domain D, there may be such a natural choice, but in the general case, there

will not be (see Exercise 17.21 below).

Example 17.12. Having already seen two examples of UFDs, it is perhaps

a good idea to look at an example of an integral domain that is not a UFD.

√

Consider the subring Z[ ’3] of the complex numbers, which consists of all

√

complex numbers of the form a + b ’3, where a, b ∈ Z. As this√ a subring

is

of the ¬eld C, it is an integral domain (one may also view Z[ ’3] as the

quotient ring Z[X]/(X2 + 3)). √

Let us ¬rst determine the units in Z[ ’3]. For a, b ∈ Z, we have N (a +

√

b ’3) =√2 + 3b2 , where N is the usual norm map on C (see Example 9.5).

a √

If ± ∈ Z[ ’3] is a unit, then there exists ± ∈ Z[ ’3] such that ±± = 1.

Taking norms, we obtain

1 = N (1) = N (±± ) = N (±)N (± ).

√

Since the norm of an element of√ ’3] is a non-negative integer, this

Z[

implies that N (±) = 1. If ± = a+b ’3, with a, b ∈ Z, then N (±) = a2 +3b2 ,

and it is clear that N (±) = 1 if and only if ± = ±1. We conclude that the

√

only units in Z[ ’3] are ±1. √

Now consider the following two factorizations of 4 in Z[ ’3]:

√ √

4 = 2 · 2 = (1 + ’3)(1 ’ ’3). (17.10)

We claim √ that 2 is irreducible. For suppose, say, that 2 = ±± , for

±, ± ∈ Z[ ’3], with neither a unit. Taking norms, we have 4 = N (2) =

N (±)N (± ), and therefore, N (±) = N (± ) = 2”but this is impossible, since

17.8 Unique factorization domains (—) 385

integers a and √such that a2 + 3b2 = 2. By the same reasoning,

there are no√ b √ √

since N (1 + ’3) = N (1 ’ ’3) = 4, we see that 1 + ’3 and 1 ’ ’3 are √

both irreducible. Further, it is clear that 2 is not associate to either 1+ ’3

√

or 1 ’ ’3, and so the two factorizations of 4 in (17.10) are fundamentally

di¬erent. 2

For a, b ∈ D, we call d ∈ D a common divisor of a and b if d | a and

d | b; moreover, we call such a d a greatest common divisor of a and

b if all other common divisors of a and b divide d. We say that a and b

are relatively prime if the only common divisors of a and b are units.

It is immediate from the de¬nition of a greatest common divisor that it is

unique, up to multiplication by units, if it exists at all. Unlike in the case of

Z and F [X], in the general setting, greatest common divisors need not exist;

moreover, even when they do, we shall not attempt to “normalize” greatest

common divisors, and we shall speak only of “a” greatest common divisor,

rather than “the” greatest common divisor.

Just as for integers and polynomials, we can generalize the notion of a

greatest common divisor in an arbitrary integral domain D from two to any

number of elements of D, and we can also de¬ne a least common multiple

of any number of elements as well.

Although these greatest common divisors and least common multiples

need not exist in an arbitrary integral domain D, if D is a UFD, they will

always exist. The existence question easily reduces to the question of the

existence of a greatest common divisor and least common multiple of a and

b, where a and b are non-zero elements of D. So assuming that D is a UFD,

we may write

r r

pf i ,

pei

a=u and b = v

i i

i=1 i=1

where u and v are units, p1 , . . . , pr are non-associate irreducibles, and the

ei and fi are non-negative integers, and it is easily seen that

r

pmin(ei ,fi )

i=1

is a greatest common divisor of a and b, while

r

pmax(ei ,fi )

i=1

is a least common multiple of a and b.

It is also evident that in a UFD D, if c | ab and c and a are relatively

386 More rings

prime, then c | b. In particular, if p is irreducible and p | ab, then p | a or

p | b. From this, we see that if p is irreducible, then the quotient ring D/pD

is an integral domain, and so the ideal pD is a prime ideal (see discussion

above Exercise 9.28).

In a general integral domain D, we say that an element p ∈ D is prime

if for all a, b ∈ D, p | ab implies p | a or p | b (which is equivalent to saying

that the ideal pD is prime). Thus, if D is a UFD, then all irreducibles are

primes; however, in a general integral domain, this may not be the case.

Here are a couple of simple but useful facts whose proofs we leave to the

reader.

Theorem 17.26. Any prime element in D is irreducible.

Proof. Exercise. 2

Theorem 17.27. Suppose D satis¬es part (i) of De¬nition 17.25. Also,

suppose that all irreducibles in D are prime. Then D is a UFD.

Proof. Exercise. 2

Exercise 17.21. (a) Show that the “is associate to” relation is an equiv-

alence relation.

(b) Consider an equivalence class C induced by the “is associate to”

relation. Show that if C contains an irreducible element, then all

elements of C are irreducible.

(c) Suppose that for every equivalence class C that contains irreducibles,

we choose one element of C, and call it a distinguished irreducible.

Show that D is a UFD if and only if every non-zero element of D can

be expressed as u · pe1 · · · per , where u is a unit, p1 , . . . , pr are distin-

r

1

guished irreducibles, and this expression is unique up to a reordering

of the pi .

√

Exercise 17.22. Show that the ring Z[ ’5] is not a UFD.

Exercise 17.23. Let D be a UFD and F its ¬eld of fractions. Show that

(a) every element x ∈ F can be expressed as x = a/b, where a, b ∈ D are

relatively prime, and

(b) that if x = a/b for a, b ∈ D relatively prime, then for any other

a , b ∈ D with x = a /b , we have a = ca and b = cb for some c ∈ D.

Exercise 17.24. Let D be a UFD and let p ∈ D be irreducible. Show that

there is no prime ideal Q of D with {0D } Q pD.

17.8 Unique factorization domains (—) 387

17.8.1 Unique factorization in Euclidean and principal ideal

domains

Our proofs of the unique factorization property in both Z and F [X] hinged

on the division with remainder property for these rings. This notion can be

generalized, as follows.

De¬nition 17.28. D is said to be a Euclidean domain if there is a “size

function” S mapping the non-zero elements of D to the set of non-negative

integers, such that for a, b ∈ D with b = 0, there exist q, r ∈ D, with the

property that a = bq + r and either r = 0 or S(r) < S(b).

Example 17.13. Both Z and F [X] are Euclidean domains. In Z, we can

take the ordinary absolute value function | · | as a size function, and for F [X],

the function deg(·) will do. 2

Example 17.14. Recall again the ring

Z[i] = {a + bi : a, b ∈ Z}

of Gaussian integers from Example 9.22. Let us show that this is a Euclidean

domain, using the usual norm map N on complex numbers (see Example 9.5)

for the size function. Let ±, β ∈ Z[i], with β = 0. We want to show the

existence of ξ, ρ ∈ Z[i] such that ± = βξ + ρ, where N (ρ) < N (β). Suppose

that in the ¬eld C, we compute ±β ’1 = r + si, where r, s ∈ Q. Let m, n be

integers such that |m ’ r| ¤ 1/2 and |n ’ s| ¤ 1/2”such integers m and n

always exist, but may not be uniquely determined. Set ξ := m + ni ∈ Z[i]

and ρ := ± ’ βξ. Then we have