We describe a construction known as the ring of fractions that has elements of the form . It is also known as the ring of quotients, but it sounds too similar to the quotient ring, a completely different concept.
Proof. and are obvious. Assume . Then such that and . Then Thus . If has no zero divisors and , then . Since is not a zero divisor, . ◻
Let be a multiplicative subset of a commutative ring and the equivalence relation from earlier. The equivalence class of is denoted . The set of all equivalence classes is denoted . Verify that
for all
If , then consists of a single equivalence class.
Proof. (i) If
,
, then
such that
Multiply the first equation by
and the second by
. Add to get
Therefore
. If we instead multiply the first equation by
and the second by
, then add, we get
Thus
. Hence addition and multiplication on
is well-defined.
(ii). If
has no zero divisors and
, then
. Thus
. Since
iff
or
, it follows that
is an integral domain.
(iii). If
, the multiplicative inverse of
is
. ◻
The ring is called the ring of fractions of by . When is an integral domain and the set of all nonzero elements, is called the fraction field or quotient field of . If is a nonzero commutative ring and is the set of all elements of that are not zero divisors of , then is called the total ring of fractions of .
Proof. (i). If
, then
whence
is well-defined. Verify that
is a ring homomorphism and that
is the multiplicative inverse of
.
(ii). If
. Since
,
.
(iii).
is a monomorphism by (ii). If
,
. Thus
is an isomorphism. ◻
It is customary to identify an integral domain with its image under and consider as a subring of its quotient field.
Proof. Verify that given by is a well-defined homomorphism of rings such that . If is another homomorphism such that , then is a unit in . . Thus . Let be the category whose objects are all where is a commutative ring with identity and a homomorphism of rings such that is a unit in for every . Define a morphism in from to to be a homomorphism of rings such that . Verify that is a category and that a morphism in is an equivalence iff is an isomorphism of rings. Then by the preceding work we have shown is a universal object in , whence is completely determined up to isomorphism. ◻
Proof. Let and apply the theorem to to obtain a homomorphism such that . is a monomorphism. Since is identified with , . Last statement follows when is the inclusion map. ◻