Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $X$ be a projective variety and let $\tilde{X}$ be the blow-up of $X$ at a subscheme $Z$. Let $F$ be the exceptional divisor of $\tilde{X}$. I wonder:

What is the number of irreducible components of $F$?

Note that this number depends strongly on the scheme structure on $Z$. For example, when $Z$ is a line in $\mathbb{P}^2$ with an embedded point, $F$ has two components, whereas the blow-up of a line has only one. So is it true in general that the number of components of $F$ is at least the number of associated primes of $Z$? (I am mostly interested in a lower bound for this number.)

share|improve this question
    
The blow up of a line in $\mathbb{P}^2$ is an isomorphism. There is no exceptional divisor. Do you mean the pull-back of the scheme $Z$? You could count its components... –  Karl Schwede Aug 26 '11 at 13:45
add comment

2 Answers

I had a minor thought. Did you ever look the paper Multiplicity of the special fiber of blowups by Corso,Polini, Vasconcelos.

In particular, they give an upper bound on the number (I realize that you are interested in lower bounds, but maybe some of the ideas are related / or could be useful) by in particular, bounding the multiplicity at the origin of the Rees algebra fiber ring (in other words, if you compute the blow-up by computing Proj $R[It]$, then mod out by an ideal from $R$, you get some graded ring corresponding to the fiber over the ideal you modded. Then you can study the blow-up by studying properties of the ring and I think the multiplicity they study should give you an upper bound on the number of components).

Of course, the number of minimal associated primes gives you some bound on the components of the pre-image of $Z$.

share|improve this answer
add comment

I'd say the number can be arbitrary large if $X$ is not assumed smooth along $Z$. Take the simplest example: hypersurface singularity $\{f_p+f_{p+1}=0\}\subset\Bbb C^n$, where $f_p$ is totally reducible and $f_{p+1}$ is generic enough. So that the singularity is isolated. Blowup the origin. The exceptional divisor (=the projectivization of the tangent cone) is the hyperplane arrangement, $\{f_p=0\}\subset\Bbb P^{n-1}$.

(Of course, you can compactify the hypersurface, if you insist on projective varieties)

share|improve this answer
add comment

Your Answer

 
discard

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.