$P \mapsto f^{-1}(P) : \operatorname{Spec} B \to \operatorname{Spec} A$ is a continuous map between Zariski topologies.

Question

Let $A, B$ be commutative rings with $1$.

Let $f: A \to B$ be a ring homomomorphism. For $P \in \operatorname{Spec} B$, $f^{-1}(P) \in \operatorname{Spec} A$, so that $g : \operatorname{Spec} B \to \operatorname{Spec} A$ defined by $g(P) = f^{-1}(P)$ is a function. The goal is to show that for any $V(I) \subset \operatorname{Spec} A$ a closed set, that $g^{-1}(V(I)) = V(f(I)B)$. Since then $g^{-1}$ takes closed sets to closed sets and so is continuous.

So far I have:

$$ P \in g^{-1}(V(I)) \\ \iff g(P) = f^{-1}(P) \in V(I) \\ \iff ? $$

I know that if $P \supset f(I)B$ then $P \supset f(I)$ since $P$ has to be a proper ideal. Not sure where that plugs into the proof though.

Answer 1

Thanks @Thorgott for teaching me.

The step I was missing was if $I \subset f^{-1}(\mathfrak{p})$ then $f(I) \subset ff^{-1}(\mathfrak{p}) \subset \mathfrak{p}$.

Thus, we have:

$$ g^{-1}(V(I)) = \{ P \in \operatorname{Spec} B: g(P) \in V(I) \} \\ = \{ P \in \operatorname{Spec} B: f^{-1}(P) \in V(I) \} \\ = \{ P \in \operatorname{Spec} B: I \subset f^{-1}(P) \} \\ = \{ P \in \operatorname{Spec} B: f(I) \subset ff^{-1}(P) \subset P \} \\ = V(f(I)B) = V(IB) $$

The last equality is just by notational definition of $IB \equiv f(I)B$ as used in Matsumura's text. That's the extension of an ideal along $f$. And $A \cap J \equiv f^{-1}(J)$ is the contraction of $J$ along $f$. All just notational.