1
$\begingroup$

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.

$\endgroup$

1 Answer 1

3
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ 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 $B \to A \to A/\operatorname{im}(\phi)$? $\endgroup$
    – Ray
    Aug 5, 2021 at 8:40
  • $\begingroup$ @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 $c=1\cdot c=\sum g_i^mh_ic=0$. $\endgroup$ Aug 5, 2021 at 8:53
  • $\begingroup$ @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. $\endgroup$ Sep 26, 2021 at 4:32
  • $\begingroup$ @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 $S^{-1}B$. $\endgroup$ Sep 26, 2021 at 5:09
  • $\begingroup$ @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. $\endgroup$ Sep 26, 2021 at 15:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .