2
$\begingroup$

An integral domain $A$ is called normal if it equals its integral closure in its field of fractions K. I can appreciate the condition of a ring being equal to its integral closure inside some ring, but why is it especially important that a ring equals its integral closure in its field of fractions?

Evidently we need to have some natural way of assigning to each ring a larger ring in which to check the condition of being integrally closed. However, the analogous concept for fields of algebraic closure does not depend on an embedding into a larger ring or field. How does one explain this discrepancy?

In number theory there are examples of rings which are integrally closed in their fields of fractions but not in field extensions, e.g. $\mathbb{Z} \subset \mathbb{Q} \subset \mathbb{Q}( \sqrt 2 )$. Are there any rings $A$ which are integrally closed in any ring $B \supset A$? (This is true for algebraically closed fields, is it not?) Why is this not a better definition for integrally closed?

I can see that integrally closed rings have a plethora of nice properties, so obviously it is a good definition, but can anyone give me a little more confidence that this was the correct definition to have chosen? I would be happy with algebraic explanations from commutative algebra or number theory or geometric explanations from algebraic geometry.

$\endgroup$
  • 2
    $\begingroup$ The correct definition of "normal domain" is "an integral domain for which its localization at every prime ideal is integrally closed". Notice that this implies that the ring itself is integrally closed. $\endgroup$ – Manos May 27 '13 at 22:34
  • 2
    $\begingroup$ @Manos: This is unnecessarily complicated. An integral domain is integrally closed iff every localization of it is integrally closed. So why should this be the "correct definition"? Usually this way one defines normality of rings which are locally integral domains. But for integral domains we don't need to use localizations. $\endgroup$ – Martin Brandenburg May 28 '13 at 10:14
  • 1
    $\begingroup$ @PrimeRibeyeDeal: There are just not enough universally integrally closed domains (this is how I would call integral domains which are integrally closed in any ring extension) appearing in practice. $\endgroup$ – Martin Brandenburg May 28 '13 at 10:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.