CHAPTER 2
Integral Domains: ED, PID and UFDs
2.1. Domains
In this whole chapter, R will denote a commutative ring. We will write R∗ := R \ {0}
to refer to the set of all nonzero elements of R.
D EFINITION 2.1.1. Let R be a ring, a ∈ R. We say that a is a unit if there is some
b ∈ R such that ab = 1 = ba. When such an element b exists, it is called the inverse of a,
and denoted by b = a−1 . The set of all units of the ring R is denoted by U (R)
E XERCISE 2.1.2. Show that U (R) is a multiplicative group.
D EFINITION 2.1.3. Let R be a nontrivial ring. We say that R is a field if every nonzero
element of R is a unit, i.e. if for each a ∈ R∗ there is some b ∈ R such that ab = ba = 1.
R EMARK 2.1.4. For historical reasons, normally one only considers commutative
fields. There are examples of noncommutative rings such that every nonzero element is
invertible, for example this happens for the quaternion ring. Some books call those rings
noncommutative fields, but it is much more common to find them under the denomination
of division rings.
E XAMPLES 2.1.5. i) Q, R and C are all fields.
ii) Zp is a field
� whenever p is a prime�number.
iii) R(x) =
f (x)
g(x) |
f, g ∈ R[x], g �= 0 , the field of rational functions.
D EFINITION 2.1.6. Let R be a nontrivial ring. We say that R is an integral domain
(ID for short) if whenever a, b ∈ R are such that ab = 0, then necessarily a = 0 or b = 0.
In other words, if whenever a, b �= 0 then one has ab �= 0.
E XAMPLES 2.1.7.
i) The ring of integers Z.
ii) Every field F is an integral domain.
iii) If R is an ID, and S ≤ R is a subring of R, then S is also an ID. In particular, any
subring of a field is an integral domain.
iv) If R is an integral domain, then the polynomial ring R[x] is also an integral domain.
Integral domains have very nice arithmetical properties similar to some well known
ones that we find in the ring of integers. An example is the following:
P ROPOSITION 2.1.8 (Cancellation law). Let R be an ID, and let a, b, c ∈ R such that
ab = ac. If a �= 0, one has b = c.
P ROOF. From ab = ac one gets 0 = ab−ac = a(b−c); as a �= 0 it must be b−c = 0,
and thus b = c.
�
We know that every subring of a field is an integral domain. Actually, the converse is
also true, i.e. for any integral domain R we can find a field Q such that R is a subring of
Q.
T HEOREM 2.1.9 (Field of fractions). Let R be an integral domain, then there is a field
Q satisfying the following properties:
9
10
2. INTEGRAL DOMAINS: ED, PID AND UFDS
(1) R ≤ Q subring,
(2) Every q ∈ Q can be written as q = ab−1 for some a, b ∈ R, b �= 0.
The field Q is unique (up to isomorphism) and receives the name of field of fractions (or
field of quotients) of R.
P ROOF. The proof is constructive, giving an explicit description of the field Q. Actually, the construction mimics the procedure of constructing the rational numbers from
integers. More precisely, we define Q as
Q := {(a, b)| a ∈ R, b ∈ R∗ } / ∼,
where we define the equivalence relation
(a, b) ∼ (c, d) ⇐⇒ ad = bc.
If the equivalence class of a pair (a, b) under the above equivalence relation is denoted by
(a, b), we define the operations in Q as
(a, b) + (c, d) := (ad + bc, bd)
(a, b)(c, d) := (ac, bd).
1
With these operations, Q is a field.
�
D EFINITION 2.1.10. Let R be a nonzero ring. We say that R is simple if its only
ideals are the zero ideal {0} and the total ideal R.
P ROPOSITION 2.1.11. A (commutative) ring R is simple if and only if it is a field.
P ROOF. Assume that R is simple, and let a ∈ R∗ . Since R is simple, one gets
(a) = R, thus 1 ∈ (a), and hence there must exist some b ∈ R such that 1 = ab, and
hence a ∈ U (R); thus, every nonzero element of R is a unit.
Conversely, if R is a field, and 0 �= I � R, let a ∈ I a nonzero element of I. Since
R is a field, a must be a unit, so there is b ∈ R such that ab = 1, but since a ∈ I, the
absorbency property ensures that 1 = ab ∈ I, and thus I = R, so R is simple.
�
D EFINITION 2.1.12. Let R be a ring and I � R an ideal. We say that I is a maximal
ideal if I �= R is a proper ideal and whenever I ⊆ J � R one has J = I or J = R. In
other words, I is maximal, with respect to inclusion, among proper ideals, that is, there is
no proper ideal J such that I � J.
P ROPOSITION 2.1.13. Let R be a ring and I � R an ideal. The quotient ring R/I is
a field if and only if I is maximal.
P ROOF. R/I is a field if and only if it is simple, if and only if R/I �= 0 and for all
K � R/I, either K = 0 or K = R/I. By the correspondence theorem, each ideal K of
R/I must be of the form J/I for I ⊆ J � R, so R/I is a field if and only if J/I = 0 or
J/I = R/I for all J ⊇ I, or equivalently if and only if for all J containing I, either J = I
or J = R, i.e. if and only if I is maximal.
�
D EFINITION 2.1.14. An ideal I � R is called a prime ideal if it is a proper ideal of R
such that whenever ab ∈ I then either a ∈ I or b ∈ I.
P ROPOSITION 2.1.15. Let I � R for a ring R, then R/I is an integral domain if and
only if I is a prime ideal.
P ROOF. R/I ID if and only if R/I �= 0 and for all a, b ∈ R/I such that ab = 0 then
either a = 0 or b = 0. This latest conditions are equivalent to I �= R and whenever ab ∈ I,
then either a ∈ I or b ∈ I, i.e. if and only if I is a prime ideal.
�
1I will include a fully detailed proof of this theorem in the way it was stated on the exercise sheet. Lecture
notes will be upgraded after Reading Week to include that detailed proof.
2.3. PRIMES AND IRREDUCIBLES
11
C OROLLARY 2.1.16. Every maximal ideal is a prime ideal.
P ROOF. If I maximal, then R/I is a field, in particular R/I is an ID, and thus I is
prime.
�
2.2. Ideals and divisibility
D EFINITION 2.2.1. Let R a ring, a, b ∈ R. We say that a divides b, or that b is a
multpiple of a, or that b is divisible by a, and write a|b if there is some r ∈ R such that
b = ar.
Divisibility can be nicely described in terms of ideals, since a|b if and only if b = ar
for some r ∈ R, if and only if b ∈ (a), if and only if (b) ⊆ (a).
D EFINITION 2.2.2. Let a, b ∈ R for some ring R, we say that b is associated to a, or
that a and b are associates if there is some unit u ∈ U (R) such that b = ua. If a and b are
associates, we will write a ∼ b.
P ROPOSITION 2.2.3. Let R be an integral domain, a, b ∈ R. The following properties
hold:
(1) a ∼ b if and only if a|b and b|a, if and only if (a) = (b).
(2) a ∼ 1 if and only if a ∈ U (R), if and only if (a) = (1) = R.
(3) a ∼ 0 if and only if a = 0, if and only if (a) = (0) = 0.
(4) “Being associates” is an equivalence relation on R.
P ROOF.
(1). ⇒ If a ∼ b then b = au for some u ∈ U (R), so a|b, but also a = bu−1 , so b|a.
⇐ If a|b, b|a there are some c, d ∈ R such that b = ac, a = bd, and thus one gets
b = ac = bdc. By the cancellative law it follows that cd = 1, so both c, d ∈ U (R) and we
get a ∼ b. Also, a|b and b|a if and only if (a) ⊆ (b) and (b) ⊆ (a), if and only if (a) = (b).
(2). If a ∼ 1 then a = u1 = u ∈ U (R). Conversely, if a ∈ U (r), as a = a1 we get a ∼ 1.
(3). a ∼ 0 if and only if (a) = (0), if and only if a = 0.
(4). By (1), a ∼ b if and only if (a) = (b), so being associates is an equivalence relation.
�
E XAMPLES 2.2.4.
(1) U (Q) = Q∗ , or more generally for any field F one has U (F) = F∗ , and the only
classes of associates are {0} and F∗ .
(2) Let R be an ID, then U (R[x]) = U (R). So for instance (x + 1) ∼ 2(x + 1) in
Q[x], but not in Z[x], as U (Z[x]) = U (Z) = {−1, 1}.
(3) U (Zn ) = {a ∈ {1, . . . , n − 1}| gcd(a, n) = 1}.
2.3. Primes and irreducibles
D EFINITION 2.3.1. Let R be a (commutative) ID. An element a ∈ R is said to be a
prime, or a prime element of R if:
(1) a �= 0, a ∈
/ U (R), i.e. a is neither 0 nor a unit of R.
(2) Whenever b, c ∈ R are such that a|bc, then either a|b or a|c.
The above conditions can be rewritten in terms of the principal ideal generated by a as
follows:
(1’) (0) � (a) � R
(2’) If bc ∈ (a), then either b ∈ (a) or c ∈ (a).
In other words, for a ∈ R∗ , we have that a is prime if and only if the principal ideal (a) is
a prime ideal.
D EFINITION 2.3.2. Let R be an ID, a ∈ R. We say that b ∈ R is a proper divisor of
a if b|a and b is neither a unit or an associate of a, i.e. if (a) � (b) � R.
12
2. INTEGRAL DOMAINS: ED, PID AND UFDS
D EFINITION 2.3.3. Let R be an ID. We say that an element a ∈ R is an irreducible
element (also called an atom) if
(1) a �= 0, a ∈
/ U (R),
(2) a has no proper divisors, i.e. if b|a then either b ∈ U (R) or b ∼ a.
In terms of ideals, a is irreducible if whenever (a) ⊆ (b) then either (a) = (b) or (b) = R.
In other words, a is irreducible if the principal ideal (a) is maximal among proper principal
ideals.
E XAMPLES 2.3.4.
(1) 2 and 3 are irreducible in Z (actually, in Z irreducible elements an primes are the
same thing).
(2) x2 +1 is irreducible in R[x], but not in C[x], as one has (x2 +1) = (x−i)(x+i).
(3) 6 is not irreducible in Z, as 6 = 2 · 3.
The notion of irreducibility admits several equivalent descriptions:
P ROPOSITION 2.3.5. Let a ∈ R a nonzero, nonunit element of an ID R. The following
are equivalent:
(1) a is irreducible.
(2) If a = bc for some b, c ∈ R, then either b ∈ U (R) or c ∈ U (R).
(3) If a = bc for some b, c ∈ R, then either b ∼ a or c ∼ a.
The proof is completely straightforward.
P ROPOSITION 2.3.6. Let R be an integral domain. Then every prime element a ∈ R
is also irreducible.
P ROOF. Let ∈ R be a prime, and suppose a = bc, then one has b|a, c|a. As a|bc and
a is prime, then either a|b or a|c, yielding either a ∼ b or a ∼ c.
�
2.4. Principal ideal domains
D EFINITION 2.4.1. Let R be an ID. One says that R is a principal ideal domain (PID
for short) if every ideal is a principal ideal, i.e. for any I � R there is some a ∈ R such
that I = (a).
R EMARK . If R is a PID and I = (a), then the element a is unique up to associates,
as for any integral domain one has (a) = (b) if and only if a ∼ b.
E XAMPLES 2.4.2.
(1) Let F be a field. If I � R ideal, then either I = 0 = (0) or I = R = (1), so F is
a principal ideal domain.
(2) Let I � Z, as Z is a cyclic group, every additive subgroup is also cyclic, and thus
I = (n) for some n; hence, Z is a PID.
P ROPOSITION 2.4.3. Let R be a PID, then every irreducible a ∈ R is also prime.
P ROOF. Let a ∈ R be irreducible, we will show that a is prime by showing the ideal
(a) is a maximal ideal, and thus a prime ideal. Let I be an ideal such that (a) ⊆ I. As R
is a PID, it must be I = (b) for some b ∈ R, thus we have a ∈ (b) and hence a = bc for
some c ∈ R. As a is irreducible, either b ∈ U (R), and hence (b) = R, or c ∈ U (R), in
which case a ∼ b and thus (a) = (b) = I. So, (a) is a maximal ideal and by Corollary
2.1.16 it follows that (a) is a prime ideal, and hence a is prime.
�
C OROLLARY 2.4.4. In a PID the notions of prime and irreducible are equivalent.
2.4. PRINCIPAL IDEAL DOMAINS
13
Note that from the proof of Proposition 2.4.3 it follows that if R is a PID and a ∈ R
is irreducible, then (a) is a maximal ideal and thus R/(a) is a field.
We know already that Z is a PID, and we will show later in this chapter that for any
field F the polynomial ring F[x] is also a PID. As we will see, the technique that proves
F[x] is PID is very similar to the one that allow us to prove that Z is a PID:
• There is some kind of norm or measure of “how big” nonzero elements are.
• There is a “division with remainder” with the remainder being strictly smaller
than the divisor.
These properties can be axiomatized, leading to a family of particularly nice rings.
D EFINITION 2.4.5. An Euclidean domain (ED for short) is an ID R endowed with a
map N : R∗ → N such that
ED1 For a, b ∈ R∗ , if a|b then N (a) ≤ N (b).
ED2 For any a, b ∈ R, with b �= 0, there exist q, r ∈ R such that a = bq + r and either
r = 0 or N (r) < N (a).
R EMARK 2.4.6. If (R, N ) is a ED, since for all a ∈ R one has 1|a, one must have
N (1) ≤ N (a). In other words, the norm of the unit element is the smallest amont the
norms of all the elements in the ring.
E XAMPLES 2.4.7.
(1) Z, with the map N defined as N (a) := |a| for all a �= 0. The division is the
usual one.
(2) For any field F, the polynomial ring F[x] is and ED, with norm map N (f ) :=
deg(f ). Here we use the usual division of polynomials.
(3) The ring of Gaussian integers Z[i] = {a + bi| a, b ∈ Z}, with norm N (z) =
zz = a2 + b2 for any z = a + bi ∈ Z[i], is an ED.
P ROOF. The ring Z[i] is an integral domain because we have Z[i] ⊆ C, and
C is a field.
ED1. If z|w, then w = zt, and then
N (w) = N (zt) = ztzt = zztt = N (z)N (t),
thus N (z) ≤ N (w).
ED2. Let z, w ∈ Z[i], with w �= 0. As Z[i] ⊆ Q(i) and Q(i) is a field we can
take zw−1 = a+bi in Q(i). Pick now integers u, v ∈ Z such that |a − u| ≤ 1/2,
|b − v| ≤ 1/2 (u and v are the “rounding” of a and b, respectively); the element
q = u + vi belongs to Z[i]. Now let s = (a − u) + (b − v)i ∈ Q(i), and define
r := sw. Note that
qw + r = qw + sw = (q + s)w = (a + bi)w = zw−1 w = z ∈ Z[i],
and thus r = z − qw ∈ Z[i]. Finally observe that
N (s)N (w) = ssN (w) = ((a − u)2 + (b − v)2 )N (w)
�
�
1 1
≤
+
N (w) < N (w),
4 4
so q and r satisfy the required properties.
N (r)
=
�
P ROPOSITION 2.4.8. If R is an ID with a map N : R∗ → N satisfying ED2, then R
is a PID. In particular, every Euclidean Domain is a PID.
P ROOF. Let I � R be a (nonzero) ideal. Then there is some a ∈ I, a �= 0, such that
N (a) is minimal. Let b ∈ I; by ED2, there are some q, r ∈ R such that b = aq + r
and wither r = 0 or N (r) < N (a), but r = b − qa ∈ I (as both a, b ∈ I) and since
N (a) is minimal, we must have r = 0, and hence b = qa is a multiple of a, henceforth
I = (a).
�
14
2. INTEGRAL DOMAINS: ED, PID AND UFDS
C OROLLARY 2.4.9. The rings Z, F[x], Z[i] are all PDIs.
P ROPOSITION 2.4.10. Let R be an ED, a ∈ R∗ , then a ∈ U (R) if and only if N (a) =
N (1).
P ROOF.
⇒ Let a ∈ U (R), then there is some b ∈ R such that ab = 1, thus a|1 and by ED1,
N (a) ≤ N (1). But one always has 1|a and thus N (1) ≤ N (a). Henceforth N (a) = N (1).
⇐ Using the division algorithm, write 1 = aq + r. If it were r �= 0 one would have
N (r) < N (a) = N (1) but that is not possible, since 1|r and thus N (1) ≤ N (r); thus, it
must be r = 0, and hence 1 = aq, implying a ∈ U (R).
�
E XAMPLES 2.4.11.
(1) U (Z) = {u ∈ Z | |u| = 1} = {1, −1},
(2) U (F[x]) = �{f ∈ F[x] | deg f = deg 1 =� 0} = F∗ ,
(3) U (Z[i]) = a + bi ∈ Z[i] | a2 + b2 = 1 = {1, −1, i, −i}.
2.5. Unique Factorization Domains
D EFINITION 2.5.1. An ID R is said to be a unique factorization domain (UFD for
short) if every nonzero, nonunit element a ∈ R∗ \ U (R) can be written as a product of
irreducible elements a = p1 · · · ps , and such a expression is unique up to reordering of the
factors and associates.
P ROPOSITION 2.5.2. Let R be an ID. The following are equivalent:
(1) R is a UFD.
(2) Every a ∈ R∗ \ U (R) can be written as a product of primes.
(3) Every irreducible in R is prime, and every a ∈ R∗ \ U (R) can be written as a
product of irreducibles,
P ROOF.
1 ⇒ 3 By definition of UFD, every a ∈ R∗ \ U (R) can be written as a product of
irreducibles. Let a ∈ R be an irreducible element, and suppose that a|bc, then there is
some d ∈ R such that ad = bc. If bc = 0 then either b = 0 or c = 0, and trivially a|0, so
either a|b or a|c. So let us assume bc �= 0. If either b or c are units, we immediately
get�that
�
a divides
the
other
one,
so
we
can
also
assume
b,
c
nonunits.
Write
b
=
b
,
c
=
cj ,
i
�
d = dk where bi , cj , dk irreducibles. Then we get
ad1 · · · dr = b1 · · · bs c1 · · · ct .
Since R is a UFD, it must be a ∼ bi or a ∼ cj for some i or j; in the first case, a divides
b, in the second a divides c, thus a is prime.
3 ⇒ 2 Obvious.
2 ⇒ 1 Let a ∈ R∗ \ U (R). By (2), a = p1 · · · pr for pi primes. Every prime is also
irreducible, so a is a product of irreducibles. Suppose p1 · · · pr = qi · · · qs for some irreducibles q1 , . . . , qs . We will show by induction that r = s, and after some relabelling we
get pi ∼ qi .
• For r = 1, one has a = p1 irreducible; if p1 = q1 · · · qs , as p1 is irreducible we
must have s = 1 and q1 = p1 .
• Assume now (inductive hypothesis) the property holds for r − 1, with r > 1, and
any s, and let a = p1 · · · pr = q1 · · · qs .
One has pr |q1 · · · qs , and as pr is prime then pr divides some qj . After
relabelling we can assume pr |qs ; then, as qs is irreducible, it must be qs = upr
for some u ∈ U (R), i.e. pr ∼ qs , and hence
p1 · · · pr = q1 · · · qs = q1 · · · qs−1 pr u.
2.6. CHAIN CONDITIONS
15
Applying the cancellation law, one gets p1 · · · pr−1 = q1 · · · qs−1 u, and the result
follows by the inductive hypothesis.
�
2.6. Chain conditions
Our goal now is to show that every PID is also a UFD. We have already shown that in a
PID every ireducible is prime, so we only have to check that every element a ∈ R∗ \ U (R)
in a PID can be written as a product of irreducibles.
D EFINITION 2.6.1. A ring R is said to satisfy the ascending chain condition on
principal ideals (ACC on principal ideals) if whenever we have an ascending chain of
principal ideals
I1 ⊆ I2 ⊆ · · · ⊆ In ⊆ In+1 ⊆ · · ·
then there is some N ∈ N such that In = IN for all n ≥ N .
R EMARK 2.6.2. For some rings, the ascending chain condition can be satisfied for
all ideals, not just principal ones. When this happens, we say that the ring is noetherian.
Noetherian rings are an important topic in other areas of mathematics such as Number
Theory and Algebraic Geometry.
P ROPOSITION 2.6.3. Let R be a ring satisfying the ACC on (principal) ideals, and let
S be a nonempty family of (principal) ideals of R, then S has a maximal element, i.e. there
exists some J ∈ S such that for all I ∈ S satisfying J ⊆ I one has I = J.
P ROOF. By contradiction, assume that S does not have such a maximal element. As
S is not empty, let I1 ∈ S. By our assumption, I1 is not maximal in S, so there is some
I2 ∈ S such that I1 � I2 . Again, I2 cannot be maximal in S so there must exist I3 ∈ S
such that I1 � I2 � I3 . By proceeding along the same lines, we end up providing an
infinite chain
I1 � I2 � · · · � In � · · ·
of (principal) ideals of R, contradicting the ACC for (principal) ideals. Thus, if R satisfies
ACC, then S must have a maximal element.
�
R EMARK 2.6.4. The converse of the result is also true (the proof is an easy exercise).
We have only stated the result for principal ideals, but the same proof works verbatim for
any family of ideals if R is noetherian.
E XAMPLE 2.6.5. Let R be a UFD, a ∈ R∗ \ U (R), and write a = a1 · · · as for some
irreducible elements ai ∈ R. Suppose that (a) ⊆ (b), then b|a = a1 · · · as . As R is a UFD,
it must be b ∼ ai1 · · · ait for some 1 ≤ i1 < · · · < it ≤ s, thus there are only a finite
number of principal ideals containing (a), namely the ones generated by some product of
the irreducibles appearing in the factorization of a. Thus, every UFD satisfies the ACC on
principal ideals.
P ROPOSITION 2.6.6. Let R be an ID. If R satisfies the ACC on principal ideals, then
every nonzero, nonunit element of R can be written as a product of irreducibles.
P ROOF. Suppose there is an element a ∈ R∗ \ U (R) which is not a product of irreducibles, then the family
S := {(a) | a is not a product of irreducibles}
is non-empty and since R satisfies ACC for principal ideals, by Proposition 2.6.3 we can
pick a such that (a) is maximal in S. In particular, a cannot be irreducible, so there are
some b, c ∈ R∗ \ U (R) such that a = bc, where b, c are proper divisors of a, and thus
one gets (a) � (b) � R and (a) � (c) � R. Since (a) is maximal in S, bith b and c
must be written as a product of irreducibles, b = b1 · · · bs , c = c1 · · · cs , but then we get
a = b1 · · · bs c1 · · · ct product of irreducibles, which is a contradiction.
�
16
2. INTEGRAL DOMAINS: ED, PID AND UFDS
P ROPOSITION 2.6.7. Any PID satisfies the ACC on (principal) ideals.
P ROOF. Let I1 ⊆ I2 ⊆ · · · an ascending chain of ideals of R. Consider I =
We claim that I � R is an ideal of R:
�
n≥0 In .
• 0 ∈ I1 ⊆ I, hence 0 ∈ I and I1 is satisfied.
• Let a, b ∈ I, then a ∈ Ir , b ∈ Is for some r, s, and then a, b ∈ Imax{r,s} , hence
a + b ∈ Imax{r,s} ⊆ I, thus I is an additive subgroup of R.
• Let a ∈ I, r ∈ R, then a ∈ In for some n, and since In is an ideal, ar ∈ In ⊆ I,
hence the absorbency property is also satisfied, and thus I is an ideal of R.
Now, since R is a PID, the ideal�
I must be principal, i.e. there must exist some a ∈ R such
that I = (a). But then a ∈ I = In , so there must be some N ∈ N such that a ∈ IN , and
hence IN ⊇ (a) = I, and it follows IN = I. Consequently, for all n ≥ N one has
(a) ⊆ IN ⊆ In ⊆ I = (a),
and thus In = IN for all n ≥ N , so R satisfies ACC on (principal) ideals.
�
C OROLLARY 2.6.8.
ED ⇒ PID ⇒ UFD.
2.7. GCD, LCM and factorization
D EFINITION 2.7.1. Let R be a UFD, and let a, b ∈ R. An element d ∈ R is said
to be a greatest common divisor (gcd for short) of a and b if the following conditions are
satisfied:
(1) d|a, d|b,
(2) For any e ∈ R such that e|a and e|b one has e|d.
These conditions can be restated in terms of ideals as follows:
(1’) (a) ⊆ (d), (b) ⊆ (d),
(2’) For any e ∈ R such that (a) ⊆ (e) and (b) ⊆ (e) one has (d) ⊆ (e).
Note that if both d and d� are gcd’s of a and b, by the second property one gets (d) ⊆
(d ) and (d� ) ⊆ (d), thus (d) = (d� ) and thus d ∼ d� . Henceforth, the greatest common
divisor (if it exists at all!) is unique up to associates.
�
R EMARK . In some particular UFDs we can make canonical choices for gcd’s. For
instance in the ring of integers Z we can always pick the positive gcd, or in a polynomial
ring F[x] with coefficients in a field we can choose the monic one. For more general UFDs,
there is no way of making a canonical choice.
E XAMPLES 2.7.2.
(1) If a = 0, then for any b ∈ R one has gcd(a, b) = b.
(2) If a ∈ U (R), for any b ∈ R one has gcd(a, b) = 1 ∼ a.
αr
1
(3) If R is a UFD, and a, b ∈ R∗ \ U (R), then we can write a = upα
1 · · · pr and
β1
βr
b = vp1 · · · pr where u, v ∈ U (R) are units, pi ∈ R are distinct primes, and
αi , βi ≥ 0. In this case, the element d = pγ11 · · · pγr r is a greatest common divisor
of a and b. In particular, gcd’s always exist in a UFD.
(4) Let R be a PID (and hence a UFD), a, b ∈ R. The ideal (a) + (b) = {ra +
sb | r, s ∈ R} must be principal, so there must exist d ∈ R such that (a) +
(b) = (d). Then d is a greatest common divisor of a and b, and moreover, since
d ∈ (d) = (a) + (b) there must exist h, k ∈ R such that d = ha + kb. In other
words, we have a Bézout’s identity for gcd’s in a PID.
2.8. PROPERTIES OF POLYNOMIAL RINGS OVER DOMAINS
17
(5) Let R be an ED with norm map N (in particular, R is a PID); then we know
that gcd(a, b) exists for all a and b and can be written as d = ah + bk for some
h, k ∈ R. In this case, both d, h and k can be explicitly computed by repeated
use of the division algorithm. The resulting algorithm is exactly the same as the
well-known Euclidean algorithm for computing the gcd of integers.
D EFINITION 2.7.3. Let R be an ID, a, b ∈ R. An element e ∈ R is a least common
multiple (lcm) of a and b if the following properties are satisfied:
(1) a|e, b|e,
(2) For any f ∈ R such that a|f and b|f one has e|f .
Or in terms of ideals:
(1’) (e) ⊆ (a), (e) ⊆ (b),
(2’) For any f ∈ R such that (f ) ⊆ (a) and (f ) ⊆ (b) one has (f ) ⊆ (e).
P ROPOSITION 2.7.4.
(1) If a = 0, then for any b ∈ R one has lcm(a, b) = 0.
(2) If a ∈ U (R), then for any b ∈ R one has lcm(a, b) = b.
αt
1
(3) If R is a UFD, a, b ∈ R∗ \U (R) elements that can be written as a = upα
1 · · · pt
β1
βt
and b = vp1 · · · pt , where u, v ∈ U (R) are units, pi ∈ R are distinct primes,
and αi , βi ≥ 0, and denote δi := max{αi , βi }. Then lcm(a, b) = pγ11 · · · pγt t .
(4) If R is a UFD, a, b ∈ R, then (a) ∩ (b) = (lcm(a, b)).
D EFINITION 2.7.5. Let R be a UFD, we say that elements a, b ∈ R are coprime if
gcd(a, b) = 1, i.e. if (a) + (b) = R.
P ROPOSITION 2.7.6. Let R be a UFD, r ∈ R. The one has gcd(ra1 , . . . , rak ) =
r gcd(a1 , . . . , ak ). In particular, if d = gcd(a1 , . . . , ak ) then one has gcd( ad1 , . . . , adk ) =
1.
2.8. Properties of polynomial rings over domains
We already know some properties of the ring of polynomials for certain types of rings
of coeeficients. For instance:
(1) If F is a field, then F[x] is NOT a field. In fact, the polynomial ring R[x] is never
a field.
(2) If R is an ID, then R[x] is also an ID.
(3) If F is a field, then F[x] is a ED, and hece a PID.
Our goal in this section is to show that R is a UFD if and only if R[x] is also a UFD.
The “only if” part is fairly trivial, but the “if” part, i.e. showing that if R is a UFD then
R[x] is also a UFD, is trickier. The main idea that we will use for this proof will be to go
from the ring R[x] to the ring Q[x], where Q is the field of fractions of R, and then take
advantage of the fact that Q[x] is a PID and in particular a UFD.
Note that a polynomial f ∈ R[x] might be irreducible in Q[x] but not in R[x], for
instance f (x) = 2x2 + 4 is irreducible in Q[x], but in Z[x] we have f (x) = 2(x2 + 2).
This happens because 2 is a unit in Q but not in Z. To look at irreducibles f ∈ R[x] we
will need to chack that f does not have a non-unit constant factor in R.
D EFINITION 2.8.1. Let R be a UFD, 0 �= f = a0 + a1 x + · · · + an xn ∈ R[x]. The
polynomial f is said to be primitive if gcd(a0 , . . . , an ) = 1, i.e. if there isn’t any prime
p ∈ R such that p|ai for all i = 0, . . . , n.
E XAMPLES 2.8.2.
(1) If f = xn + an−1 xn−1 + · · · + a0 is monic, then f is also primitive.
(2) The polynomial 3+4x+2x2 is primitive in Z[x], but 4+4x−2x2 +10x3 +20x4
is not, since gcd(4, 4, −2, 10, 20) = 2.
18
2. INTEGRAL DOMAINS: ED, PID AND UFDS
(3) If f ∈ R[x] is irreducible, then f is primitive.
�
�
L EMMA 2.8.3. Let R be a UFD with field of fractions Q = ab | a, b ∈ R, b �= 0 ,
and let f ∈ Q[x] a non-zero polynomial. The f can be written as f = λf˜ where λ ∈ Q∗
and f˜ ∈ R[x] is primitive. Moreover λ and f˜ are unique up to multiplication by units in
R.
P ROOF. Let f = ab00 + · · · + abnn xn ∈ Q[x]. Take r = b0 · · · bn ∈ R∗ , and a�i = ai r/bi .
Let d = gcd(a�0 , . . . , a�n ), and denote ci := a�i /d ∈ R. We then have
d
f = (c0 + · · · + cn xn )
r
∗
Obviously d/r ∈ Q . By Proposition 2.7.6, gcd(c0 , . . . , cn ) = 1, thus f˜ := c0 +· · ·+cn xn
is primitive.
Assume now that f = λf˜ = µg̃ where f˜ and g̃ are primitive. Write λ = a/b, µ = c/d,
where a, b, c, d ∈ R∗ , and let f˜ = a0 + · · · + an xn , g̃ = b0 + · · · + bn xn . From the equality
λf˜ = µg̃ we get
ad(a0 + · · · + an xn ) = bc(b0 + · · · + bn xn ),
thus adai = bcbi for all i = 0, . . . , n. One then has
ad
∼
ad gcd(a0 , . . . , an )
=
gcd(ada0 , . . . , adan )
=
gcd(bcb0 , . . . , bcbn )
=
bc gcd(b0 , . . . , bn )
∼
bc,
where we are using the fact f˜ and g̃ are primitive. Since the gcd is only defined up to a unit
in R, we get that there must exist u ∈ U (R) such that bc = uad, and therefore µ = uλ.
Now, since we have adai = bcbi = uadbi , and a, d �= 0, we obtain bi = u−1 ai , and
thus g̃ = u−1 f˜, so λ and f˜ are unique up to multiplication by a unit of R.
�
D EFINITION 2.8.4. If f ∈ Q[x] is written as f = λf˜ as in the previous lemma, the
element λ ∈ Q∗ is called the content of f , and denoted by λ = c(f ), and the polynomial
f˜ is called the primitive part of f .
E XAMPLE 2.8.5. Let f =
21x .
2
4
3
8
+ 21
x + 2x2 ∈ Q[x], then c(f ) =
2
21
and f˜ = 14 + 4x +
P ROPOSITION 2.8.6. Let R be a UFD with field of fractions Q, f ∈ Q[x], f �= 0. The
following properties hold:
� = f˜.
i) If λ ∈ Q∗ , then c(λf ) = λc(f ) and λf
ii) f ∈ R[x] if and only if c(f ) ∈ R.
iii) f ∈ R[x] is primitive if and only if c(f ) = 1.
iv) If f, g ∈ R[x] are primitive, and f ∼ g in Q[x], then f ∼ g ∈ R[x].
P ROOF. i) f = c(f )f˜, where f˜ primitive. Thus λf = λc(f )f˜. By uniqueness of the
� = f˜.
content and the primitive part, c(λf ) = λc(f ) and λf
n
ii) ⇒ If f = a0 + · · · + an x , and d = gcd(a0 , · · · , an ), then f = d( ad0 + · · · + adn xn ),
and gcd(a0 /d, . . . , an /d) = 1, thus c(f ) = d ∈ R.
⇐ If c(f ) ∈ R, as f˜ ∈ R[x], then obviously f = c(f )f˜ ∈ R[x].
iii) If f is primitive, then f˜ = f = c(f )f˜, thus c(f ) = 1. Conversely, if c(f ) = 1, then
f = c(f )f˜ = f , and thus f is primitive.
iv) Let f, g be primitive, then c(f ) = c(g) = 1. If f ∼ g in Q[x] there is some λ ∈
U (Q[x]) = Q∗ such that g = λf , but then
1 = c(g) = c(λf ) = λc(f ) = λ,
2.8. PROPERTIES OF POLYNOMIAL RINGS OVER DOMAINS
19
�
so λ = 1 (up to units in R), and thus f ∼ g in R[x].
T HEOREM 2.8.7 (Gauss’ lemma). Let R be a UFD, if f, g ∈ R[x]∗ are primitive, then
f g is also primitive.
P ROOF. Write f = a0 + �
· · · + an xn , g = b0 + · · · + bm xm ; then f g = c0 + · · · +
m+n
cm+n + x
, where ci =
j+k=i aj bk = a0 bi + · · · ai b0 . Suppose that f g is not
primitive, then there must exist p ∈ R prime such that p|ci for all i. Now, since f and g are
primitive, they must have a first coefficient which is not divisible by p, i.e. there are some
i, j such that p|a0 , . . . , p|ai−1 , p � ai , and p|bo , . . . , p|bj−1 , p � bj . But then one has
ci+j = a0 bi+j + · · · ai−i bj+1 +ai bj + ai+1 bj−1 + · · · + ai+j b0 .
�
��
�
�
��
�
A
B
Now, p divides A because it divides a0 , . . . , ai−1 , and p divides B because it divides
b0 , . . . , bj−1 . By assumption p divides ci+j , and as ai bj = ci+j − A − B, one gets that
p divides ai bj , since p is prime, it must divide either ai or bj which is a contradiction.
Henceforth, f g must be primitive.
�
P ROPOSITION 2.8.8. Let R be a UFD with field of fractions Q, and let f, g ∈ Q[x]∗ ,
then one has f�g = f˜g̃ and c(f g) = c(f )c(g).
P ROOF. As f = c(f )f˜ and g = c(g)g̃, one gets f g = c(f )f˜c(g)g̃ = c(f )c(g)f˜g̃. By
Gauss’ Lemma, f˜g̃ is primitive, so it is the primitive part of f g, i.e. f�g = f˜g̃, and by the
uniqueness of the content-primitive part decomposition we get c(f g) = c(f )c(g).
�
By using the previous results, we can describe the irreducibles in R[x] in terms of the
irreducibles in Q[x]:
P ROPOSITION 2.8.9. Let R be a UFD with field of fractions Q, and let f ∈ R[x],
f=
� 0. The following properties hold:
i) If deg f = 0 (so f ∈ R∗ ) then f is irreducible in R[x] if and only if it is irreducible in
R.
ii) If deg f ≥ 1, then f is irreducible in R[x] if and only if f is primitive and is irreducible
in Q[x].
P ROOF.
i) If f is not irreducible in R[x], then f = gh for some g, h ∈ R[x]. As deg f = 0 we must
have deg g = deg h = 0 as well, but then g and h are also elemets of R, and thus f is not
irreducible in R. The converse statement is immediate.
ii) If f is irreducible in R[x], then f is primitive. Assume f is not irreducible in Q[x], then
we can write f = gh for some g, h ∈ Q[x], and then we have
f = f˜ = c(g)c(h)g̃ h̃.
As by Gauss’ Lemma f˜g̃ is primitive, we get c(g)c(h) = 1, and hence f = g̃ h̃ with
g̃, h̃ ∈ R[x], contradicting the irreducibility of f in R[x]. Hence, f must be irreducible in
Q[x]. The converse statement is obvious.
�
T HEOREM 2.8.10. If R is a UFD, then R[x] is also a UFD.
P ROOF. Let f ∈ R[x] a nonzero, nonunit element of R[x]. If deg f = 0, then f ∈ R,
and since R is a UFD we can write f = p1 · · · ps where pi are irreducible in R. By
Proposition 2.8.9, pi are also irreducible in R[x].
If deg f ≥ 1, as R[x] ⊆ Q[x] we can look at f as an element in Q[x]. SInce Q is a
field, Q[x] is a PID (in particular a UFD), and hence we can write f = f1 · · · fk where fi
are irreducible in Q[x]. By taking content-primitive part we get
f = c(f )f� = c(f1 ) · · · c(fk )f˜1 · · · f˜k ,
20
2. INTEGRAL DOMAINS: ED, PID AND UFDS
where now each f˜i is primitive and belongs to R[x]. As fi are irreducible in Q[x] and
f˜i ∼ fi in Q[x], we have f˜i irreducible in Q[x]; by proposition 2.8.9, we obtain that
f˜i are irreducible in R[x]. Now, as f ∈ R[x] then c(f ) ∈ R, so we can write c(f ) =
p1 · · · ps ,with pi irreducible in R (and thus in R[x]). Putting everything together, we can
write
f = p1 · · · ps f˜1 · · · f˜k ,
where each term is irreducible in R[x]. Thus, every nonzero, nonunit element of R[x] can
be written as a product of irreducibles.
Now we need to show uniqueness of the factorization. Suppose that
f = p1 · · · ps f 1 · · · f k = q 1 · · · q s � g 1 · · · g k �
where pi , qj are irreducibles in R and fi , gj are irreducibles in R[x] of degree bigger than
0. As every irreducible is primitive, taking content and primitive part we get
p1 · · · p s = q 1 · · · q s � ,
and
f1 · · · fk = g 1 · · · g k � .
Now, using the fact that R is a UFD we obtain that s = s� and (after possibly some
reordering) pi ∼ qi for all i = 1, . . . , s. Now since all fi and gj are irreducible in R[x], by
2.8.9 they are also irreducible in Q[x], so we have two equal decompositions as products
of irreducibles f1 · · · fk = g1 · · · gk� in Q[x]. Since Q is a field, Q[x] is a UFD, so we have
k = k � and, up to reordering, fi ∼ gi in Q[x]. As fi and gi are primitive, by proposition
2.8.6 fi ∼ gi in R[x].
�
R EMARK 2.8.11. By repeated use of the previous theorem, we obtain that for any
UFD R the multivariate polynomial ring R[x1 , . . . , xn ] s also a UFD.