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Does this concept is defined for every birational morphism? What is the precise meaning? Thank you for your comments.

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The exceptional set is defined for every birational morphism $\pi : Y \to X$. This is defined as follows. Set $\Sigma \subset X$ to be the smallest closed subset of $X$ outside of which $\pi : (Y \setminus \pi^{-1}(\Sigma)) \to (X \setminus \Sigma)$ is an isomorphism.

In this case the exceptional set is defined to be $$E = \pi^{-1}(\Sigma).$$ Now, $\Sigma$ itself is just a set, and not a scheme, so often people will give $E$ the reduced scheme structure and consider it as a closed subscheme of $Y$.

In many applications (perhaps even most), $E$ is actually a divisor, in other words, it is a union of subvarieties of codimension 1. In this case, $E$ is called the exceptional divisor. Indeed, if you obtained $\pi : Y \to X$ by blowing up $\Sigma$, and say $X$ is a normal variety with $\Sigma \subseteq X$ a codimension $\geq 2$ subset, then $E$ is indeed a divisor. However:

Negative result: There are many examples when $E$ is not a divisor. This is particularly common for small resolutions (resolutions of singularities for which the exceptional set has no divisorial components), which are in many cases desirable. For example, they are crepant resolutions.

The most common example of a small resolution is probably the following. $$X = V(xy - uv) \subseteq \mathbb{A}^4.$$ Note $X$ has dimension 3.

In this case, consider blowing up the ideal $(x,u) \subseteq O_X$. This is the ideal of a divisor on $X$. I'm not going to do this for you, but it is a good exercise. In this case, the blowup gives you a resolution of singularities of $X$, but the exceptional set is 1-dimensional.

Positive result: If $X$ is smooth (or even factorial) and $\pi : Y \to X$ is birational with $Y$ quasi-projective, then the exceptional set is always a divisor. Sándor explains this quite nicely in THIS QUESTION.

Positive result #2: If you are willing to change your variety $Y$ by taking a further blow-up, then you can always turn the exceptional set into a divisor. For example, simply take $\rho : Y' \to Y$ to be the (normalized) blow up of $E$.

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Thanks for your excellent answer and patient a lot! –  MZWang Apr 8 '12 at 5:17
    
Dear Karl, can you give me some references? –  MZWang Apr 12 '12 at 10:07
    
Sure, you could look at Hartshorne's Algebraic Geometry, that will at least talk about exceptional sets, the basics for positive result #2 can also be found there in the section on blowing up. I don't know a standard reference for a small resolutions though, but a search of the arXiv or mathscinet will bring something up. –  Karl Schwede Apr 12 '12 at 11:17
    
Thank you!I will check Hartshorne. –  MZWang Apr 17 '12 at 8:37

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