What is the total ring of fractions?

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 localization $S^{-1}R$ is defined as the collection of formal fractions $r/s$ where the numerator $r$ is an element of the commutative ring $R$ and the denominator $s$ is an element of the multiplicatively closed subset $S$.

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

A subset $S$ of a ring $R$ is multiplicatively closed if it is stable under multiplication and contains the multiplicative identity. Specifically, if two elements are taken from $S$, their product also sits in $S$, and the identity element belongs to $S$.

Conditions: $S$ is a subset of the commutative ring $R$.; Products of elements of $S$ remain in $S$.; The identity element $1$ belongs to $S$.

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 $R$.

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