If $\phi:A\to B$ is a ring homomorphism, why does there exist $\psi:\text{spec}(A)\to \text{spec}(B)$?

Question

Let $\phi:A\to B$ be a ring homomorphism, where $A$ and $B$ are commutative rings. We know that if $q$ is a prime ideal in $B$, then $\phi^{-1}(q)$ is a prime ideal in $A$. Hence, there exists a mapping $\phi^{*}: \operatorname{Spec}(B)\to\operatorname{Spec}(A)$.

My book (Commutative Algebra by Atiyah and MacDonald, Problem 21 on page 13) then talks about $\phi^{*-1}:\operatorname{Spec}(A)\to\operatorname{Spec}(B)$. Why does this mapping exist? The image of a prime ideal under a ring homomorphism may not be a prime ideal, unless the mapping is surjective.

Answer 1

Your skepticism is justified; such a map does not exist, in general.

Perhaps your book is working in a case where $\phi^*$ is known to be bijective, or perhaps your book is speaking of $\phi^{*{-1}}$ of some subset of $\text{Spec}(A)$ (if $f$ is a map of sets $X \to Y$, then $f^{-1}$ makes sense as a map from the power set of $Y$ to the power set of $X$).

Answer 2

You read incorrectly. $\phi^{*-1}$ is not an application $\operatorname{Spec} A \to \operatorname{Spec} B$; it is the "preimage" operation from subsets of $\operatorname{Spec} A$ to subsets of $\operatorname{Spec} B$. It is induced by $\phi^* : \operatorname{Spec} B \to \operatorname{Spec} A$ just like any map of sets $f : X \to Y$ induces a map $f^{-1} : P(Y) \to P(X)$, defined by $$f^{-1}(U) = \{ x \in X : f(x) \in U \}.$$

For some context, the equation written in the book is (among others) $\phi^{*-1}(V(\mathfrak{a})) = V(\mathfrak{a}^e)$, where $\mathfrak{a}$ is a prime ideal of $A$ and $\phi : A \to B$. Here $V(\mathfrak{a})$ is a subset of $\operatorname{Spec} A$, not an element of it.