A <multiplicative subset> $S\subseteq R$ contains $1$ and satisfies $s,t\in S\Rightarrow st\in S$. The <localization of a ring> consists of fractions $r/s$ modulo the relation
$$
\frac r s=\frac{r'}{s'}
\quad\Longleftrightarrow\quad
\text{some }u\in S\text{ satisfies }u(rs'-r's)=0.
$$
Its structure map $\iota:R\to S^{-1}R$ makes every $s\in S$ invertible. The <universal property of localization> says that if $f:R\to T$ is any ring homomorphism for which every $f(s)$ is a unit, there is one unique homomorphism $\widetilde f:S^{-1}R\to T$ satisfying $\widetilde f\iota=f$, namely
$$
\widetilde f(r/s)=f(r)f(s)^{-1}.
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 101 / 2 / i / Solution

{title2=$S^{-1}R$}
{wiki=Localization_(commutative_algebra)}

For a multiplicatively closed set $S$, the localization $S^{-1}R$ formally makes every element of $S$ invertible. Elements are represented by fractions $r/s$.


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 Incoming links: Localization of a ring