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

Revision as of 17:09, 2 January 2009


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 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.

References

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