Why Read This Book? - Index of

9.8 Integral Domains 317
Corollary 9.8.18 In a domain, (a) = (b) if and only if a and b are associates.
If a is a proper divisor of b, then by Exercises 9.8.5 and 9.8.14, a and b are not
associates. So (b) ⊆ (a) by Exercise 9.8.17, but (a) �= (b) by Corollary 9.8.18. Thus
we have the following.
Corollary 9.8.19 If a is a proper divisor of b in a domain, then (b) is a proper
subset of (a).
Now let’s look at the relationship between prime elements and prime ideals.
EXERCISE 9.8.20 If D is a domain and (p) is a prime ideal, then p is prime in D.
But what about the converse? If p is a prime element of a domain, must it
generate a prime ideal? In the integers the answer is yes. Thanks to Exercise 2.5.11,
if p is a prime number, then (p) is a prime ideal in the integers. But in the integers,
Exercise 2.5.11 depends on the existence of greatest common divisors, which itself
depends on theWOP. Strangely, in some domains a prime element can generate an
ideal that is not prime, and a prime element p might satisfy p | ab, while p divides
neither a nor b. The next exercise illustrates these possibilities. We provide the
solution to the first part and leave the rest to you. We will return to this example
in Section 9.9.
EXERCISE 9.8.21 The ring Z[ √ −5] is a domain because it is a subring of the
complex numbers and contains 1. However, you can show the following.
(a) 3 is prime in Z[ √ −5].
Solution First note
To show 3 is prime in Z[ √ −5], suppose
9 = 3 · 3 = (2 + � −5)(2 − � −5) (9.50)
3 = (a + b � −5)(c + d � −5) (9.51)
where a, b, c, d are integers. We show that one of these two factors must be
a unit, which by Theorem 9.4.3 must be ±1. Multiplying out the factors and
using the definition of equality in Z[ √ −5] yields
ac − 5bd = 3 and ad + bc = 0 (9.52)
Proceeding as in the proof of Theorem 9.4.3, we square each equation, multiply
the latter by 5, add and factor to have
(a 2 + 5b 2 )(c 2 + 5d 2 ) = 9 (9.53)