# Localisation (ring theory)

(Redirected from Field of fractions)  Main Article Discussion Related Articles  [?] Bibliography  [?] External Links  [?] Citable Version  [?] This editable Main Article is under development and subject to a disclaimer. [edit intro]

In ring theory, the localisation of a ring is an extension ring in which elements of the base ring become invertible.

## Construction

Let R be a commutative ring and S a non-empty subset of R closed under multiplication. The localisation $S^{-1}R$ is an R-algebra in which the elements of S become invertible, constructed as follows. Consider the set $R\times S$ with an equivalence relation $(x,s)\sim (y,t)\Leftrightarrow xt=ys$ . We denote the equivalence class of (x,s) by x/s. Then the quotient set becomes a ring $S^{-1}R$ under the operations

${\frac {x}{s}}+{\frac {y}{t}}={\frac {xt+ys}{st}}$ ${\frac {x}{s}}\cdot {\frac {y}{t}}={\frac {xy}{st}}.$ The zero element of $S^{-1}R$ is the class $0/s$ and there is a unit element $s/s$ . The base ring R is embedded as $x\mapsto {\frac {xs}{s}}$ .

### Localisation at a prime ideal

If ${\mathfrak {p}}$ is a prime ideal of R then the complement $S=R\setminus {\mathfrak {p}}$ is a multiplicatively closed set and the localisation of R at ${\mathfrak {p}}$ is the localisation at S, also denoted by $R_{\mathfrak {p}}$ . It is a local ring with a unique maximal ideal — the ideal generated by ${\mathfrak {p}}$ in $R_{\mathfrak {p}}$ .

## Field of fractions

If R is an integral domain, then the non-zero elements $S=R\setminus \{0\}$ form a multiplicatively closed subset. The localisation of R at S is a field, the field of fractions of R. A ring can be embedded in a field if and only if it is an integral domain.