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. ◻
Proof. Note that , Let where . Then . Thus . ◻
Proof. If , and hence . Conversely, if , whence . Since , . Thus . ◻
is called the extension of in . If is an ideal in , then is an ideal in . is called the contraction of in .
Proof. (i). If
. Thus
whence
.
(ii). Since
, every element of
is of the form
where
. Thus
whence
. Conversely, if
, then
whence
. Thus
and
.
(iii).
is an ideal such that
by the previous theorem. If
then
with
.
. Since
and
,
whence
or
. Thus
or
. Thus
is prime.
by (i). If
, then
. Thus
with
.
,
. Since
. ◻
Proof. By the preceding lemma, is injective. Let be a prime ideal of and let . Since , it suffices to show that is prime. If . Since is prime, or . Thus either or . ◻
Let be a commutative ring with identity and a prime ideal of . Then is multiplicative. is called the localization of at and is denoted . If is an ideal in , then the ideal in is denoted .
Proof. The prime ideals contained in are precisely those disjoint from . (i) follows from the previous theorem. If is a maximal ideal of , then is prime. Whence for some prime ideal of with . But implies . Since , we must have . Thus is the unique maximal ideal in . ◻
Proof. If
is an ideal of
and
, then
. Thus
iff
consists only of nonunits. These facts imply
(ii)(iii) and
(iii)(i).
(i)(ii). If
is a nonunit, then
. Thus
is contained in the unique maximal ideal of
. ◻