Basis Theorem
21.1
Unique Factorization Domains (Factorial Rings)
We saw in Section 20.5 that if K is a field, then every nonnull polynomial in K[X] can
be factored as a product of irreducible factors, and that such a factorization is essentially
unique. The same property holds for the ring K[X1, . . . , Xn] where n ≥ 2, but a different
proof is needed.
The reason why unique factorization holds for K[X1, . . . , Xn] is that if A is an integral
domain for which unique factorization holds in some suitable sense, then the property of
unique factorization lifts to the polynomial ring A[X]. Such rings are called factorial rings,
or unique factorization domains. The first step if to define the notion of irreducible element
in an integral domain, and then to define a factorial ring. If will turn out that in a factorial
ring, any nonnull element a is irreducible (or prime) iff the principal ideal (a) is a prime
ideal.
Recall that given a ring A, a unit is any invertible element (w.r.t. multiplication). The
set of units of A is denoted by A∗. It is a multiplicative subgroup of A, with identity 1. Also,
given a, b ∈ A, recall that a divides b if b = ac for some c ∈ A; equivalently, a divides b iff
(b) ⊆ (a). Any nonzero a ∈ A is divisible by any unit u, since a = u(u−1a). The relation “a
divides b,” often denoted by a | b, is reflexive and transitive, and thus, a preorder on A−{0}.
Definition 21.1. Let A be an integral domain. Some element a ∈ A is irreducible if a = 0,
a /
∈ A∗ (a is not a unit), and whenever a = bc, then either b or c is a unit (where b, c ∈ A).
Equivalently, a ∈ A is reducible if a = 0, or a ∈ A∗ (a is a unit), or a = bc where b, c /
∈ A∗
(a, b are both noninvertible) and b, c = 0.
Observe that if a ∈ A is irreducible and u ∈ A is a unit, then ua is also irreducible.
Generally, if a ∈ A, a = 0, and u is a unit, then a and ua are said to be associated. This
is the equivalence relation on nonnull elements of A induced by the divisibility preorder.
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CHAPTER 21. UFD’S, NOETHERIAN RINGS, HILBERT’S BASIS THEOREM
The following simple proposition gives a sufficient condition for an element a ∈ A to be
irreducible.
Proposition 21.1. Let A be an integral domain. For any a ∈ A with a = 0, if the principal
ideal (a) is a prime ideal, then a is irreducible.
Proof. If (a) is prime, then (a) = A and a is not a unit. Assume that a = bc. Then, bc ∈ (a),
and since (a) is prime, either b ∈ (a) or c ∈ (a). Consider the case where b ∈ (a), the other
case being similar. Then, b = ax for some x ∈ A. As a consequence,
a = bc = axc,
and since A is an integral domain and a = 0, we get
1 = xc,
which proves that c = x−1 is a unit.
It should be noted that the converse of Proposition 21.1 is generally false. However, it
holds for factorial rings, defined next.
Definition 21.2. A factorial ring or unique factorization domain (UFD) (or unique factor-
ization ring) is an integral domain A such that the following two properties hold:
(1) For every nonnull a ∈ A, if a /
∈ A∗ (a is not a unit), then a can be factored as a product
a = a1 · · · am
where each ai ∈ A is irreducible (m ≥ 1).
(2) For every nonnull a ∈ A, if a /
∈ A∗ (a is not a unit) and if
a = a1 · · · am = b1 · · · bn
where ai ∈ A and bj ∈ A are irreducible, then m = n and there is a permutation σ of
{1, . . . , m} and some units u1, . . . , um ∈ A∗ such that ai = uibσ(i) for all i, 1 ≤ i ≤ m.
Example 21.1. The ring Z of integers if a typical example of a UFD. Given a field K, the
polynomial ring K[X] is a UFD. More generally, we will show later that every PID is a UFD
(see Theorem 21.12). Thus, in particular, Z[X] is a UFD. However, we leave as an exercise
to prove that the ideal (2X, X2) generated by 2X and X2 is not principal, and thus, Z[X]
is not a PID.
First, we prove that condition (2) in Definition 21.2 is equivalent to the usual “Euclidean”
condition.
21.1. UNIQUE FACTORIZATION DOMAINS (FACTORIAL RINGS)
567
√
There are integral domains that are not UFD’s. For example, the subring Z[ −5] of C
√
consisting of the complex numbers of the form a + bi 5 where a, b ∈ Z is not a UFD.
Indeed, we have
√
√
9 = 3 · 3 = (2 + i 5)(2 − i 5),
√
√
and it can be shown that 3, 2 + i 5, and 2 − i 5 are irreducible, and that the units are ±1.
√
The uniqueness condition (2) fails and Z[ −5] is not a UFD.
√
Remark: For d ∈ Z with d < 0, it is known that the ring of integers of Q( d) is a UFD iff d
is one of the nine primes, d = −1, −2, −3, −7, −11, −19, −43, −67 and −163. This is a hard
theorem that was conjectured by Gauss but not proved until 1966, independently by Stark
and Baker. Heegner had published a proof of this result in 1952 but there was some doubt
about its validity. After finding his proof, Stark reexamined Heegner’s proof and concluded
that it was essentially correct after all. In sharp contrast, when d is a positive integer, the
√
problem of determining which of the rings of integers of Q( d) are UFD’s, is still open. It
√
can also be shown that if d < 0, then the ring Z[ d] is a UFD iff d = −1 or d = −2. If
√
d ≡ 1 (mod 4), then Z[ d] is never a UFD. For more details about these remarkable results,
see Stark [96] (Chapter 8).
Proposition 21.2. Let A be an integral domain satisfying condition (1) in Definition 21.2.
Then, condition (2) in Definition 21.2 is equivalent to the following condition:
(2 ) If a ∈ A is irreducible and a divides the product bc, where b, c ∈ A and b, c = 0, then
either a divides b or a divides c.
Proof. First, assume that (2) holds. Let bc = ad, where d ∈ A, d = 0. If b is a unit, then
c = adb−1,
and c is divisible by a. A similar argument applies to c. Thus, we may assume that b and c
are not units. In view of (1), we can write
b = p1 · · · pm and c = pm+1 · · · qm+n,
where pi ∈ A is irreducible. Since bc = ad, a is irreducible, and b, c are not units, d cannot
be a unit. In view of (1), we can write
d = q1 · · · qr,
where qi ∈ A is irreducible. Thus,
p1 · · · pmpm+1 · · · pm+n = aq1 · · · qr,
where all the factors involved are irreducible. By (2), we must have
a = ui p
0
i0
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CHAPTER 21. UFD’S, NOETHERIAN RINGS, HILBERT’S BASIS THEOREM
for some unit ui ∈ A and some index i
0
0, 1 ≤ i0 ≤ m + n. As a consequence, if 1 ≤ i0 ≤ m,
then a divides b, and if m + 1 ≤ i0 ≤ m + n, then a divides c. This proves that (2 ) holds.
Let us now assume that (2 ) holds. Assume that
a = a1 · · · am = b1 · · · bn,
where ai ∈ A and bj ∈ A are irreducible. Without loss of generality, we may assume that
m ≤ n. We proceed by induction on m. If m = 1,
a1 = b1 · · · bn,
and since a1 is irreducible, u = b1 · · · bi−1bi+1bn must be a unit for some i, 1 ≤ i ≤ n. Thus,
(2) holds with n = 1 and a1 = biu. Assume that m > 1 and that the induction hypothesis
holds for m − 1. Since
a1a2 · · · am = b1 · · · bn,
a1 divides b1 · · · bn, and in view of (2 ), a1 divides some bj. Since a1 and bj are irreducible,
we must have bj = uja1, where uj ∈ A is a unit. Since A is an integral domain,
a1a2 · · · am = b1 · · · bj−1uja1bj+1 · · · bn
implies that
a2 · · · am = (ujb1) · · · bj−1bj+1 · · · bn,
and by the induction hypothesis, m − 1 = n − 1 and ai = vibτ(i) for some units vi ∈ A and
some bijection τ between {2, . . . , m} and {1, . . . , j − 1, j + 1, . . . , m}. However, the bijection
τ extends to a permutation σ of {1, . . . , m} by letting σ(1) = j, and the result holds by
letting v1 = u−1.
j
As a corollary of Proposition 21.2. we get the converse of Proposition 21.1.
Proposition 21.3. Let A be a factorial ring. For any a ∈ A with a = 0, the principal ideal
(a) is a prime ideal iff a is irreducible.
Proof. In view of Proposition 21.1, we just have to prove that if a ∈ A is irreducible, then the
principal ideal (a) is a prime ideal. Indeed, if bc ∈ (a), then a divides bc, and by Proposition
21.2, property (2 ) implies that either a divides b or a divides c, that is, either b ∈ (a) or
c ∈ (a), which means that (a) is prime.
Because Proposition 21.3 holds, in a UFD, an irreducible element is often called a prime.
In a UFD A, every nonzero element a ∈ A that is not a unit can be expressed as a
product a = a1 · · · an of irreducible elements ai, and by property (2), the number n of factors
only depends on a, that is, it is the same for all factorizations into irreducible factors. We
agree that this number is 0 for a unit.
Remark: If A is a UFD, we can state the factorization properties so that they also applies
to units:
21.1. UNIQUE FACTORIZATION DOMAINS (FACTORIAL RINGS)
569
(1) For every nonnull a ∈ A, a can be factored as a product
a = ua1 · · · am
where u ∈ A∗ (u is a unit) and each ai ∈ A is irreducible (m ≥ 0).
(2) For every nonnull a ∈ A, if
a = ua1 · · · am = vb1 · · · bn
where u, v ∈ A∗ (u, v are units) and ai ∈ A and bj ∈ A are irreducible, then m = n,
and if m = n = 0 then u = v, else if m ≥ 1, then there is a permutation σ of {1, . . . , m}
and some units u1, . . . , um ∈ A∗ such that ai = uibσ(i) for all i, 1 ≤ i ≤ m.
We are now ready to prove that if A is a UFD, then the polynomial ring A[X] is also a
UFD.
The fact that nonnull and nonunit polynomials in A[X] factor as products of irreducible
polynomials is rather easy to prove. First, observe that the units of A[X] are just the units of
A. If f (X) is a polynomial of degree 0 that is not a unit, the fact that A is a UFD yields the
desired factorization of f (X). If f (X) has degree m > 0 and f (X) is reducible, then f (X)
factors as the product of two nonunit polynomials g(X), h(X). Let fm be the coefficient of
degree m in f . We have
f (X) = g(X)h(X),
and if both g(X) and h(X) have degree strictly less than m, by induction, we get a factor-
ization of f (X) as a product of irreducible polynomials. Otherwise, either g(X) or h(X) is
a constant. Consider the case where g(X) is a constant, the other case being similar. Then,
g(X) = b is not a unit, and b factors as a product b = b1 · · · bn of irreducible elements bi,
where n only depends on b. Since
fm = bhm,
where hm be the coefficient of degree m in h, we see that hm is a product of p of the bi’s, up
to units, and thus, p < m. Again, we conclude by induction. More formally, we can proceed
by induction on (m, n), where m is the degree of f (X) and n is the number of irreducible
factors in fm.
For the uniqueness of the factorization, by Proposition 21.2, it is enough to prove that
condition (2 ) holds. This is a little more tricky. There are several proofs, but they all involve
a pretty Lemma due to Gauss.
First, note the following trivial fact. Given a ring A, for any a ∈ A, a = 0, if a divides
every coefficient of some nonnull polynomial f (X) ∈ A[X], then a divides f(X). If A is an
integral domain, we get the following converse.
Proposition 21.4. Let A be an integral domain. For any a ∈ A, a = 0, if a divides a
nonnull polynomial f (X) ∈ A[X], then a divides every coefficient of f(X).
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CHAPTER 21. UFD’S, NOETHERIAN RINGS, HILBERT’S BASIS THEOREM
Proof. Assume that f (X) = ag(X), for some g(X) ∈ A[X]. Since a = 0 and A is an
integral ring, f (X) and g(X) have the same degree m, and since for every i (0 ≤ i ≤ m)
the coefficient of Xi in f (X) is equal to the coefficient of Xi in ag(x), we have fi = agi, and
whenever fi = 0, we see that a divides fi.
Lemma 21.5. (Gauss’s lemma) Let A be a UFD. For any a ∈ A, if a is irreducible and a
divides the product f (X)g(X) of two polynomials f (X), g(X) ∈ A[X], then either a divides
f (X) or a divides g(X).
Proof. Let f (X) = fmXm + · · · + fiXi + · · · + f0 and g(X) = gnXn + · · · + gjXj + · · · + g0.
Assume that a divides neither f (X) nor g(X). By the (easy) converse of Proposition 21.4,
there is some i (0 ≤ i ≤ m) such that a does not divide fi, and there is some j (0 ≤ j ≤ n)
such that a does not divide gj. Pick i and j minimal such that a does not divide fi and a
does not divide gj. The coefficient ci+j of Xi+j in f (X)g(X) is
ci+j = f0gi+j + f1gi+j−1 + · · · + figj + · · · + fi+jg0
(letting fh = 0 if h > m and gk = 0 if k > n). From the choice of i and j, a cannot divide
figj, since a being irreducible, by (2 ) of Proposition 21.2, a would divide fi or gj. However,
by the choice of i and j, a divides every other nonnull term in the sum for ci+j, and since a
is irreducible and divides f (X)g(X), by Proposition 21.4, a divides ci+j, which implies that
a divides figj, a contradiction. Thus, either a divides f (X) or a divides g(X).
As a corollary, we get the following proposition.
Proposition 21.6. Let A be a UFD. For any a ∈ A, a = 0, if a divides the product
f (X)g(X) of two polynomials f (X), g(X) ∈ A[X] and f(X) is irreducible and of degree at
least 1, then a divides g(X).
Proof. The Proposition is trivial is a is a unit. Otherwise, a = a1 · · · am where ai ∈ A is
irreducible. Using induction and applying Lemma 21.5, we conclude that a divides g(X).
We now show that Lemma 21.5 also applies to the case where a is an irreducible polyno-
mial. This requires a little excursion involving the fraction field F of A.
Remark: If A is a UFD, it is possible to prove the uniqueness condition (2) for A[X] directly
without using the fraction field of A, see Malliavin [72], Chapter 3.
Given an integral domain A, we can construct a field F such that every element of F
is of the form a/b, where a, b ∈ A, b = 0, using essentially the method for constructing the
field Q of rational numbers from the ring Z of integers.
Proposition 21.7. Let A be an integral domain.
(1) There is a field F and an injective ring homomorphism i : A → F such that every
element of F is of the form i(a)i(b)−1, where a, b ∈ A, b = 0.
21.1. UNIQUE FACTORIZATION DOMAINS (FACTORIAL RINGS)
571
(2) For every field K and every injective ring homomorphism h : A → K, there is a
(unique) field homomorphism h : F → K such that
h(i(a)i(b)−1) = h(a)h(b)−1
for all a, b ∈ A, b = 0.
(3) The field F in (1) is unique up to isomorphism.
Proof. (1) Consider the binary relation
on A × (A − {0}) defined as follows:
(a, b)
(a , b ) iff ab = a b.
It is easily seen that
is an equivalence relation. Note that the fact that A is an integral
domain is used to prove transitivity. The equivalence class of (a, b) is denoted by a/b. Clearly,
(0, b)
(0, 1) for all b ∈ A, and we denote the class of (0, 1) also by 0. The equivalence class
a/1 of (a, 1) is also denoted by a. We define addition and multiplication on A × (A − {0})
as follows:
(a, b) + (a , b ) = (ab + a b, bb ),
(a, b) · (a , b ) = (aa , bb ).
It is easily verified that
is congruential w.r.t. + and ·, which means that + and · are
well-defined on equivalence classes modulo
. When a, b = 0, the inverse of a/b is b/a, and
it is easily verified that F is a field. The map i : A → F defined such that i(a) = a/1 is an
injection of A into F and clearly
a = i(a)i(b)−1.
b
(2) Given an injective ring homomorphism h : A → K into a field K,
a
a
=
iff ab = a b,
b
b
which implies that
h(a)h(b ) = h(a )h(b),
and since h is injective and b, b = 0, we get
h(a)h(b)−1 = h(a )h(b )−1.
Thus, there is a map h : F → K such that
h(a/b) = h(i(a)i(b)−1) = h(a)h(b)−1
for all a, b ∈ A, b = 0, and it is easily checked that h is a field homomorphism. The map h
is clearly unique.
(3) The uniqueness of F up to isomorphism follows from (2), and is left as an exercise.
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CHAPTER 21. UFD’S, NOETHERIAN RINGS, HILBERT’S BASIS THEOREM
The field F given by Proposition 21.7 is called the fraction field of A, and it is denoted
by Frac(A).
In particular, given an integral domain A, since A[X1, . . . , Xn] is also an integral do-
main, we can form the fraction field of the polynomial ring A[X1, . . . , Xn], denoted by
F (X1, . . . , Xn), where F = Frac(A) is the fraction field of A. It is also called the field
of rational functions over F , although the terminology is a bit misleading, since elements of
F (X1, . . . , Xn) only define functions when the dominator is nonnull.
We now have the following crucial lemma which shows that if a polynomial f (X) is
reducible over F [X] where F is the fraction field of A, then f (X) is already reducible over
A[X].
Lemma 21.8. Let A be a UFD and let F be the fraction field of A. For any nonnull
polynomial f (X) ∈ A[X] of degree m, if f(X) is not the product of two polynomials of
degree strictly smaller than m, then f (X) is irreducible in F [X].
Proof. Assume that f (X) is reducible in F [X] and that f (X) is neither null nor a unit.
Then,
f (X) = G(X)H(X),
where G(X), H(X) ∈ F [X] are polynomials of degree p, q ≥ 1. Let a be the product of
the denominators of the coefficients of G(X), and b the product of the denominators of
the coefficients of H(X). Then, a, b = 0, g1(X) = aG(X) ∈ A[X] has degree p ≥ 1,
h1(X) = bH(X) ∈ A[X] has degree q ≥ 1, and
abf (X) = g1(X)h1(X).
Let c = ab. If c is a unit, then f (X) is also reducible in A[X]. Otherwise, c = c1 · · · cn,
where ci ∈ A is irreducible. We now use induction on n to prove that
f (X) = g(X)h(X),
for some polynomials g(X) ∈ A[X] of degree p ≥ 1 and h(X) ∈ A[X] of degree q ≥ 1.
If n = 1, since c = c1 is irreducible, by Lemma 21.5, either c divides g1(X) or c divides
h1(X). Say that c divides g1(X), the other case being similar. Then, g1(X) = cg(X) for
some g(X) ∈ A[X] of degree p ≥ 1, and since A[X] is an integral ring, we get
f (X) = g(X)h1(X),
showing that f (X) is reducible in A[X]. If n > 1, since
c1 · · · cnf(X) = g1(X)h1(X),
c1 divides g1(X)h1(X), and as above, either c1 divides g1(X) or c divides h1(X). In either
case, we get
c2 · · · cnf(X) = g2(X)h2(X)
21.1. UNIQUE FACTORIZATION DOMAINS (FACTORIAL RINGS)
573
for some polynomials g2(X) ∈ A[X] of degree p ≥ 1 and h2(X) ∈ A[X] of degree q ≥ 1. By
the induction hypothesis, we get
f (X) = g(X)h(X),
for some polynomials g(X) ∈ A[X] of degree p ≥ 1 and h(X) ∈ A[X] of degree q ≥ 1,
showing that f (X) is reducible in A[X].
Finally, we can prove that (2 ) holds.
Lemma 21.9. Let A be a UFD. Given any three nonnull polynomials f (X), g(X), h(X) ∈
A[X], if f (X) is irreducible and f (X) divides the product g(X)h(X), then either f (X)
divides g(X) or f (X) divides h(X).
Proof. If f (X) has degree 0, then the result follows from Lemma 21.5. Thus, we may assume
that the degree of f (X) is m ≥ 1. Let F be the fraction field of A. By Lemma 21.8, f(X)
is also irreducible in F [X]. Since F [X] is a UFD (by Theorem 20.16), either f (X) divides
g(X) or f (X) divides h(X), in F [X]. Assume that f (X) divides g(X), the other case being
similar. Then,
g(X) = f (X)G(X),
for some G(X) ∈ F [X]. If a is the product the denominators of the coefficients of G, we
have
ag(X) = q1(X)f (X),
where q1(X) = aG(X) ∈ A[X]. If a is a unit, we see that f(X) divides g(X). Otherwise,
a = a1 · · · an, where ai ∈ A is irreducible. We prove by induction on n that
g(X) = q(X)f (X)
for some q(X) ∈ A[X].
If n = 1, since f (X) is irreducible and of degree m ≥ 1 and
a1g(X) = q1(X)f (X),
by Lemma 21.5, a1 divides q1(X). Thus, q1(X) = a1q(X) where q(X) ∈ A[X]. Since A[X]
is an integral domain, we get
g(X) = q(X)f (X),
and f (X) divides g(X). If n > 1, from
a1 · · · ang(X) = q1(X)f(X),
we note that a1 divides q1(X)f (X), and as in the previous case, a1 divides q1(X). Thus,
q1(X) = a1q2(X) where q2(X) ∈ A[X], and we get
a2 · · · ang(X) = q2(X)f(X).
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CHAPTER 21. UFD’S, NOETHERIAN RINGS, HILBERT’S BASIS THEOREM
By the induction hypothesis, we get
g(X) = q(X)f (X)
for some q(X) ∈ A[X], and f(X) divides g(X).
We finally obtain the fact that A[X] is a UFD when A is.
Theorem 21.10. If A is a UFD then the polynomial ring A[X] is also a UFD.
Proof. As we said earlier, the factorization property (1) is easy to prove. Assume that f (X)
has degree m and that its coefficient fm of degree m is the product of n irreducible elements
(where n = 0 if fm is a unit). We proceed by induction on the pair (m, n), using the
well-founded ordering on pairs, i.e.,
(m, n) ≤ (m , n )
iff either m < m , or m = m and n < n . If f (X) is a nonnull polynomial of degree 0 which
is not a unit, then f (X) ∈ A, and f(X) = fm = a1 · · · an for some irreducible ai ∈ A, since
A is a UFD. If f (X) has degree m > 0 and is reducible, then
f (X) = g(X)h(X),
where g(X) and h(X) have degree p, q ≤ m and are not units. If p, q < m, then (p, n1) <
(m, n) and (q, n2) < (m, n), where n1 is the number of irreducible factors in gp and n2 is the
number of irreducible factors in hq, and by the induction hypothesis, both g(X) and h(X)
can be written as products of irreducible factors. If p = 0, then g(X) = g0 is not a unit, and
since
fm = g0hm,
hm is a product of n2 irreducible elements where n2 < n. Since (m, n2) < (m, n), by the
induction hypothesis, h(X) can be written as products of irreducible polynomials. Since
g0 ∈ A is not a unit, g0 can also be factored as a product of irreducible elements. The case
where q = 0 is similar.
Property (2 ) follows by Lemma 21.9. By Proposition 21.2, A[X] is a UFD.
As a corollary of Theorem 21.10 and using induction, we note that for any field K, the
polynomial ring K[X1, . . . , Xn] is a UFD.
For the sake of completeness, we shall prove that every PID is a UFD. First, we review
the notion of gcd and the characterization of gcd’s in a PID.
Given an integral domain A, for any two elements a, b ∈ A, a, b = 0, we say that d ∈ A
(d = 0) is a greatest common divisor (gcd) of a and b if
(1) d divides both a and b.
21.1. UNIQUE FACTORIZATION DOMAINS (FACTORIAL RINGS)
575
(2) For any h ∈ A (h = 0), if h divides both a and b, then h divides d.
We also say that a and b are relatively prime if 1 is a gcd of a and b.
Note that a and b are relatively prime iff every gcd of a and b is a unit. If A is a PID,
then gcd’s are characterized as follows.
Proposition 21.11. Let A be a PID.
(1) For any a, b, d ∈ A (a, b, d = 0), d is a gcd of a and b iff
(d) = (a, b) = (a) + (b),
i.e., d generates the principal ideal generated by a and b.
(2) (Bezout identity) Two nonnull elements a, b ∈ A are relatively prime iff there are some
x, y ∈ A such that
ax + by = 1.
Proof. (1) Recall that the ideal generated by a and b is the set
(a) + (b) = aA + bA = {ax + by | x, y ∈ A}.
First, assume that d is a gcd of a and b. If so, a ∈ Ad, b ∈ Ad, and thus, (a) ⊆ (d) and
(b) ⊆ (d), so that
(a) + (b) ⊆ (d).
Since A is a PID, there is some t ∈ A, t = 0, such that
(a) + (b) = (t),
and thus, (a) ⊆ (t) and (b) ⊆ (t), which means that t divides both a and b. Since d is a gcd
of a and b, t must divide d. But then,
(d) ⊆ (t) = (a) + (b),
and thus, (d) = (a) + (b).