Why is localization considered a generalization of the field of fractions construction?

Conditions

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

Reasoning, step by step

1. Start with a general commutative ring $R$ and a multiplicatively closed subset $S$.
2. Define the localization $S^{-1}R$ as fractions $r/s$.
3. Note that if $R$ is an integral domain and $S$ is the set of nonzero elements, $S^{-1}R$ is exactly the field of fractions of $R$.

Example

The lecturer says this is already familiar in the case where $R$ is an integral domain, and then taking $S$ to be $R$ without zero gives the field of fractions of $R$. The board writes "If $R$ is an integral domain" and then "$S \subset R_0 \Rightarrow S^{-1}R =$ field of fractions".

Common misconceptions

• Thinking that forming fractions like $r/s$ only makes sense when the base ring is an integral domain.
• Assuming that localization always produces a field.

Watch the explanation

Connected concepts

Explore next

• What is the definition of the localization $S^{-1}R$ of a commutative ring R at a multiplicatively closed set S?
• Under what condition does an element $a \in R$ become zero in $S^{-1}R$?