Closedness and going up property

Question

Let $f: A\rightarrow B$ be a homomorphism of commutative unital rings. The problem is to show that if $f$ has going-up property and $\text{Spec }B$ is Noetherian topological space then $f^*: \text{Spec }B \rightarrow \text{Spec }A$ is a closed map. (Atiyah and Macdonald, Introduction to Commutative Algebra, Chapter 6, Exercise 11)

According to the Exercise 5.10 of Atiyah, going-up is equivalent to the following condition. For any prime $\mathfrak{q}\subset B, f^*(V(\mathfrak{q})) = V(\mathfrak{p})$, where $\mathfrak{p} = \mathfrak{q}^c$.

I was thinking of the following. Let $Z$ be the set of all closed subsets of $\text{Spec }B$ the image of which under $f^*$ is not closed. Then the set $Z$ has the minimal elements, because $\text{Spec }B$ is Noetherian. However I do not see any way out here.

Conversely, let $K$ be the set of all closed subsets of $\text{Spec }B$ the image of which under $f^*$ is closed. The set $K$ is non-empty because $f^*(V(\mathfrak{q})) = V(\mathfrak{p})$ for any prime $\mathfrak{q}$. The set $K$ has minimal elements, too. Suppose for some ideal $I$, the corresponding closed set $V(I)$ is minimal in $K$, it means that for any ideal $J$, such that $V(J)\subset V(I)$ the $f^*(V(J))$ is not closed. However, $I\subset \mathfrak{m}$ for some maximal ideal $\mathfrak{m}$ in $B$. Then $V(\mathfrak{m})\subset V(I)$, it follows that $f^*(V(\mathfrak{m}))$ is not closed, a contradiction. Therefore the minimal elements of $K$ are of the form $V(\mathfrak{m})$. However, after this I do not see the way out of here, too.

Any help is appreciated.

Answer 1

Let $V(I) \subseteq \text{Spec}(B)$ be a closed set. Since $\text{Spec}(B)$ is a Noetherian space, $V(I)$ is a finite union of irreducible closed sets, necessarily of the form $V(q_i)$, where $q_i \in \text{Spec}(B)$. Then

$$f^*(V(I)) = f^*(\bigcup_{i=1}^n V(q_i)) = \bigcup_{i=1}^n f^*(V(q_i)) = \bigcup_{i=1}^n V(q_i^c)$$ is a finite union of closed sets, hence is closed.