312 A. ALGEBRAIC PRELIMINARIES
A.2 RINGS AND FIELDS OF FRACTIONS
Throughout this section “ring” means a commutative ring with identity.
Suppose R is a ring.An element a ∈ R is an absolute nondivisor of zero if b ∈ R,ab = 0 implies
that b = 0. In every ring, there are absolute nondivisors of zero; for example, all the units are of this
type. But there may be others as well.
Examples A.2.1 In the ring Z
, a matrix M is an absolute nondivisor of zero if and only if
the determinant of M is nonzero. Thus, a nonunit matrix can be an absolute nondivisor of zero
(cf. Problem A.1.8).
Now consider the ring C[0, 1] deﬁned in Example A.1.5. Suppose a function x(·) belonging
to this ring vanishes at only a ﬁnite number of points. If y ∈ C[0, 1] and xy = 0, then x(t)y(t) ≡ 0,
which means that y(t) = 0 for all except a ﬁnite number of values of t. However, since y(·) is
continuous, this implies that y(t) ≡ 0, or that y is the zero element of the ring. Hence, x is an
absolute nondivisor of zero.
If a is an absolute nondivisor of zero and ab = ac, then b = c; in other words, absolute
nondivisors of zero can be “cancelled.”
Of course,if R is a domain, then every nonzero element of R is an absolute nondivisor of zero.
A set M in a ring R is said to be a multiplicative system if a, b ∈ M implies that ab ∈ M.Itis
saturated if a ∈ R
,b ∈ R,ab ∈ M implies that a ∈ M,b ∈ M.
Fact A.2.2 The set N of absolute nondivisors of zero in a ring R is a multiplicative system.
Proof. Suppose a, b ∈ N,y ∈ R, and aby = 0. Then, since a, b ∈ N, it follows successively that
aby = 0 ⇒ by = 0 ⇒ y = 0. Hence, ab ∈ N.
Suppose R is a ring, M is a multiplicative system in R containing 1, and M is a subset of N
(the set of absolute nondivisors of zero in R). We now begin a construction which will ultimately
result in a ring L which contains R as a subring, and in which every element of M is a unit.
Consider the set R ×M, and deﬁne a binary relation ∼ on R × M as follows: (a, b) ∼
(c, d) ⇐⇒ ad = bc.The relation ∼ is an equivalence relation: Clearly ∼ is reﬂexive and symmetric.
To show that it is transitive,suppose (a, b) ∼ (c, d) and (c, d) ∼ (e, f ).Then ad = bc and cf = de.
Multiplying the ﬁrst equation by f and the second one by b gives adf = bcf = bde.Nowd ∈ N
since d ∈ M and M ⊆ N. Hence, d can be cancelled in the above equation to give af = be, i.e.,
(a, b) ∼ (e, f ).
Since ∼is an equivalence relation, the set R × M can be partitioned into disjoint equivalence
∼. The set of equivalence classes R × M/ ∼ is denoted by L. The set L therefore
consists of fractions (a, b),ora/b in more familiar terms, where we agree to treat two fractions a/b