Why Read This Book? - Index of

320 Chapter 9 Rings
on coefficients in Z[t] that do not apply in Q[t]. To show that Z[t] is a UFD, it helps
to view it in the context of Q[t].
Since a UFD is by definition an integral domain, we do not need a theorem
claiming that a UFD is a domain. But not all domains are UFDs. For example,
Z[ √ −5] is a domain, and Eq. (9.50) shows that 9 has two distinct factorizations
into prime elements. Thus Z[ √ −5] is not a UFD.
Definition 9.9.1 Suppose D is a domain, and let a and b be nonzero elements
of D. Suppose g is a domain element with the following characteristics:
(D1) g | a and g | b.
(D2) If h is any element of D with the properties that h | a and h | b, then it must
be that h | g also.
Then g is called a greatest common divisor of a and b and is denoted gcd(a, b).
In Section 2.5, we said that a practical way you might find gcd(a, b) in the
positive integers is by breaking down a and b into their prime factorizations, taking
the appropriate number of 2s, 3s, and so on, and building the gcd from that.
We did not prove that such a trick produces a positive integer with properties
D1–D2 because the WOP provided an easier and more useful way. Alas, in a
general UFD, the existence of gcd(a, b), unique up to association, must be proved
by exploiting the unique prime factorizations of a and b in that somewhat sloppy
way. We state the theorem here, followed by some details of the proof that might
make the notation minimally sloppy. You will write the complete proof as an
exercise.
Theorem 9.9.2 Suppose D is a UFD and a and b are nonzero elements of D.
Then there exists g ∈ D that satisfies D1–D2, and if g1 and g2 both satisfy D1–D2,
then g1 and g2 are associates.
Here are some suggestions on how to prove Theorem 9.9.2 and keep the notation
from getting outrageously complicated. If we break a and b down into prime
factorizations, then let {p1,p2,...,pn} be all the primes that appear in either a
or b, we can write
a = p α1
1 pα2
2 ...pαn n and b = p β1
1 pβ2
2 ...pβn n
(9.54)
where some of the αk and βk might be zero. If we let γk = min{αk,βk} for all
1 ≤ k ≤ n, you can show that g = p γ1
1 pγ2
2 ...pγn n satisfies D1–D2. Since γk =
min{αk,βk}, we know that γk ≤ αk and γk ≤ βk for all k. This should make it easy
to show that g has property D1. To show that g has property D2, suppose h | a
and h | b. Then there exist k1,k2 ∈ D such that hk1 = a and hk2 = b. If the unique
factorization of h is written as h = q δ1
1 qδ2
2 ...qδm m , then the prime factorizations of
hk1 and hk2 must agree with Eqs. (9.54), so that some possible reordering of the pk