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

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?

In general no, for the following reason:

Fact: every overring of a valuation domain
within its field of fractions is itself
a valuation domain.

Hence for example the integral closure
of Z in an algebraic number field never
contains a valuation domain.

The same is true for every non-local
integrally closed ring within an algebraic
function field of one variable over a field
K, that itself is algebraic over a finite
field.

The only chance to do what you want to do
is to use valuation domains within the given
domain R with strictly smaller fields of
fractions. Moreover in general the induced
field extensions need to be non-algebraic
for the following reasons:

1. the integral clsoure of a valuation domain
in an algebraic extension of its field of
fractions is a so-called Prüfer domain, that
is all of its localizations are valuation
domains.

2. Every overring of a Prüfer domain within
its field of fractions is itself a Prüfer domain.

H
.

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