Let $X$ be a smooth algebraic surface over $\mathbb{C}$, and $Y \xrightarrow{\phi} X$ the blowup at a (reduced) point with exceptional divisor $E$. Then, the we have the following universal property: Every morphism from $Y$ to an algebraic variety $Z$ that contracts $E$ to a point factors through $X$ (Beauville, Algebraic Surfaces p.17) I'm not sure to think of this (univ. prop. of blow. down) as a property of smooth surfaces over $\mathbb{C}$ or more generally. Here are some natural questions come to mind:

Is there a more general universal property of blowing down along these lines. I don't expect there to be an answer for arbitrary blowups (say of noeth. schemes), however I would like to replace the field $\mathbb{C}$ by other algebraically closed fields like $\overline{\mathbb{F}_p}$, and if the formalism allows it, even $\mathbb{Q}$. There are a couple of cases:

1.) I can't even figure out a universal property if we stay in the context schemes smooth over $\mathbb{C}$, and blow-up at smooth, irreducible subvarieties $Z$.

2.) Let $X, Y, \phi, E$ as above (in particular, surfaces). Assume that $X$ is reduced, but possibly singular, and allow $\phi$ to be a blow-up at an arbitrary (possibly non-reduced) point. Is there a universal property that blowing-down along the exceptional divisor satisfies in this case? (The way I see it, one issue is to replace "collapsing $E$ to a point" by something else.)

3.) I'm very much interested in the most general version of the universal property one can formulate. If there is a reference where this is covered (EGA?) I would love to see it. I also don't have "good" reasons to believe that a universal property doesn't hold in complete generality (say in the context of noetherian schemes). If you're convinced that this is the case, I'd love to hear your reasoning.

  • 1
    $\begingroup$ ("of blowing down along these lines " - I love it!) For (1), it seems like the thing to realize is that $E$ comes as a (projective space) bundle over $Z$, and instead of saying "any map that contracts $E$ to a point" we'd say "any map that contracts each of these subvarieties of $E$ to points". It's easy to miss that extra structure when $Z$ is itself a point. $\endgroup$ Jan 23, 2013 at 4:47
  • $\begingroup$ Allen, that's hilarious :) thanks for pointing it out, and thanks for your comment. $\endgroup$
    – LMN
    Jan 23, 2013 at 5:05
  • 4
    $\begingroup$ EGA II, 8.11.1 (applies to the blow-up $f:Y \rightarrow X$ of a normal noetherian scheme $X$ at a closed point of codimension $\ge 2$, using any target scheme $Z$; normality ensures $O_X = f_{\ast}(O_Y)$) $\endgroup$
    – user30379
    Jan 23, 2013 at 12:28
  • 1
    $\begingroup$ @pranavak, thanks. It's interesting that the required condition remains "to be constant along the fibers" - something checked purely at the level of topological spaces. It's even incredibly more general than just blowing up. It applies to any of the morphisms (with connected fibers) coming from stein factorization. $\endgroup$
    – LMN
    Jan 24, 2013 at 14:21
  • 1
    $\begingroup$ ...it also applies to essentially arbitrary blowups of normal, noeth. integral schemes. $\endgroup$
    – LMN
    Jan 24, 2013 at 14:31

2 Answers 2


Artin, "Algebraization of Formal Moduli, II".

  • 2
    $\begingroup$ jstor.org/stable/1970602 Do you mean lemma 3.7 there? It is quite different from what the OP expected... $\endgroup$
    – quim
    Jan 24, 2013 at 9:58
  • $\begingroup$ ...glad to know I wasn't totally wrong for being confused. (In any case, I'm always very appreciative for your comments Jason!) $\endgroup$
    – LMN
    Jan 24, 2013 at 18:46

There is indeed a "universal property of blowing up". See Proposition II.7.14 on page 164 of [Hartshorne]. But this is slightly different from what you are asking. In your situation it would say that if $\psi: W\to X$ is such that the preimage of the point you blow up to get $Y$ is a divisor in $W$, then $\psi$ factors through $\phi$.

For the kind of property you want, there are several issues to consider (and I am not claiming that this list is exhaustive):

  1. You would probably have to assume at least that $X$ is (semi)normal, other wise you could have a situation where $Z$ is the normalization of $X$.

  2. You have to be careful if you want to do this for blowing up positive dimensional subschemes. First of all, suppose you wanted your assumption be about contracting the exceptional divisor. Then you'd have to assume that the image of $E$ in $Z$ is not smaller dimensional than in $X$ for sure, but even that is not enough. Here is an example to show some of the difficulties:
    Let $C$ be the cone over a product. For simplicity you may assume that it is over $\mathbb P^1\times \mathbb P^1$ but the only thing that is somewhat relevant is that the two factors be the same dimension. So, then let $X$ be the blow up of $C$ along the cone over one of the "rulings", $Z$ the blow up of $C$ along the other "ruling" and $Y$ the blow up of $C$ at the vertex. Then you have the situation you describe yet there is no morphism between $X$ and $Z$.

  3. I guess the conclusion is something like what Allen suggests in his comment: you need to assume something like contracting each fiber. However, then you won't really get a universal property of blow ups, but a universal property of morphisms mapping to a normal variety that would essentially say something along the lines that if you fix what gets contracted, then the morphism is determined up to a degree one birational map of the target, so if it is normal, then the morphism is determined. (I am not making a precise statement here, but I think it can be made precise).

  • 2
    $\begingroup$ Proposition II.7.14 does not actually present a functor on schemes that is represented by the blowing up. However, it is not hard to write down such a functor (and I think I did that in a previous MO comment). $\endgroup$ Jan 23, 2013 at 17:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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