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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 $s \in S$.; $sa = 0$.

Localization is considered a generalization because it introduces fraction-like elements into a general commutative ring, whereas the field of fractions construction only applies to integral domains. When $R$ is an integral domain and $S$ is chosen to be the set of nonzero elements, the localization $S^{-1}R$ recovers the familiar field of fractions.

Conditions: $R$ is a commutative ring.; In the special case, $R$ is an integral domain and $S$ is the set of nonzero elements of $R$.

Localization is considered a generalization because it introduces fraction-like elements into a general commutative ring, whereas the field of fractions construction only applies to integral domains. When $R$ is an integral domain and $S$ is chosen to be the set of nonzero elements, the localization $S^{-1}R$ recovers the familiar field of fractions.

Conditions: $R$ is a commutative ring.; In the special case, $R$ is an integral domain and $S$ is the set of nonzero elements of $R$.