Given a commutative ring $ R $ and a multiplicatively closed subset $ S $ of $ R $, there are two ways to consturct $ S^{-1}R $:

define an equivalence relation $ \sim $ on $ R\times S $ and then take $ S^{-1}R := (R\times S)/\sim $.

Let $ R[S] $ be the free commutative algebra over $ R $ generated by $ S $ and $ i:S\to R[S] $ the canonical embedding and $ I $ the ideal of $ R[S] $ generated by $ i(s)s-1 $ for each $ s $ in $ S $ and then take $ S^{-1}R := R[S]/I $.

Suppose $ S $ dosen't contain zero divisors, if one works in the first way, it's easy to prove that the canonical map $ j:R\to R^{-1}S $ is injective. But if one works in the second way, then it's equivalent to state that $ I \cap R = \{ 0 \}$. Can we directly prove this without considering $ (R\times S)/\sim $ ?

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