In Atiyah and MacDonald's Commutative Algebra, Theorem 5.16 gives the usual notion of "going down" for a ring extension $A\subseteq B$, where a prime $\mathfrak{q}$ of $B$ lies above/over a prime $\mathfrak{p}$ of $A$ if $\mathfrak{q}\cap A=\mathfrak{p}$.
Later in Exercise 5.10, they generalize to say a morphism $f\colon A\to B$ of rings has the going down property if the same "going down" notion on chains of primes applies to $B$ and its subring $f(A)$. But now I wonder, are we considering chains of primes in $B$ lying over chains of primes in $A$ or $f(A)$, and what is the corresponding notion of "lying over" for primes and in this case? Do we intend to consider
- (Notion A) a prime $\mathfrak{q}$ of $B$ lies over a prime $\mathfrak{p}$ of $A$ if $f^{-1}(\mathfrak{q})=\mathfrak{p}$,
or do we say
- (Notion B) a prime $\mathfrak{q}$ of $B$ lies over a prime $\mathfrak{p}$ of $f(A)$ if $\mathfrak{q}\cap f(A)=\mathfrak{p}$?
The reason I'm confused is because in the original case of $A\subseteq B$, we have the inclusion morphism $i\colon A\to B$, and in such a case $i^{-1}(\mathfrak{q})=\mathfrak{q}\cap A$, so contracting under the morphism is the same as taking the intersection with the subring.
The latter notion seems like the most "verbatim" generalization of the original case where $A\subseteq B$, but the former behaves well with the notion of contraction, and the associated morphism $f^\ast\colon\operatorname{Spec}(B)\to\operatorname{Spec}(A)$. However, later in the book in Exercise 7.24, we have a situation where we wish to show $E:=f^\ast(\operatorname{Spec}(B))$ is open in $\operatorname{Spec}(A)$ if $f$ satisfies the going down property. The hint there says that if $\mathfrak{p}\in E$ and $\mathfrak{p}'\subseteq\mathfrak{p}$, then the going down property says $\mathfrak{p}'\in E$ as well. Assuming Notion A, this property is essentially the definition. But assuming Notion B, it seems incorrect since there may be primes in $A$ that are not contractions of primes in $f(A)$.
