Let ϕ r t be a unital homomorphism from r to a

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Let ϕ : R T be a unital homomorphism from R to a unital ring T that maps the multiplicatively closed subset S of R to the group of units of T , and let r, r 0 R and s, s 0 S satisfy r/s = r 0 /s 0 . Then there exists some element u of S such that us 0 r = usr 0 . Then ϕ ( u ) ϕ ( s 0 ) ϕ ( r ) = ϕ ( u ) ϕ ( s ) ϕ ( r 0 ). 42
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But ϕ ( u ), ϕ ( s ) and ϕ ( s 0 ) are units of T . It follows that ϕ ( r ) ϕ ( s ) - 1 = ϕ ( r 0 ) ϕ ( s 0 ) - 1 . There is thus a well-defined function ˆ ϕ : S - 1 R T satisfy- ing ˆ ϕ ( r/s ) = ϕ ( r ) ϕ ( s ) - 1 for all r R and s S . Moreover ˆ ϕ (( r 1 /s 1 ) + ( r 2 /s 2 )) = ˆ ϕ (( s 2 r 1 + s 1 r 2 ) / ( s 1 s 2 )) = ϕ ( s 2 r 1 + s 1 r 2 ) ϕ ( s 1 s 2 ) - 1 = ( ϕ ( s 2 ) ϕ ( r 1 ) + ϕ ( s 1 ) ϕ ( r 2 )) ϕ ( s 1 ) - 1 ϕ ( s 2 ) - 1 = ϕ ( r 1 ) ϕ ( s 1 ) - 1 + ϕ ( r 2 ) ϕ ( s 2 ) - 1 = ˆ ϕ ( r 1 /s 1 ) + ˆ ϕ ( r 2 /s 2 ) and ˆ ϕ (( r 1 /s 1 )( r 2 /s 2 )) = ˆ ϕ (( r 1 r 2 ) / ( s 1 s 2 )) = ϕ ( r 1 r 2 ) ϕ ( s 1 s 2 ) - 1 = ϕ ( r 1 ) ϕ ( r 2 ) ϕ ( s 1 ) - 1 ϕ ( s 2 ) - 1 = ˆ ϕ ( r 1 /s 1 ) ˆ ϕ ( r 2 /s 2 ) for all r 1 , r 2 R and s 1 , s 2 S , and ˆ ϕ (1 R / 1 R ) = 1 T , where 1 T is the multiplicative identity element of T . Also ˆ ϕ ( λ ( r )) = ˆ ϕ ( r/ 1 R ) = ϕ ( r ) for all r R . It follows that ˆ ϕ : S - 1 R T is a unital homomorphism that satisfies ˆ ϕ λ = ϕ . Now let ψ : S - 1 R T be a unital homomorphism that satisfies ψ λ = ϕ , and let r and s be elements of R and S respectively. Then ( r/s )( s/ 1 R ) = r/ 1 R . Moreover ψ ( r/ 1 R ) = ψ ( λ ( r )) = ϕ ( r ) and ψ ( s/ 1 R ) = ψ ( λ ( s )) = ϕ ( s ). Therefore ψ ( r/s ) ϕ ( s ) = ϕ ( r ), and thus ψ ( r/s ) = ϕ ( r ) ϕ ( s ) - 1 = ˆ ϕ ( r/s ). This shows that ˆ ϕ : S - 1 R T is the unique unital homomorphism from S - 1 R T that satisfies ˆ ϕ λ = ϕ . Let R be a unital commutative ring. An element x of R is said to be a zero divisor if there exists some non-zero element y of R for which xy = 0 R . The multiplicative identity element 1 R of R is not a zero divisor. Definition An element x of a unital commutative ring R is said to be regular if it is not a zero divisor. The multiplicative identity element 1 R is regular. Let x and y be regular elements of R . Then x and y are non-zero elements of R , because the zero element of R is a zero divisor. Moreover xyz 6 = 0 R for all non-zero elements z if R . It follows that xy is a regular element of R . Thus the set R reg of regular elements of the unital commutative ring R is a multiplicatively closed set. We can therefore form the corresponding ring of fractions Q ( R ), where Q ( R ) = R - 1 reg R . 43
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Definition Let R be a unital commutative ring. The total ring of fractions Q ( R ) of R is the ring R - 1 reg R , where R reg is the set consisting of all regular elements of R . Lemma 2.42 Let R be a unital commutative ring, and let S be a multiplica- tively closed subset of S , where 0 R 6∈ S . Let λ : R S - 1 R be the natural homomorphism that maps each element r of R to r/ 1 R . Then the homomor- phism λ is injective if and only if every element of the set S is a regular element of R . Proof Suppose that u S is a zero divisor. Then there exists some non- zero element r of R for which ur = 0 R . But then λ ( r ) = 0 R / 1 R , and thus r ker λ . Thus if S contains a zero divisor then the homomorphism λ is not injective. Conversely if the homomorphism λ is not injective then there exists some non-zero element r of R satisfying r/ 1 R = 0 R / 1 R . There must then exist some element u
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  • Fall '16
  • Jhon Smith
  • Algebra, Integers, Prime number, Integral domain, Ring theory, Principal ideal domain

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