = X p
− 1,
(X − 1)
= ((X − 1) ) = (X − 1) =
i
i=0
because the binomial coefficients for 1 ≤ i ≤ p − 1 vanish modulo p and because
(−1)p = −1 mod p for all p, including p = 2.
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CYCLOTOMIC POLYNOMIALS
Again let p be prime, and let e ≥ 1. We show that the cyclotomic polynomial Φpe
is irreducible in Z[X], thereby making it irreducible in Q[X] by Gauss’s Lemma.
Recall that
e
e−1
Xp − 1
Φpe (X) = Φp (X p ) = pe−1
−1
X
That is,
e−1
e
(X p
− 1)Φpe (X) = X p − 1 in Z[X],
and so reducing the coefficients modulo p gives
(X p
e−1
e
− 1)Φpe (X) = X p − 1
in (Z/pZ)[X],
and then by the lemma,
(X − 1)p
e−1
Φpe (X) = (X − 1)p
and therefore
Φpe (X) = (X − 1)p
e−1
(p−1)
e
in (Z/pZ)[X],
in (Z/pZ)[X].
Also,
Φpe (1) = Φp (1e ) = Φp (1) = p 6= 0 in Z/p2 Z.
The previous two displays show precisely that Φpe (X) is irreducible in Z[X] by
Schönemann’s Criterion and hence is irreducible in Q[X] by Gauss’s Lemma.
6. Irreducibility of General Cyclotomic Polynomials
In contrast to the previous section, quickly showing the irreducibility of the
general cyclotomic polynomial Φn (X) in Z[X] (and hence in Q[X]) requires an
argument that extends beyond Z and Q. Again we establish a preliminary result.
Lemma 6.1. Let n > 1 be an integer and let p - n be prime. Then the cyclotomic
polynomial Φn (X) has no repeated factors modulo p.
Proof. Indeed, Φn (X) divides X n − 1, which is coprime to its derivative nX n−1
modulo p because p - n. Hence X n − 1 has no repeated factors modulo p, and hence
neither does Φn (X).
Let n > 1. Let ζn = e2πi/n . The complex roots of Φn (X) are
{ζne : gcd(e, n) = 1}.
Any root of Φn (X) can be obtained starting from ζn and repeatedly raising to
various primes p - n. Suppose that
Φn (X) = g(X)h(X)
in Z[X].
We may assume that g(ζn ) = 0. To show that all roots of Φn are roots of g it
suffices to show that:
For any root ρ of g and for any prime p - n, also ρp is a root of g.
To establish the displayed statement, let ρ be a root of g and let p - n be prime.
We need to show that h(ρp ) 6= 0. If instead h(ρp ) = 0 then work modulo p and
compute
(h(ρ))p = h(ρp ) = 0 mod p and so h(ρ) = 0 mod p.
Also, certainly g(ρ) = 0 mod p. It follows that g(X) and h(X) share a factor
k(X) in (Z/pZ)[X]; indeed, a relation G(X)g(X) + H(X)h(X) = 1 would yield
the impossibility 0 = 1 upon substituting ρ for X. Thus Φn (X) = g(X)h(X)
CYCLOTOMIC POLYNOMIALS
7
has a repeated factor k(X)2 in (Z/pZ)[X]. This contradicts the lemma. And so,
h(ρp ) 6= 0, forcing g(ρp ) = 0, and the argument is complete.
7. Reducibility of Cyclotomic Polynomials Modulo p
Proposition 7.1. Let p be prime, let e ≥ 1, and let m ≥ 1 with gcd(p, m) = 1.
Then
e
Φpe m (X) = Φm (X)ϕ(p ) in (Z/pZ)[X].
Here ϕ is Euler’s totient function, so that ϕ(pe ) = pe (1 − 1/p).
Proof. Working modulo p, pre-compute
e
e
e
Xp m − 1
(X m − 1)p
m
=
− 1)ϕ(p ) in (Z/pZ)[X].
e−1 m
e−1 = (X
p
m
p
X
−1
(X − 1)
Now, breaking the proper divisors of pe m into two groups gives
e
Φpe m (X) =
Xp m − 1
Q
e−1 m
p
(X
− 1) d|m Φpe d (X)
d<m
ϕ(pe )
m
− 1)
d|m Φpe d (X)
(X
=Q
in (Z/pZ)[X].
d<m
If m = 1 then the product in the denominator is 1, and we have shown
Φpe (X) = (X − 1)ϕ(p
e
)
= Φ1 (X)ϕ(p
e
)
in (Z/pZ)[X].
If m > 1 then by induction on m, the product in the denominator is
Y
Y
e
Φpe d (X) =
Φd (X)ϕ(p ) in (Z/pZ)[X],
d|m
d<m
d|m
d<m
and so overall,
e
(X m − 1)ϕ(p )
ϕ(pe )
Φpe m (X) = Q
e ) = Φm (X)
ϕ(p
d|m Φd (X)
in (Z/pZ)[X].
d<m