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Why should the normalization of a ring correspond to the intersection of valuation rings containing it? I am looking for a geometric explanation, if possible.

I understand that normalization at a point of a variety corresponds to making sure every bounded function in a neighborhood of that point extends over the point. But I'm having trouble translating that statement into the algebraic one.

There is some discussion of this issue by Kollár on page 20 here, but don't see how "definition" 1.23 connects to the preceding discussion. And from there, it seems like a stretch to get to the correct definition (using inclusions instead of arbitrary homomorphisms, and using arbitrary valuation rings instead of just DVRs).

Edit: As pointed out in the comments, any Noetherian valuation ring is a DVR (or field). In geometric applications, everything is Noetherian, so that helps a bit. I would still appreciate an answer explaining Kollár's remarks about the geometric content of the DVR viewpoint on normalization a bit more.

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  • $\begingroup$ In the geometric setting I think that the valuation rings are DVRs. $\endgroup$ – user26857 Jan 31 '16 at 21:18
  • $\begingroup$ @user26857 Hi. That is very interesting -- could you say why that might be the case? $\endgroup$ – user4571 Jan 31 '16 at 21:19
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    $\begingroup$ This holds for noetherian integral domains, and I guess the geometric rings you want to consider are so. $\endgroup$ – user26857 Jan 31 '16 at 21:21
  • $\begingroup$ @user26857 Great. Thank you for pointing that out. $\endgroup$ – user4571 Jan 31 '16 at 21:36

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