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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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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