Re: Smallest integral domain to have the same qoutient field.

On 25.10.2006 11:04, Jose Capco wrote:
All right, I get the idea. I do have a stupid question however.. its
not entirely related to the original post though. We know from Krull
that an integrally closed domain is just the intersection of the
valuation rings containing the domain and whose qoutient fields agree
with the quotient field of the domain (in fact the integral closure of
any domain could be formed in such a way, I suppose). Now however, I'd
like to approach the integrally closed domain from below.... i.e., I
want to know if it can be regarded as the say direct colimit of
valuation rings that are inside the domain (with usually transition
maps between the rings just being the inclusion map..).. am I making
any sense here?

Sincerely,
Jose Capco


I am not sure that it works to exhaust the integral closure by valuation
rings. Since in the situation of Serre's intersection theorem what
valuation rings are available lying *below* the integral closure. If I
remember correctly the valuations to be considered are induced by primes
of the integral closure, hence the associated valuation ring contains
the integral closure.

However I could imagine that the integral closure of the integral domain
R in its quotient field Q(R) is represented as direct limit of rings in
Q(R) containing R which are finitely generated R-modules.

HTH.

J.
.

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