All rings in this question will be commutative, and ring homomorphisms are required to send $1$ to $1$. Suppose $\phi:A\to B$ is a ring homomorphism, $X=\operatorname{Spec} A$ and $Y=\operatorname{Spec} B$. The map $\phi$ induces a continuous map $\phi^*:Y\to X$ given by sending each prime ideal in $B$ to its contraction in $A$. In Atiyah and Macdonald, there is an exercise asking for a proof of the identity $\overline{\phi^*(V(\mathfrak b))}=V(\mathfrak b^c)$ for all ideals $\mathfrak b$ in $B$. Is there an intuition behind why the identity holds? It seems that this theorem has some geometric content to it, but none of the proofs I have seen feel particularly geometric. I am also curious if there is an analogue of the identity in classical algebraic geometry, where $V(\mathfrak a)$ denotes the vanishing locus of $\mathfrak a$, rather than the set of prime ideals including $\mathfrak a$.
Intuition behind the identity $\overline{\phi^*(V(\mathfrak b))}=V(\mathfrak b^c)$
abstract-algebraalgebraic-geometrycommutative-algebraintuitionring-theory
Related Solutions
First, I will make some statements without proof. For an ideal $\mathfrak{a} \unlhd A$, $\sqrt{\mathfrak{a}} = \cap_{\substack{p \in Spec(A) \\ p \supset a}} p$. Also, $V(\mathfrak{a}) \subset Spec(A)$ is a closed set for any $\mathfrak{a} \unlhd A$. Lastly, for $\phi : A \rightarrow B$ a ring homomorphism, $\mathfrak{a} \unlhd A$, we have $(\phi^{*})^{-1}(V(\mathfrak{a})) = V(\mathfrak{a}^{e})$, where $\mathfrak{a}^{e} \unlhd B$ is the ideal generated by the image of $\mathfrak{a}$. We want to show $(\overline{\phi^{*}(V(\mathfrak{b}))}) = V(\mathfrak{b}^{c})$.
$\boxed{ \subseteq }$ Suppose that $x \in \phi^{*}(V(\mathfrak{b}))$. Then there is $y \in V(\mathfrak{b})$, $\phi^{*}(y) = x \implies x = y^{c}$. But $y \supseteq \mathfrak{b} \implies x = y^{c} \supseteq \mathfrak{b}^{c} \implies x \in V(\mathfrak{b}^{c})$. Since $\mathfrak{b}^{c} \unlhd A$, $V(\mathfrak{b}^{c})$ is closed, which implies $\overline{\phi^{*}(V(\mathfrak{b}))} \subseteq V(\mathfrak{b}^{c})$
$\boxed{ \supseteq }$ Observe that $\overline{\phi^{*}(V(\mathfrak{b}))} \subseteq Spec(A)$ is closed, so $\overline{\phi^{*}(V(\mathfrak{b}))} = V(\mathfrak{a})$, for some $\mathfrak{a} \unlhd A$. Then: $$ V(\mathfrak{a}^{e}) = (\phi^{*})^{-1}(V(\mathfrak{a})) = (\phi^{*})^{-1}(\overline{\phi^{*}(V(\mathfrak{b}))}) \supseteq V(\mathfrak{b}) $$ which implies that $\mathfrak{a}^{e} \subseteq \sqrt{\mathfrak{b}}$. Let $t \in \mathfrak{a}$, then $\phi(t) \in \mathfrak{a}^{e} \subseteq \sqrt{\mathfrak{b}}$. Then $\phi(t^{n}) = \phi(t)^{n} \in \mathfrak{b}$ for some $n \in \mathbb{N}$. So $t^{n} \in \mathfrak{b}^{c} \implies t \in \sqrt{\mathfrak{b}^{c}} \implies \mathfrak{a} \subseteq \sqrt{\mathfrak{b}^{c}}$. So: $$ \overline{\phi^{*}(V(\mathfrak{b}))} = V(\mathfrak{a}) \supseteq V(\sqrt{\mathfrak{b}^{c}}) = V(\mathfrak{b}^{c}) $$ which completes the proof. Let me know if you have questions.
One correct claim that can be proven is that if $f:A\to B$ is a homomorphism of rings so that $f^{-1}$ reflects containment of ideals (ie $f^{-1}(I)\subset f^{-1}(J)$ implies $I\subset J$), then the induced map on spectra $\def\Spec{\operatorname{Spec}}\Spec B\to\Spec A$ is a homeomorphism on to its image. Here's how:
- The map $\varphi:\Spec B\to \Spec A$ is continuous: the preimage of some $V(I)$ for $I\subset A$ is $V(f(I)B)$, so the preimage of a closed set is closed (this is general for any map of rings).
- The map $\varphi$ is injective: if $\mathfrak{q}_1,\mathfrak{q}_2\subset B$ are prime ideals with $f^{-1}(\mathfrak{q}_1)=f^{-1}(\mathfrak{q}_2)=\mathfrak{p}$ a prime ideal of $A$, then since $f^{-1}$ reflects containment we see that $\mathfrak{q}_1=\mathfrak{q}_2$. Therefore $\varphi$ is a bijection on to its image.
- The map $\varphi:\Spec B\to \varphi(\Spec B)$ is closed. Suppose $\mathfrak{p}\subset A$ is in the image of $\varphi$, that is, there exists a prime ideal $\mathfrak{q}\subset B$ with $f^{-1}(\mathfrak{q})=\mathfrak{p}$. Further, suppose that $\mathfrak{p}$ is in the closure of $\varphi(V(I))$ for some ideal $I\subset B$. As $\overline{\varphi(V(I))}=V(f^{-1}(I))$ for any map of rings (where the closure is taken in $\Spec A$), we see that $f^{-1}(I)\subset f^{-1}(\mathfrak{q})$, and as $f^{-1}$ reflects containment we must have $I\subset\mathfrak{q}$. Therefore $\mathfrak{q}\in V(I)$ and $\varphi(V(I))$ is closed in $U$. (This looks to be approximately what you've written, but I don't find your presentation to be so clear - maybe that's my fault.)
- A continuous closed bijection is a homeomorphism, so $\varphi:\Spec B\to \varphi(\Spec B)$ is a homeomorphism.
Note that if we drop this assumption about reflecting containment, we get counterexamples: if $R$ is a discrete valuation ring with maximal ideal $\mathfrak{m}$, then the map $f:R\to R/\mathfrak{m}\times \operatorname{Frac}(R)$ is a continuous bijection which is not a homeomorphism and $f^{-1}$ does not reflect containment. (This is noted in the other answer and its comments.)
As far as a characterization of what sort of maps of rings have this property other than (compositions of) localizations and quotients, I'm not sure that there is a well-known description. Any such map will be an epimorphism of rings, and there's a seminar by Samuel about these that I'm aware of but have not dug in to much. Some of the examples here should satisfy your assumptions and also give you an idea that this is maybe not such a simple condition to classify.
Best Answer
We'll play a little fast and loose with what "vanishing" means here, but you can make all of this work no matter what definition of "vanishing" you pick, and you did say that you were looking for intuition anyways.
Suppose $f$ is a function on $X$. The function $f$ vanishes on $\phi^*(V(\mathfrak{b}))$ if and only if the pullback of $f$ vanishes on $V(\mathfrak{b})$. As the map $A\to B$ is given by pullback and the functions vanishing on $V(\mathfrak{b})$ are exactly $\mathfrak{b}$, this means that $f$ vanishing on $\phi^*(V(\mathfrak{b}))$ is equivalent to the image of $f$ in $B$ lying in $\mathfrak{b}$, or $f\in\mathfrak{b}^c$. So the functions vanishing on $\phi^*(V(\mathfrak{b}))$ are exactly $\mathfrak{b}^c$. As $V(I(S))=\overline{S}$, this gives that $\overline{\phi^*(V(\mathfrak{b}))}= V(\mathfrak{b}^c)$ and we're done.