MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

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

Take scheme morphism $f: X\to Y$ and suppose $f$ surjective. If $y \in Y$ can one find affine open $V \subset Y$ containing $y$ and affine open $U \subset X$ such $f(U) = V$ ? Thank you.

Later: Very good answer of Kevin shows it is not true. Is there hypothese which make it true ? For example $X$ irreducible and/or $f$ faithfuly flat ?

share|cite|improve this question
Take Y=Spec k, with k a field. Let f:X---> Y be a morphism. (It is automatically surjective.) Then f satisfies your condition if and only if X is affine. What you are asking for is the "relativization" of this. That is, your morphism f will satisfy your condition if it is "affine". Let me emphasize here that the morphism needs to be affine (and not the schemes necessarily.) Finite morphisms are affine. When they are surjective they are called (branched) covers. It is not true that every surjective affine morphism is finite. Consider for example the projection of A^n to A^1. – Ariyan Javanpeykar Feb 13 '11 at 21:12
@Ariyan: if $Y$ is one point, any any non-empty affine open subset of $X$ maps surjectively to $Y$. So $f$ always satisfies the required property in this case. – Qing Liu Feb 13 '11 at 21:23
To see that you are asking about "affine morphisms" you could look at chap. II, exercise 5.17 of Hartshorne's Algebraic geometry. (Not completely sure about the exercise number because I don't have access to the book right now.) – Ariyan Javanpeykar Feb 13 '11 at 21:23
Oops. Sorry. I guess I should have looked better. – Ariyan Javanpeykar Feb 13 '11 at 21:29
Ok so you're not (really) asking about affine morphisms because U doesn't need to be the inverse image of V. Again apologies. Thank you Qing Liu. – Ariyan Javanpeykar Feb 13 '11 at 21:31
up vote 6 down vote accepted

If $f$ is open (e.g. $f$ finite type and flat over noetherian $Y$), then your condition is trivially satisfied: let $V'$ be any affine open neighborhood of $y$ and let $U'$ be an affine open subset of $X$ such that $y\in f(U')\subseteq V'$. Take a principal open subset $V'_h$ such that $y\in V'_h\subseteq f(U')$, then $V:=V'_h$ and $U:=U'_h$ are what you want.

A counterexample with $X$ irreducible and $f$ projective : consider $Y$ the affine plan, $y$ the origin and $f : X\to Y$ the blowing-up of $y$. For any affine open subset $V$ containing $y$, $f^{-1}(V) \to V$ is the blowing-up of $y$. If $f(U)=V$, then the complement of $U$ in $f^{-1}(V)$ is finite because $f$ is an isomorphism out of $y$. By Zariski's extension theorem, $O_X(U)=O_X(f^{-1}(V))=O_Y(V)$. This is impossible as $U$ is affine, because $U$ would be the image of a section $V\to f^{-1}(V)$, hence closed in (and then equal to) $f^{-1}(V)$.

share|cite|improve this answer
Wonderfull, Professor! This is very amusing because source of question is your book! In example 3.5 of 4.3.1 you say that normalization of integral scheme is flat only if scheme was already normal but you do not give proof and say to do exercise 1.2.10 (which is affine case) . I can do exercise just with faithfully flat assumption: if $A\subset B$ is faithfully flat ring extension and $B$ is integrally closed domain, then $A$ is also integrally closed ( no finiteness or noether assumption and I do not assume that $B$ is integral closure of $A$). Is that correct? – evgeniamerkulova Feb 13 '11 at 22:29
So now we deduce thanks to your wonderfull answer following result, because we can reduce to affine case. If $f:X \to Y$ is faithfully flat morphism of integral schemes and $X$ is normal, then also $Y$ is normal.( Maybe should add $f$ locally of finite presentation to be sure $f$ is open?). Is that true, Professor? – evgeniamerkulova Feb 13 '11 at 22:43
Your conclusion is correct, but you don't openess of $f$ : if $y\in Y$ and $x\in f^{-1}(y)$, then $O_{Y,y}\to O_{X,x}$ is faithfully flat. – Qing Liu Feb 14 '11 at 10:15
Professor Liu thank you for answer, but I am not sure I understand it correctly because a word lacks between "don't" and "openess"! For local morphism of local rings flat= faithfully flat, right? – evgeniamerkulova Feb 14 '11 at 12:33
Sorry, you don't need openess of $f$. Yes for local rings, flat=faithfully flat. – Qing Liu Feb 14 '11 at 17:48

Consider the disjoint union Spec$(\mathbb{Q})\coprod_{p}$Spec$(\mathbb{F}_p)$ with its canonical map to Spec$(\mathbb{Z})$. This is bijective on points, but the preimage of any open in Spec$(\mathbb{Z})$ won't even be compact, much less affine.

share|cite|improve this answer
Thank you very much: this is perfect answer! – evgeniamerkulova Feb 13 '11 at 19:39
Mr. Ventullo I am very sorry that cannot accept two answers and that I can give you only one vote: your answer merits more ! – evgeniamerkulova Feb 14 '11 at 12:36
Oh, that's all right. This example is pretty pathological; Qing Liu's example shows that this can happen in "geometric" situations. – Kevin Ventullo Feb 15 '11 at 5:10

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.