MA3412S2_Hil2014.pdf

# Proof it is clear from the definition of the relation

• Notes
• 38

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Proof It is clear from the definition of the relation S that ( r, s ) S ( r, s ) for all ( r, s ) R × S . Moreover elements ( r, s ) and ( r 0 , s 0 ) of R × S satisfy ( r, s ) S ( r 0 , s 0 ) if and only if ( r 0 , s 0 ) S ( r, s ). Thus the relation S on R × S is both reflexive and symmetric. Let ( r, s ), ( r 0 , s 0 ) and ( r 00 , s 00 ) be elements of R × S . Suppose that ( r, s ) S ( r 0 , s 0 ) and ( r 0 , s 0 ) S ( r 00 , s 00 ). Then there exist elements u and v of S such that us 0 r = usr 0 and vs 00 r 0 = vs 0 r 00 . Then ( uvs 0 )( s 00 r ) = ( vs 00 )( us 0 r ) = ( vs 00 )( usr 0 ) = ( us )( vs 00 r 0 ) = ( us )( vs 0 r 00 ) = ( uvs 0 )( sr 00 ) , and uvs 0 S , and therefore ( r, s ) S ( r 00 , s 00 ). Thus the relation S on R × S is transitive. It follows that this relation is an equivalence relation on R × S . Lemma 2.39 Let S be a multiplicatively closed subset of a unital commu- tative ring R , and let S be the equivalence relation on R × S defined so that elements ( r, s ) and ( r 0 , s 0 ) of R × S satisfy ( r, s ) S ( r 0 , s 0 ) if and only if there exists some element u of S for which us 0 r = usr 0 . Let ( r 1 , s 1 ) , ( r 0 1 , s 0 1 ) , ( r 2 , s 2 ) and ( r 0 2 , s 0 2 ) be elements of R × S satisfying ( r 1 , s 1 ) S ( r 0 1 , s 0 1 ) and ( r 2 , s 2 ) S ( r 0 2 , s 0 2 ) . Then ( s 2 r 1 + s 1 r 2 , s 1 s 2 ) S ( s 0 2 r 0 1 + s 0 1 r 0 2 , s 0 1 s 0 2 ) and ( r 1 r 2 , s 1 s 2 ) S ( r 0 1 r 0 2 , s 0 1 s 0 2 ) . Proof There exist elements u 1 and u 2 of S such that u 1 s 0 1 r 1 = u 1 s 1 r 0 1 and u 2 s 0 2 r 2 = u 2 s 2 r 0 2 . Then ( u 1 u 2 )( s 0 1 s 0 2 )( s 2 r 1 + s 1 r 2 ) = ( u 2 s 2 s 0 2 )( u 1 s 0 1 r 1 ) + ( u 1 s 1 s 0 1 )( u 2 s 0 2 r 2 ) = ( u 2 s 2 s 0 2 )( u 1 s 1 r 0 1 ) + ( u 1 s 1 s 0 1 )( u 2 s 2 r 0 2 ) = ( u 1 u 2 )( s 1 s 2 )( s 0 2 r 0 1 + s 0 1 r 0 2 ) , and u 1 u 2 S , and therefore ( s 2 r 1 + s 1 r 2 , s 1 s 2 ) S ( s 0 2 r 0 1 + s 0 1 r 0 2 , s 0 1 s 0 2 ) . 40

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Also ( u 1 u 2 )( s 0 1 s 0 2 )( r 1 r 2 ) = ( u 1 s 0 1 r 1 )( u 2 s 0 2 r 2 ) = ( u 1 s 1 r 0 1 )( u 2 s 2 r 0 2 ) = ( u 1 u 2 )( s 1 s 2 )( r 0 1 r 0 2 ) , and therefore ( r 1 r 2 , s 1 s 2 ) S ( r 0 1 r 0 2 , s 0 1 s 0 2 ) , as required. Let S be a multiplicatively closed subset of a unital commutative ring R , and let S be the equivalence relation on R × S defined so that elements ( r, s ) and ( r 0 , s 0 ) of R × S satisfy ( r, s ) S ( r 0 , s 0 ) if and only if there exists some element u of S for which us 0 r = usr 0 . Then the equivalence relation S partitions the set R × S into equivalence classes. We denote by r/s the equivalence class of an element ( r, s ) of R × S . Then r, r 0 R and s, s 0 S satisfy r/s = r 0 /s 0 if and only if there exists some element u of S for which us 0 r = urs 0 . It follows from Lemma 2.39 that there are well-defined operations of addition and multiplication defined on S - 1 R , where ( r 1 /s 1 ) + ( r 2 s 2 ) = ( s 2 r 1 + s 1 r 2 ) / ( s 1 s 2 ) and ( r 1 /s 1 )( r 2 /s 2 ) = ( r 1 r 2 ) / ( s 1 s 2 ) . If 0 R S then S - 1 R is the zero ring, consisting of a single element. We now show that if 0 R 6∈ S then S - 1 R is a unital commutative ring. Proposition 2.40 Let S be a multiplicatively closed subset of a unital com- mutative ring R , where 0 R 6∈ S , and let S be the equivalence relation on R × S defined so that elements ( r, s ) and ( r 0 , s 0 ) of R × S satisfy ( r, s ) S ( r 0 , s 0 ) if and only if there exists some element u of S for which us 0 r = usr 0 .
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• Fall '16
• Jhon Smith
• Algebra, Integers, Prime number, Integral domain, Ring theory, Principal ideal domain

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