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Answers for “环局部化中分数的等价关系是什么?”

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Meet the concept

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Two fractions r1/s1r_1/s_1 and r2/s2r_2/s_2 in the localization S−1RS^{-1}R are defined to be equivalent if there exists an element tt in the multiplicatively closed set SS such that t(r1s2−s1r2)=0t(r_1s_2 - s_1r_2) = 0. This relation ensures that the fractions represent the same element in the localization.

Conditions: RR is a commutative ring.; S⊂RS \subset R is a multiplicatively closed subset.; r1,r2∈Rr_1, r_2 \in R and s1,s2∈Ss_1, s_2 \in S.

Understand why

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The naive condition r1s2=s1r2r_1s_2 = s_1r_2 is insufficient because in a general commutative ring, elements can be annihilated by nonzero elements of SS. The refined condition requires the existence of some t∈St \in S such that t(r1s2−s1r2)=0t(r_1s_2 - s_1r_2) = 0, which accounts for zero-divisor phenomena and ensures the equivalence relation is well-defined.

Conditions: RR is a general commutative ring (not necessarily an integral domain).; S⊂RS \subset R is multiplicatively closed.

Meet the concept

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Two fractions r1/s1r_1/s_1 and r2/s2r_2/s_2 are equal in the localization S−1RS^{-1}R exactly when there exists an element t∈St \in S such that t(r1s2−s1r2)=0t(r_1s_2 - s_1r_2) = 0. This refines the naive cross-multiplication rule to handle zero-divisor phenomena in general commutative rings.

Conditions: RR is a commutative ring.; S⊂RS \subset R is multiplicatively closed.; r1,r2∈Rr_1, r_2 \in R and s1,s2∈Ss_1, s_2 \in S.