Ring of fractions and localization

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

  2. for all

  3. 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. ◻