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The exercise 5.10 of Atiyah's Commutative Algebra gives the definition of going-up map:

A ring homomorphism $f:A\rightarrow B$ is said to have the going-up (resp. the going-down property) if the conclusion of going-up theorem (resp. the going-down theorem) holds for $B$ and its subring $f(A)$.

I think this definition is unnatural and the definition should be

....... if the conclusion of going-up theorem (resp. the going-down theorem) holds for $B$ and $A$.

It seems these two definitions are different. Why the book chooses the "unnatural" definition? Also one part of exercise is

i) $f$ has the going-down property

ii) For any prime ideal $q$ of $B$, if $p=q^{c}$, then $f^{*}: \mathrm{Spec}(B_{q})\to\mathrm{Spec}(A_{p})$ is surjective

Prove that i) $\Leftrightarrow$ ii).

It is trivial if using the definition which I think. But I do not know how to prove if using the book's definition. I am very confused now.

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Mike
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    Your alternative definition has nothing to do with the map $f$, so how could it possibly be a definition of a property of $f$? Besides, the setting you (wrongly) assume you have is $A \subseteq B$, only then can you talk about the going-up property for $B \vert A$. Now we generalize this setting naturally to rings that are related via a ring homomorphism $f\colon A \to B$, by talking about the ring extension $f(A) \subseteq B$. This even gives the prior definition in the special case $A \subseteq B$ by taking $f$ to be the inclusion of $A$ into $B$. – Mickey Jan 12 '19 at 10:37
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    Atiyah and Macdonald defined those properties for ring extensions. When you have a homomorphism $f:A\to B$ then the natural way to (re)define the properties is to let them hold for the ring extension $f(A)\subset B$. – user26857 Oct 13 '19 at 21:37
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    Btw, you didn't tell us what are your "natural" definitions. – user26857 Oct 13 '19 at 21:38

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