Section 23 – Factorization of polynomials
over a field
Instructor: Yifan Yang
Spring 2007
Notation
Throughout this section, the letter F will always denote a field.
Division algorithm for F [x]
Theorem (23.1, Division algorithm for F [x])
Let F be a field. Let f (x) and g(x) be polynomials in F [x].
Suppose that g(x) is not the zero polynomial. Then there exist
unique polynomials q(x) and r (x) such that
1. f (x) = g(x)q(x) + r (x),
2. r (x) = 0 or deg r (x) < deg g(x).
Proof.
• Assume that g(x) = bm x m + · · · + b0 , where m ≥ 0 and
bm 6= 0. Consider the set
S = {f (x) − g(x)s(x) : s(x) ∈ F [x]}.
• If 0 ∈ S, then ∃ s(x) ∈ F [x] such that f (x) = g(x)s(x).
Then we can take q(x) = s(x), r (x) = 0, and we are done.
Proof of Theorem 23.1, continued
• Assume that 0 6∈ S. Let r (x) be an element of minimal
degree in S. We have f (x) = g(x)q(x) + r (x) for some
q(x) ∈ F [x]. We need to show that deg r (x) < deg g(x).
• Suppose that r (x) = ck x k + · · · + c0 with k ≥ m and
ck 6= 0. Consider f (x) − g(x)(q(x) + ck /bm x k −m ).
• We have
f (x)−g(x)q(x) − (ck /bm )x k −m g(x) = r (x) − (ck /bm )x k −m g(x)
= (ck x k + · · · + c0 ) − (ck /bm )(bm x k + · · · ),
whose degree is less than r (x). This contradicts to the
assumption that r (x) is of minimal degree in S. Thus,
deg r (x) must be less than deg g(x).
Proof of Theorem 23.1, continued
We now show that q(x) and r (x) are unique.
• Suppose that
f (x) = g(x)q1 (x) + r1 (x)
f (x) = g(x)q2 (x) + r2 (x),
where ri (x) either are the zero polynomial or satisfy
deg ri (x) < deg g(x).
• Then we have r1 (x) − r2 (x) = g(x)(q2 (x) − q1 (x)).
• Now r1 (x) − r2 (x) is either zero or a polynomial of degree
< deg g(x). However, if the right-hand side is not zero,
then the degree is at least deg g(x).
• Thus, the only possibility is that r1 (x) = r2 (x), and
q1 (x) = q2 (x). This completes the proof.
Example
Example. Let F = Z5 , f (x) = x 4 + x 2 + 3x + 2, and
g(x) = x 2 + 2x + 3. Let us find the polynomials q(x) and r (x).
Solution.
x2
x 2 + 2x + 3 x 4
x4
+ 3x
2x 3
+
3x 3
3x 3
+2
+x 2
+ 3x 2
+ 3x 2
+ x2
2x 2
2x 2
+3x
+ 3x
+ 4x
+ 4x
+ 4x
Thus, we find q(x) = x 2 + 3x + 2 and r (x) = 0.
+1
+1
+1
0
Remark
Note that the condition that F is a field is crucial. For example,
assume that F = Z instead. Let f (x) = x + 1 and g(x) = 2x.
Then it is impossible to find q(x), r (x) ∈ Z[x] such that
f (x) = g(x)q(x) + r (x) with r (x) = 0 or deg r (x) < deg g(x).
Zeros of f (x) ∈ F [x]
Corollary (23.3)
An element a ∈ F is a zero of f (x) ∈ F [x] if and only if x − a is
a factor of f (x) in F [x].
Proof.
• Suppose that f (a) = 0. By Theorem 23.1, we have
f (x) = (x − a)g(x) + r (x) for some q(x) ∈ F [x], where
r (x) = 0 or deg r (x) < deg(x − a) = 1, i.e., or r (x) = c is a
constant polynomial.
• Then we have 0 = f (a) = 0g(a) + c. That is, r (x) is the
zero polynomial. This proves that f (a) = 0 ⇒ (x − a)|f (x).
• The proof of the converse statement is easy.
Zeros of f (x) ∈ F [x]
Corollary (23.5)
A nonzero polynomial f (x) ∈ F [x] of degree n can have at most
n zeros in F .
Proof.
• We will prove by induction on the degree of f (x).
• When f (x) has degree 0, i.e., when f (x) = c for some
c 6= 0 ∈ F , it is clear that f (x) has no zeros.
• Now assume that the statement holds for all polynomials of
degree n ≤ k . We will prove that the statement also holds
for polynomials f (x) of degree k + 1.
Proof of Corollary 23.5, continued
• If f (x) has no zeros in F , we are done. Otherwise, assume
that a ∈ F is a zero of f (x).
• By Corollary 23.3, f (x) = (x − a)g(x) for some polynomial
g(x) of degree k .
• Now if b ∈ F is a zero of f (x), then (b − a)g(b) = 0.
• Since F has no zero divisors, this implies that b − a = 0 or
g(b) = 0, i.e., b = a or b is a zero of g(x).
• By the induction hypothesis, the polynomial g(x) has at
most k zeros. It follows that f (x) has at most k + 1 zeros.
• By the principle of mathematical induction, we conclude
that the statement holds for all polynomials.
The multiplicative group F ×
Corollary (23.6)
Let F be a field. If G is a finite subgroup of the multiplicative
group F × of all nonzero elements in F . Then G is cyclic. In
particular, if F is a finite field, then F × is cyclic.
Example
1. In Z5 we have 22 = 4, 23 = 3, and 24 = 1. Thus, Z×
5 = h2i
is cyclic.
2. We have |Z×
7 | = 6. Thus, the multiplicative order of an
2
element in Z×
7 is 1, 2, 3, or 6. Now 3 = 2 6= 1 and
3
3 = 6 6= 1. Therefore, the multiplicative order of 3 is 6,
and Z×
7 = h3i is cyclic.
Proof of Corollary 23.6
• Recall that by the fundamental theorem for finitely
generated abelian groups (Theorem 11.12), the finite
abelian group G is isomorphic to Zpe1 × · · · × Zpek for some
k
1
primes pi and integers ei ≥ 1.
• Since the maximal order of an element in Zpe1 × · · · × Zpek
1
is lcm(p1e1 , . . . , pkek ), the group G is cyclic if and only if pi
are all distinct.
k
• Observe also that the order of every element in G is a
divisor of m = lcm(p1e1 , . . . , pkek ). In other words, every
element of G is a zero of x m − 1.
• If G is not cyclic, then m is strictly less than |G|.
• Now the polynomial x m − 1 has degree m, but has at least
|G| > m zeros. This contradicts to Corollary 23.5.
In-class exercises
Find q(x) and r (x) for the following pairs of polynomials in the
specified rings of polynomials.
1. F = Z2 , f (x) = x 5 + x 3 + x + 1, g(x) = x 3 + x 2 + 1.
2. F = Z3 , f (x) = x 5 + x 3 + x + 1, g(x) = x 3 + x 2 + 1.
3. F = Z5 , f (x) = x 5 + x 3 + x + 1, g(x) = x 3 + x 2 + 1.
Find a generator for each of the following groups.
1. Z×
13 .
2. Z×
17 .
Irreducible polynomials
Definition
A nonconstant polynomial f (x) ∈ F [x] is irreducible over F or is
an irreducible polynomial in F [x] if f (x) cannot be expressed as
a product g(x)h(x) of two polynomials in F [x] with
0 < deg g(x), deg h(x) < deg f simultaneously.
If a polynomial f (x) ∈ F [x] is nonconstant and is not irreducible
over F , then it is reducible over F .
Example
1. x 2 + 1 is irreducible over R.
2. x 2 + 1 is reducible over C since x 2 + 1 = (x − i)(x + i) in C.
Remarks
• Observe that if f (x) ∈ F [x] is irreducible over F , then in
f (x) = g(x)h(x) in F [x], one of g(x) and h(x) has degree
equal to deg f (x), and the other has degree 0. Since the
nonzero constant polynomials in F [x] are precisely the
units in F [x], the definition of irreducible polynomials can
also be given as “f (x) is irreducible over F if in any
factorization f (x) = g(x)h(x) in F [x], one of g(x) and h(x)
is a unit”.
• Note that whether a polynomial f (x) is irreducible or not
depends on which field we are talking about. For example,
x 2 + 1 is irreducible over R, but is reducible over C. Also,
√
x 2 − 2 is irreducible over Q, but reducible over R or Q( 2).
Polynomials of degree 2 or 3
Theorem (23.10)
Suppose that f (x) ∈ F [x] is of degree 2 or 3. Then f (x) is
reducible over F if and only if f (x) has a zero in F .
Proof.
Assume that f (x) is reducible, say, f (x) = g(x)h(x), where
0 < deg g(x), deg h(x) < deg f (x) = 2 or 3. Then one of g(x)
and h(x) is of degree 1 taking the form x − a for some a ∈ F .
Thus f (x) has a zero a in F .
The converse statement is trivial. This proves the theorem.
Remark
Remark
When the degree of f (x) is 4, it is possible that
f (x) = g(x)h(x), where each of g and h is of degree 2. To
determine whether f (x) is irreducible over F , we also need to
consider this possibility.
Examples
Example. Determine whether f (x) = x 3 + x + 1 is irreducible
over Z5 .
Solution.
• Since f (x) = x 3 + x + 1 is of degree 3, it is reducible over
Z5 if and only if it has a zero in Z5 .
• Now we have f (0) = 1, f (1) = 3, f (2) = 8 + 2 + 1 = 1,
f (3) = 27 + 3 + 1 = 1, and f (4) = 64 + 4 + 1 = 4. None of
these is equal to 0.
• Thus, f (x) is irreducible over Z5 .
Examples
Example. Determine whether f (x) = x 4 + x 2 + x + 1 is
irreducible over Z5 .
Solution.
• We have f (0) = 1, f (1) = 4, f (2) = 3, f (3) = 4, and
f (4) = 2. Thus, f (x) has no linear factors.
• Now assume that f (x) = (x 2 + ax + b)(x 2 + cx + d). Then
we have x 4 + x 2 + x + 1 =
x 4 + (a + c)x 3 + (ac + b + d)x 2 + (ad + bc)x + bd.
• Comparing the constant terms, we get
(b, d) = (1, 1), (2, 3), (3, 2), or (4, 4).
• From a + c = 0, we get a = −c. Then (ad + bc) = 1 gives
a(d − b) = 1.
• It follows that (b, d, a) = (2, 3, 1), or (3, 2, 4).
• But ac + b + d = 4 6= 1. Thus, f (x) is irreducible over Z5 .
Gauss’s lemma
Theorem (23.11, Gauss’s lemma)
Let f (x) ∈ Z[x]. Then f (x) factors into a product of two
polynomials of lower degrees r and s in Q[x] if and only if it has
such a factorization with polynomials of the same degrees r
and s in Z[x].
Proof.
See the supplemental material.
Corollary 23.12
If f (x) = x n + an−1 x n−1 + · · · + a0 is in Z[x] with a0 6= 0, and if
f (x) has a zero in Q, then the zero is in Z and satisfies m|a0 .
Example
Example. Factor f (x) = x 4 + 10x 3 + 13x 2 − 32x + 12 ∈ Z[x]
into a product of irreducible polynomials over Q.
Solution.
• We first check if it has a linear factor. By Corollary 23.12,
such a factor is of the form x − m for some integer m|12.
• By a direct computation, we find ±1, ±2, ±3, ±4, ±6, ±12
are not zeros of f (x). Thus, f (x) has no linear factors.
• Assume that f (x) = (x 2 + ax + b)(x 2 + cx + d). Then by
Gauss’s lemma, (b, d) = ±(1, 12), ±(2, 6), or ±(3, 4).
Example, continued
Example. Factor f (x) = x 4 + 10x 3 + 13x 2 − 32x + 12 ∈ Z[x]
into a product of irreducible polynomials over Q.
Solution.
• Assume that (b, d) = (1, 12). We have
x 4 + 10x 3 + 13x 2 − 32x + 12 =
x 4 + (a + c)x 3 + (ac + 13)x 2 + (12a + c)x + 12.
• Comparing the coefficients, we get a + c = 10, ac = 0, and
12a + c = −32. However, no integers can satisfy these
equations.
• Likewise, the choices (b, d) = (−1, −12) and (2, 6) do not
work either.
• When (b, d) = (−2, −6), we find a + c = 10, ac − 8 = 13,
−6a − 2c = −32. A solution is (a, c) = (3, 7). Thus,
f (x) = (x 2 + 3x − 2)(x 2 + 7x − 6).
Reduction modulo m
Lemma (Exercise 37, reduction modulo m)
Let m > 1 be a positive integer. Define φm : Z[x] → Zm [x] by
φm (an x n + · · · + a0 ) = an x n + · · · + a0 ,
where ai denotes the residue class modulo m containing ai .
Then φm is a ring homomorphism, called reduction modulo m.
Example
1. The reduction of x 2 + 4x + 7 modulo 2 is x 2 + 1.
2. The reduction of x 3 + 3x + 9 modulo 3 is x 3 .
Reduction modulo p
Idea.
• Let f (x) ∈ Z[x]. Consider the reduction f (x) modulo p,
where p is a prime.
• If f (x) = g(x)h(x) is reducible over Q, then the reduction
modulo p gives f (x) = g(x)h(x). (Exercise 37.)
• Thus, if f (x) is irreducible over Zp , then f (x) is irreducible
over Q. However, note that if f (x) is reducible over Zp , it is
still possible that f (x) is irreducible over Q. For example,
f (x) = x 2 + 1 and p = 2 with f (x) = (x + 1)2 .
Example. The polynomial x 3 + 13x + 81 is irreducible over Q
because its reduction modulo 2 is x 3 + x + 1, which is
irreducible over Z2 .
Eisenstein criterion
Theorem (23.15, Eisenstein criterion)
Let p be a prime. Suppose that f (x) = an x n + · · · + a0 is in Z[x],
and an 6≡ 0 mod p, ai ≡ 0 mod p for all i < n, but a0 6≡ 0
mod p2 . Then f (x) is irreducible over Q.
Example
The polynomial x 2 − 2 is irreducible
over Q by the Eisenstein
√
criterion. This shows that 2 is an irrational number.
Proof of Theorem 23.15
• Let f (x) denote the reduction of f (x) modulo p. By
assumption, we have f (x) = dx n for some d 6= 0 ∈ Zp .
• Now if f (x) = g(x)h(x) for some g(x), h(x) ∈ Z[x], then
g(x)h(x) = dx n .
• We claim that this implies that g(x) = bx k , h(x) = cx m for
some nonnegative integers k , m, and some a, b ∈ Zp
satisfying k + m = n and ab = d mod p.
• Assume that the claim is true for the moment. Then
g(x) = bk x k + · · · + b0 and cm x m + · · · + c0 , where
bi , cj ≡ 0 mod p for all i < k and all j < m.
• If k and m are both > 0, then the constant term a0 = b0 c0
of f (x) is divisible by p2 , contradicting to a0 6≡ 0 mod p2 .
• Therefore, one of g(x) and h(x) must be a constant
polynomial. That is, f (x) is irreducible over Q.
Proof of the claim
• Write g(x) = bk x k + · · · + b0 and h(x) = cm x m + · · · + c0 .
• Let r be the smallest integer such that br 6≡ 0 mod p, and
s be the smallest integer such that cs 6≡ 0 mod p.
P
• Now the `th coefficient a` of f (x) is equal to `i=0 bi c`−j .
• For ` = r + s, we have
a` = (b0 c` + · · · + br −1 cs+1 ) + br cs + (br +1 cs−1 + · · · + b` c0 )
≡ br cs 6≡ 0
mod p.
• Since a` ≡ 0 mod p for all ` < n, we conclude that
r + s = n, which in turn implies r = k and s = m.
Remark
Remark
In the proof above, the assumption that p is a prime is crucially
used in the step br cs 6≡ 0 mod p. Without the assumption that
p is a prime, the claim is not true at all. For example, in Z4 [x]
we have x 2 = (x + 2)(x + 2).
Cyclotomic polynomial
Corollary 23.17
Let p be a prime. Then the polynomial
Φp (x) = x p−1 + x p−2 + · · · + 1 =
xp − 1
x −1
is irreducible over Q.
Remark
• Φp (x) is called the pth cyclotomic polynomial.
• The zeros of Φp (x) are precisely the pth roots of unity,
except for 1.
• It plays the central role in Kummer’s approach to Fermat’s
Last Theorem. He proved that X n + Y n = Z n has no
non-trivial solutions if 3 ≤ n ≤ 100, with possible
exceptions n = 37, 59, 67.
Proof of Corollary 23.17
• In general, a polynomial f (x) ∈ F [x] is irreducible over F if
and only if f (x + 1) is irreducible over F . Thus, it suffices to
prove that Φp (x + 1) is irreducible over Q.
• We have (x + 1)p − 1 = x p + p1 x p−1 + · · · + px. Thus,
p p−2
p−1
Φp (x + 1) = x
+
x
+ · · · + p.
1
• The binomial coefficients
0 < k < p.
p
k
are all multiples of p when
• But the constant term is p, which is not divisible by p2 .
• Thus, by the Eisenstein criterion, Φp (x) is irreducible over
Q.
In-class exercises
Factors the following polynomials into products of irreducible
polynomials in the given fields.
1. x 3 + x + 1 over Z3 .
2. x 4 + x + 1 over Z5 .
3. x 4 − 3x 3 − 3x 2 + 11x − 6 over Q.
4. x 4 + 3x 3 + 6x 2 − 3x + 12 over Q.
Unique factorization in F [x]
Theorem (23.18)
Let p(x) be an irreducible polynomial in F [x]. If p(x)|r (x)s(x)
for r (x), s(x) ∈ F [x], then p(x)|r (x) or p(x)|s(x)| in F [x].
Proof.
Will be proved in Theorem 27.27.
Remark
In other words, irreducible polynomials in F [x] are in many
ways similar to primes in Z.
Unique factorization in F [x]
Theorem (23.20)
Every nonconstant polynomial f (x) ∈ F [x] can be factored in
F [x] into a product of irreducible polynomials. The factorization
is unique, except for order and for units.
Remarks
• The phrase “except for order” means that we consider the
factorizations f (x) = p1 (x)p2 (x) and f (x) = p2 (x)p1 (x) as
the same one.
• The phrase “except for units” means that the factorizations
p1 (x)p2 (x) and (cp1 (x))(c −1 p2 (x)) are considered as the
same, where c 6= 0 ∈ F . (Note that the nonzero elements
in F are the units in F [x].)
Proof of Theorem 23.20
We first prove by induction that every nonconstant polynomial
f (x) is a product of irreducible polynomials over F .
• Polynomials of degree 1 are clearly irreducible
polynomials.
• Suppose that every polynomial of degree ≤ n is a product
of irreducible polynomials.
• Now let f (x) be a polynomial of degree n + 1.
• If f (x) is irreducible, we are done. Otherwise, assume that
f (x) = g(x)h(x), where deg g(x), h(x) ≤ deg f (x) − 1 = n.
• By the induction hypothesis, g(x) and h(x) are products of
irreducible polynomials, and thus so is f (x) = g(x)h(x).
• We conclude that every nonconstant polynomial is a
product of irreducible polynomials.
Proof of uniqueness
• Assume that f (x) = p1 (x) . . . pr (x) and
f (x) = q1 (x) . . . qr (x) are two factorizations of f (x) into a
product of irreducible polynomials.
• By Theorem 23.18, p1 (x) divides one of qi (x).
• By rearranging the indices, we assume that p1 (x)|q1 (x),
i.e., q1 (x) = p1 (x)r1 (x) for some r1 (x) ∈ F [x].
• Since q1 (x) is irreducible, r1 (x) must be a unit, that is,
r1 (x) = u1 is a constant polynomial in F [x].
• Then p1 (x) . . . pr (x) = (u1 p1 (x))q2 (x) . . . qs (x).
• Canceling p1 (x), we get p2 (x) . . . pr (x) = u1 q2 (x) . . . qs (x).
• Continuing this way, we get qi (x) = ui pi (x) for i = 1, 2 . . .,
for some units ui ∈ F , and we have r = s.
Homework
Problems 4, 10, 14, 16, 20, 30, 34, 36, 37 of Section 23.