MA3412S2_Hil2014.pdf

# S satisfying ur 0 r and this element u is a zero

• Notes
• 38

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S satisfying ur = 0 R , and this element u is a zero divisor of R . It follows that the multiplicative subset S of R contains a zero divisor if and only if the natural homomorphism λ : R S - 1 R is not injective. The result follows. Let R be a unital commutative ring. It follows from Lemma 2.42 that the homomorphism λ 0 : R Q ( R ) from R to its total ring of fractions Q ( R ) that sends each element r of R to r/ 1 R is injective. Moreover if S is a multiplica- tively closed subset of R , and if the homomorphism λ : R S - 1 R sending each element r of R to r/ 1 R is injective, then S is a subset of the set R reg of regular elements of R , and the ring S - 1 R can therefore be embedded as a subring of Q ( R ). Thus the total ring of fractions Q ( R ) of R is the largest ring of fractions into which the ring R can be embedded. Other important rings of fractions arise through a process known as lo- calization . Let R be a unital commutative ring, let P be a prime ideal of R , and let S be the complement of P in R . It follows from the definition of prime ideals that S is a multiplicatively closed subset of R . Indeed an ideal of R is prime if and only if its complement is multiplicatively closed. We define R P = S - 1 R , where S = R \ P . The ring R P is the localization of R at the prime ideal P . Each ideal I of R determines a corresponding ideal I P of R P , where I P = { x/s R P : x I and s R \ P } . Lemma 2.43 Let R be a unital commutative ring, let P be a prime ideal of R , and let R P be the localization ( R \ P ) - 1 R of R at the prime ideal P . Then the ideal P P of R P consisting of all elements of R P that are of the form x/s for some x P and s R \ P is a maximal ideal of R P . Moreover it is the only maximal ideal of R P . 44

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Proof We show that P P is a proper ideal of R P . Suppose that there were to exist x P and s R \ P such that x/s = 1 R / 1 R . Then there would exist some element u of R \ P such that ux = us . But this is impossible, because ux would be an element of the ideal P and us would be an element of the complement R \ P of P . Therefore the identity element of R P does not belong to the ideal P P , and hence this ideal is a proper ideal of R P . If r R , s R \ P and r/s R P \ P P then r R \ P , and therefore r/s is a unit of R P with inverse s/r . It follows that no proper ideal of R P can intersect the complement of P P and therefore all proper ideals of R P are contained in P P . It follows that P P is a maximal ideal of R P , and is the only maximal ideal of R P . Let f be an element of a unital commutative ring R , and let S be the set { 1 R , f, f 2 , f 3 , . . . } of powers of f . Then S is a multiplicatively closed subset of R which therefore gives rise to a corresponding ring of fractions R f , where R f = S - 1 R . Elements of R f are represented as fractions of the form r/f k , where r R and k is some non-negative integer. If f n = 0 R for some positive integer n then R f is the zero ring.
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• Fall '16
• Jhon Smith
• Algebra, Integers, Prime number, Integral domain, Ring theory, Principal ideal domain

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