# Localisation (ring theory)

(Redirected from Field of fractions)

Main Article
Discussion
Related Articles  [?]
Bibliography  [?]
Citable Version  [?]

This editable Main Article is under development and subject to a disclaimer.

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

${\displaystyle {\frac {x}{s}}+{\frac {y}{t}}={\frac {xt+ys}{st}}}$
${\displaystyle {\frac {x}{s}}\cdot {\frac {y}{t}}={\frac {xy}{st}}.}$

The zero element of ${\displaystyle S^{-1}R}$ is the class ${\displaystyle 0/s}$ and there is a unit element ${\displaystyle s/s}$. The base ring R is embedded as ${\displaystyle x\mapsto {\frac {xs}{s}}}$.

### Localisation at a prime ideal

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

## Field of fractions

If R is an integral domain, then the non-zero elements ${\displaystyle S=R\setminus \{0\}}$ 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.