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Given commutative rings $A$ and $B$, if $B$ is an $A$-algebra, under what conditions, other than $B$ being integral over $A$, will the Going-Up property hold? Is there a condition weaker than integrality for which Going-Up holds? If not, could one direct me to a proof that shows that Going-Up implies integrality.

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up vote 2 down vote accepted

Check out exercise 10 in Chapter 5 of Atiyah Macdonald.

In that exercise it is established that going-up is equivalent to the following. If $q\subset B$ is a prime ideal and $p=q^c\subset A$ then $f^{\ast}:\text{spec} B/q\rightarrow \text{spec} A/p$ is surjective. Note this implies the statement Georges makes in b) above.

Also if the map $\text{spec} B\rightarrow \text{spec} A$ is closed then we also get going-up.

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Thanks for pointing out the problem in Atiyah-Macdonald. You refer to some comment that Georges makes here. I don't see a comment by anyone else. – Rankeya Oct 5 '11 at 21:01
There used to be...not sure where it went. – Lalit Jain Oct 6 '11 at 18:45
@Lalit: I had given a class of counter-examples (local morphisms of discrete valuation rings) showing that going-up does not imply integrality. Since your excellent reference to Atiyah-Macdonald completely solves Rankeya's question, I thought I might as well delete my answer. Sorry for the inconvenience I caused. By the way, Rankeya, I think you might acknowledge Lalit's answer by upvoting it and/or accepting it. – Georges Elencwajg Oct 10 '11 at 22:32

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