Localization (commutative algebra): Difference between revisions

Revision as of 04:46, 12 January 2006

Revision as of 04:46, 12 January 2006

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Since the product of units is a unit and since ring homomorphisms respect products, we may and will assume that ''S'' is multiplicatively closed, i.e. that for ''s'' and ''t'' in ''S'', we also have ''st'' in ''S''. For the same reason, we also assume that 1 is in ''S''.

In case ''R'' is an [[integral ___domain]] there is an easy construction of the localization. Since the only ring in which 0 is a unit is the [[trivial ring]] {0}, the localization ''R*'' is {0} if 0 is in ''S''. Otherwise, the [[field of fractions]] ''K'' of ''R'' can be used: we take ''R*'' to be the subring of ''K'' consisting of the elements of the form the ^''r''⁄_''s'' with ''r'' in ''R'' and ''s'' in ''S''. In this case the homomorphism from ''R'' to ''R*'' is the standard embedding and is injective: but that will not be the case in general. See for example [[dyadic fraction]], for the case R the integers and S the powers of 2.

For general [[commutative ring]]s, we don't have a field of fractions. Nevertheless, a localization can be constructed consisting of "fractions" with [[denominator]]s coming from ''S''; in contrast with the integral ___domain case, one can now safely 'cancel' from [[numerator]] and [[denominator]] only elements of ''S''.