Let $B$ be a ring containing $A$, and the ring extension is integral. Furthermore, $B$ is a finitely generated $A$-algebra. Then how to show that the induced map on spectra of the rings is a finite map, i.e., inverse image of finite sets is finite.
[Math] Induced map on spectra of rings
algebraic-geometrycommutative-algebra
Related Solutions
I prefer summerize and improve the answers of Qiaochu Yuan and basket!
Let $A$ and $B$ be rings, $X=\operatorname{Spec}A$ and $Y=\operatorname{Spec}B$; one can prove that the function $$ \Phi:(f,f^{\sharp})\in\operatorname{Hom}_{\bf LocRingSp}((X,\mathcal{O}_X),(Y,\mathcal{O}_Y))\to f^{\sharp}(Y)\in\operatorname{Hom}_{\bf Ring}(B,A) $$ is bijective; moreover, the morphism $(f,f^{\sharp})$ is uniquely determined by $f^{\sharp}(Y)$.
What I mean? Which is the improvement?
As you (user306194) affirm: given a morphism of rings (that is $f^{\sharp}(Y)$ in my notation) you can construct a unique morphism of locally ringed spaces $(f,f^{\sharp})$; but, given a continuous map $f$ between the support of locally ringed spaces, you can't define uniquely $f^{\sharp}$!
For example: you read the previous comment of Hoot.
Obviously, all this works in the setting of schemes.
In general, let $A$ be a ring and let $Y$ be a scheme, $X=\operatorname{Spec}A$; one can prove that the function $$ \Phi:(f,f^{\sharp})\in\operatorname{Hom}_{\bf Sch}(Y,X)\to f^{\sharp}(X)\in\operatorname{Hom}_{\bf Ring}(A,\mathcal{O}_Y(Y)) $$ is bijective.
Moreover in general: let $S$ be a scheme, let $X$ be an $S$-scheme with structure morphism $p$ and let $\mathcal{A}$ be a quasi-coherent $\mathcal{O}_S$-algebra; one can prove that the function $$ \Phi:(f,f^{\sharp})\in\operatorname{Hom}_{S-\bf Sch}(X,\operatorname{Spec}\mathcal{A})\to f^{\sharp}(\operatorname{Spec}\mathcal{A})\in\operatorname{Hom}_{\mathcal{O}_S-\bf Alg}(\mathcal{A},p_{*}\mathcal{O}_X) $$ is bijective; where $\operatorname{Spec}\mathcal{A}$ is the relative spectrum of $\mathcal{A}$.
For a reference, I council Bosch - Commutative Algebra and Algebraic geometry, Springer, sections 6.6 and 7.1.
In conclusion, I remember you that $\operatorname{Spec}(A\otimes_{\mathbb{C}}B)$ is canonically isomorphic to $\operatorname{Spec}A\times_{\operatorname{Spec}\mathbb{C}}\operatorname{Spec}B$; usually, the Cartesian product of sets is very bad in the category of schemes.
From the same book, I council you the section 7.2 with any examples and exercises 3, 4 and 5.
Unintentionally answering my own question here, although I will leave the question open to see if there is a more direct solution. But the answer to this question: Show that a morphism of algebraic sets which restricts to finite morphisms of principal sets is finite also answers my question using the following fact:
Let $\phi : R \to S$ be a ring extension such that $S$ is finitely generated as an $R$-algebra. Then $\phi$ is an integral ring extension if and only if $S$ is finitely generated as an $R$-module.
Best Answer
The fiber under $\text{Spec} (B)\to \text{Spec} (A)$ of a point $P=[\mathfrak p]\in \text{Spec} (A)$ is the spectrum of the ring $R=B\otimes_A \kappa (P)$, where $\kappa (P)=\text {Frac}\:(A/\mathfrak p)$.
Since finiteness of modules is preserved under base change, the algebra $R$ is finite as a module over the field $\kappa (P)$ (in other words is finite-dimensional!) and thus has finite spectrum by the following result which I will display for future reference :
Lemma
An algebra $R$ of finite dimension over a field $k$ has only finitely many prime ideals and these prime ideals are all maximal.
Proof
Given a prime $\mathfrak m\subset R$ the quotient $R/\mathfrak m$ is necessarily a field, so that $\mathfrak m$ is maximal.
The Chinese remainder theorem applied to the surjective morphism $R\twoheadrightarrow \prod_{\mathfrak m} R/\mathfrak m$ (where we take finitely many maximal ideals $\mathfrak m$) immediately implies that the number of those maximal ideals is finite and $\leq [R:k]$.