What would you like to understand?

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