Why Read This Book? - Index of

9.8 Integral Domains 315
EXERCISE 9.8.5 Suppose D is a domain and a and b are domain elements such
that a | b. Then the element k such that ak = b is unique.
With Exercise 9.8.5, the following term becomes meaningful.
Definition 9.8.6 Suppose D is a domain, a, b ∈ D, and a is not a unit. If a | b,
where ak = b and k is not a unit, then a is called a proper divisor of b.
In a ring in which the result of Exercise 9.8.5 does not apply, such as those in
Exercise 9.3.10, then Definition 9.8.6 cannot be unambiguously applied. For it is
possible to have ak1 = b and ak2 = b in a ring R where k1 is a unit in R but k2 is not.
Example 9.8.7 In Z2×2,
�
1
1
��
1 1
1 0
�
1
=
1
Now
while
�
1
0
�
1
is a unit in Z2×2, for
1
�
1
��
1 1
�
−1
0 1 0 1
=
� �
2 1
is not a unit in Z2×2. �
−1 1
EXERCISE 9.8.8 Show that
� �
1 2
=
1 2
� 1 1
1 1
�
1
��
−1 1
�
1
0 1 0 1
��
2
�
1
−1 1
=
� �
1 0
0 1
� �
2 1
is not a unit in Z2×2.
−1 1
(9.48)
(9.49)
Multiplicative cancellation in a domain is a logical consequence of the principle
of zero products. We could have defined a domain as a commutative ring with unity
where multiplicative cancellation holds for nonzero elements, and we could have
then shown that the principle of zero products follows from that. In a commutative
ring, the principle of zero products and multiplicative cancellation are logically
equivalent.
EXERCISE 9.8.9 A commutative ring with unity and multiplicative cancellation
is a domain.
If a commutative ring with unity has a nonzero characteristic, then the only
way it can be a domain is if the characteristic is prime.
EXERCISE 9.8.10 If a domain has nonzero characteristic n, then n is prime.
With Theorem 9.3.25 and Exercises 9.8.1 and 9.8.10, we see that Zn is a domain
if and only if n is prime. The only feature a field has that a domain might not is the