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What is the general criterion for equality of two fractions r1/s1r_1/s_1 and r2/s2r_2/s_2 in a localization?

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.

Reasoning, step by step

  1. Start with two arbitrary fractions r1/s1r_1/s_1 and r2/s2r_2/s_2.
  2. Compute the cross-product difference r1s2−s1r2r_1s_2 - s_1r_2.
  3. Check if there exists an element t∈St \in S that annihilates this difference, i.e., t(r1s2−s1r2)=0t(r_1s_2 - s_1r_2) = 0.
  4. Conclude that the fractions are equivalent if and only if such a tt exists.

Example

The board writes r1s1=r2s2\frac{r_1}{s_1} = \frac{r_2}{s_2}, then (r1s2−s1r2)=0(r_1 s_2 - s_1 r_2)=0, then ∃t∈S:t(r1s2−s1r2)=0\exists t \in S : t(r_1 s_2 - s_1 r_2) = 0. The speaker explains that equality holds not only when the cross product itself vanishes, but when there exists some t∈St \in S such that t(r1s2−s1r2)=0t(r_1 s_2 - s_1 r_2)=0.

Common misconceptions

  • Thinking two fractions in S−1RS^{-1}R are equal exactly when r1s2=s1r2r_1s_2=s_1r_2.
  • Assuming the naive cross-multiplication rule is sufficient in the general localization setting.

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