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. ◻