Localisation (Ring Theory)

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

Construction[edit]

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 ${\displaystyle (x,s)\sim (y,t)\iff xt=ys}$. 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[edit]

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[edit]

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.