MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose $X$ is a scheme over a ring $A$, $B$ is an $A$-algebra, and $X\times_AB$ is affine. I am looking for conditions on $A$ and $B$ (and perhaps the structure morphism of $X$ over $A$) that will force $X$ to be affine as well.

A trivial observation is that if $B$ is zero, $X\times_AB$ is affine ($Spec 0$), and that says nothing about $X$. One situation that I'm guessing the desired conclusion might hold is if $A$ and $B$ are fields, but I don't have a proof. Maybe it is enough to have $Spec B\rightarrow Spec A$ faithfully flat, but again I am not sure. Any input is appreciated.

share|cite|improve this question
See Hartshorne, Ex. III.4.2, p. 222 (Chevalley's theorem). – Damian Rössler Mar 21 '14 at 7:51

Suppose the structure morphism $g: X\to \operatorname{Spec}(A)$ is separated and of finite type, and $f: \operatorname{Spec}(B)\to \operatorname{Spec}(A)$ is faithfully flat; furthermore, assume $A, B, X$ are all Noetherian. Denote $X\times_A B$ as $X_B$ and let $f': X_B\to X$ be the base change of $f$ along $g$, and $g': X_B\to \operatorname{Spec}(B)$ the base change of $g$ along $f$.

By Serre's affineness criterion, it suffices to show that $R^ig_*\mathcal{F}=0$ for $i>0$ and $\mathcal{F}$ an arbitrary quasi-coherent sheaf on $X$.

By flat base change, the morphism $f^*R^ig_*\mathcal{F}\to R^i{g'}_*{f'}^*\mathcal{F}$ is an isomorphism for any quasi-coherent sheaf $\mathcal{F}$ on $X$. As $X_B$ is affine, the right hand side vanishes for $i>0$. Thus $f^*R^ig_*\mathcal{F}=0$; as $f$ is faithfully flat, this implies $R^ig_*\mathcal{F}=0$ as well, as desired.

In fact, we needn't assume $X\to \operatorname{Spec}(A)$ is separated and finite type, as these properties descend through fpqc morphisms. I haven't thought about dropping the Noetherian hypotheses.

Thus we've shown that if $A\to B$ is faithfully flat and everything is Noetherian, affineness descends.

Here's an example to see that some hypothesis on $f$ is necessary. Let $A$ be a local ring and $B$ its residue field. Let $X=\mathbb{P}^1_A\setminus\{x\}$ where $x$ is any closed point of $X$. Then $X_B$ is $\mathbb{A}^1_B$, whereas $X$ is obviously not affine (as its generic fiber is $\mathbb{P}^1$). I don't see an easy example where $f$ is flat but not faithfully flat at the moment (aside from stupid things, e.g. with $\operatorname{Spec}(A)$ disconnected).

I've just noticed that the Stacks Project gives a proof of this claim, avoiding the Noetherian hypotheses and Serre's criterion.

share|cite|improve this answer
Since you ask for a flat counterexample: let $A$ be a DVR with uniformizer $t$, let $B$ be the fraction field $A[1/t]$, let $\overline{X}$ be $\mathbb{A}^n_A = \text{Spec}A[x_1,\dots,x_n]$ for $n>1$, and let $X$ be the complement of the closed point $\langle t,x_1,\dots,x_n \rangle$. – Jason Starr Mar 21 '14 at 11:21
Ah, nice. Thanks! – Daniel Litt Mar 21 '14 at 12:05

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.