# Finite morphism is affine local on target

According to Vakil, a scheme morphism $$\pi: X \to Y$$ is finite if for every affine open subset $$\operatorname{Spec} B$$ of $$Y$$, $$\pi^{-1}(\operatorname{Spec} B) = \operatorname{Spec} A$$ with $$A$$ a finite $$B$$-algebra. Then the reader is asked to prove that if $$Y$$ admits an open cover $$\{\operatorname{Spec} B_i\}$$ such that $$\pi^{-1}(\operatorname{Spec} B_i)$$ is the spectrum of a finite $$B_i$$-algebra, then $$\pi$$ is finite (c.f. Exercise 7.3.G).

My guess is to apply the Affine Communication Lemma 5.3.2 (similar to Exercise 7.3.C), so I have to show that

(i) If $$\operatorname{Spec} B \subseteq Y$$ such that $$\pi^{-1}(\operatorname{Spec} B)$$ is the spectrum of a finite $$B$$-algebra, then $$\pi^{-1}(\operatorname{Spec} B[g^{-1}])$$ is the spectrum of a finite $$B[g^{-1}]$$-algebra.

(ii) If $$\operatorname{Spec} B \subseteq Y$$ and $$(g_1, \cdots, g_n) = B$$ such that $$\pi^{-1}(\operatorname{Spec} B[g_i^{-1}])$$ is the spectrum of a finite $$B[g_i^{-1}]$$-algebra, then $$\pi^{-1}(\operatorname{Spec} B)$$ is the spectrum of a finite $$B$$-algebra.

Well, (i) is easy, but I don’t see a way to prove (ii). Since $$\operatorname{Spec} B$$ is covered by the distinguished open subset $$D_B(g_i)$$, it follows that the restriction of $$\pi$$ to a morphism $$\pi^{-1}(\operatorname{Spec} B) \to \operatorname{Spec} B$$ is affine (by Proposition 7.3.4), so $$\pi^{-1}(\operatorname{Spec} B)$$ is the spectrum of a ring $$A$$. Since $$\operatorname{Spec} A \to \operatorname{Spec} B$$ corresponds to a ring morphism $$B \to A$$, this gives $$A$$ the structure of a $$B$$-algebra. But how can I show that it is a finitely generated $$B$$-module? I know I didn’t used all the hypothesis, But I honestly don’t see a way to put them together to get the desired conclusion.

Any help is appreciated.

For (ii) we need to show for $$\pi^{-1}(\operatorname{Spec} B)=\operatorname{Spec} A$$, $$A[g_i^{-1}]$$ is a finite $$B[g_i^{-1}]$$-algebra for all $$i$$, implies that , $$A$$ is a finite $$B$$-algebra. Let $$A[g_i^{-1}]$$ is generated by $$\{f_{i1},\dots, f_{in}\}$$ as $$B[g_i^{-1}]$$-module where $$f_{ij}\in A$$.

Consider $$\phi: B^{\oplus N}\to A$$ sending $$e_{ij}\mapsto f_{ij}.$$

Consider the cokernel $$C$$ of the map $$\phi$$. Now the $$B$$-module $$C$$ has the property that $$C[g_i^{-1}]$$ =0. Since $$g_i$$'s generate the unit ideal, we can conclude $$C=0$$.

Added: To see $$C=0$$, let $$c\in C$$. Then for each $$i$$, since $$c\in C[g_i^{-1}]=0$$, we get $$g_i^mc=0$$ for some $$m$$. Now $$g_i$$'s generate unit ideal, implies $$g_i^m$$'s generate the unit ideal. So 1 is a $$B$$-linear combination of $$g_i^m$$'s. So $$c=1\cdot c=\sum g_i^mh_ic=0$$.

To put (i) and (ii) together: Using the 'Affine communication lemma', $$\operatorname{Spec} B\subset Y$$ satisfies the property $$P$$ if $$\pi^{-1}(\operatorname{Spec} B)$$ is the spectrum of a finite $$B$$-algebra.

So you are done.

• I am actually not too familiar with module theory. So why would $C$ has that property? And I guess when you write $C[g_i^{-1}]$, you are implicitly identifying $g_i$ as an element of $C$ via the map $B \to A \to A/\operatorname{im}(\phi)$?
– Ray
Aug 5, 2021 at 8:40
• @Ray By $C[g_i^{-1}]$ I mean the localization of $C$ as a $B$-module w.r.t the multiplicative closed subset $\{1, g_i, g_i^2,\dots \}$. To see $C=0$, let $c\in C$. Then for each $i$, since $c\in C[g_i^{-1}]=0$, we get $g_i^mc=0$ for some $m$. Now $g_i$'s generate unit ideal, implies $g_i^m$'s generate the unit ideal. So 1 is a $B$-linear combination of $g_i^m$'s. So $c=1\cdot c=\sum g_i^mh_ic=0$. Aug 5, 2021 at 8:53
• @EvansGambit In your first sentence, you should be considering $A[\pi^\#(g_i)^{-1}]$, not $A[g_i^{-1}]$. The rest of the proof follows this error. Not sure if it is correct. Sep 26, 2021 at 4:32
• @user46372819 I think it is fine. $A$ is a module over the ring $B$. $S=\{1, g_i, g_i^2,\dots\}$ is a multiplicatively closed subset of $B$. Now $A[g_i^{-1}]:=S^{-1}A$ as a module over the ring $S^{-1}B$. Sep 26, 2021 at 5:09
• @EvansGambit Your statement in the comment is correct; but the assumption you are making in the first sentence of your answer is not the correct assumption. Since $\pi^{-1}(\operatorname{Spec}B_{g_i}) = \operatorname{Spec}A_{\pi^\#(g_i)}$, we know that $A_{\pi^\#(g_i)}$ is a finite $B_g$-algebra. Sep 26, 2021 at 15:23