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Two fractions $r_1/s_1$ and $r_2/s_2$ are equal in the localization $S^{-1}R$ exactly when there exists an element $t \in S$ such that $t(r_1s_2 - s_1r_2) = 0$. This refines the naive cross-multiplication rule to handle zero-divisor phenomena in general commutative rings.

Conditions: $R$ is a commutative ring.; $S \subset R$ is multiplicatively closed.; $r_1, r_2 \in R$ and $s_1, s_2 \in S$.

Two fractions $r_1/s_1$ and $r_2/s_2$ are equal in the localization $S^{-1}R$ exactly when there exists an element $t \in S$ such that $t(r_1s_2 - s_1r_2) = 0$. This refines the naive cross-multiplication rule to handle zero-divisor phenomena in general commutative rings.

Conditions: $R$ is a commutative ring.; $S \subset R$ is multiplicatively closed.; $r_1, r_2 \in R$ and $s_1, s_2 \in S$.

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

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

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

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

An element $a \in R$ becomes zero in the localized ring $S^{-1}R$ if there exists an element $s \in S$ such that the product $sa = 0$. This happens because the fraction $a/1$ can be rewritten as $sa/s$, which equals $0/s$, and thus $0$ in $S^{-1}R$.

Conditions: $R$ is a commutative ring.; $S \subset R$ is multiplicatively closed.; $a \in R$ and $s \in S$.; $sa = 0$.

An element $a \in R$ becomes zero in the localized ring $S^{-1}R$ if there exists an element $s \in S$ such that the product $sa = 0$. This happens because the fraction $a/1$ can be rewritten as $sa/s$, which equals $0/s$, and thus $0$ in $S^{-1}R$.

Conditions: $R$ is a commutative ring.; $S \subset R$ is multiplicatively closed.; $a \in R$ and $s \in S$.; $sa = 0$.