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Let $X$ be a complex, affine variety and $Z\subseteq X$ a closed subset of $X$ (i.e. a closed, reduced subscheme). Let $E$ be the exceptional divisor of the blow-up $\pi:\tilde X\to X$ of $X$ with center $Z$.

My question is: Under which conditions on $X$ and $Z$ does $E$ have the same number of irreducible components as $Z$?

For example, if $X$ is smooth and the irreducible components of $Z$ are smooth and disconnected, then the statement holds. There are examples of surfaces $X$ which are singlar at a single point $Z$ and the exceptional divisor is not irreducible, so I know that it is not always true.

I suppose that I will have to assume $X$ smooth, but I am wondering if the assumption on $Z$ can be weakened. I would be most happy if it was always true for smooth $X$ because $Z$ is a reduced subscheme, but I do not really know whether that is to be expected.

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  • $\begingroup$ Are you always blowing up the set $Z$ with the reduced scheme structure, or are you allowing other scheme structures... If the former I'm not sure there is much that can be said. $\endgroup$ – Karl Schwede May 26 '15 at 16:52
  • $\begingroup$ @KarlSchwede: Yes, I am blowing up the set $Z$ with the reduced scheme structure. I should have stated that more clearly. I am surprised that in this case, the question becomes harder than in the more general case, though. $\endgroup$ – Jesko Hüttenhain May 26 '15 at 17:05
  • $\begingroup$ @JeskoHüttenhain:It's actually not surprising. Think about the case of a quadric cone and a ruling. If you blow up the line with the reduced scheme structure, you get the same as if you blew up the point, because it is not a Cartier divisor. But if you blow up the double line, then nothing happens, because you blew up a Cartier divisor. Similar things can happen if X is smooth, but Z is singular. $\endgroup$ – Sándor Kovács May 26 '15 at 19:21
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I agree that you should assume that $X$ is smooth, otherwise giving a reasonable criterion for this seems unlikely.

Given that, let's say that $I$ is the ideal sheaf of Z inside $X$. Then, since $X$ is smooth, the preimage of $Z$ in the blow up coincides with $E$ and is isomorphic to $\mathrm{Proj}_Z \oplus_d I^d/I^{d+1}$. If this Proj is a $\mathbb P^n$-bundle, then you win and this happens for example if $Z$ is a local complete intersection in $X$, because in that case $\oplus_d I^d/I^{d+1}\simeq \mathrm{Sym}(I/I^2)$ and hence the above Proj is simply the projectivization of the conormal bundle(!) $I/I^2$ (which is indeed a bundle if $Z$ is a local complete intersection).

I suspect that one can give examples when $Z$ is reduced (obviously not a local complete intersection), but this Proj is not relatively irreducible over $Z$ in which case you get extra components.

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  • $\begingroup$ Thanks a lot! Do you think it might be enough to have the components of $Z$ meet transversally? My $Z$ might not be equidimensional, so it is not Cohen-Macualay, and that means it's not an lci. I suppose. $\endgroup$ – Jesko Hüttenhain May 26 '15 at 7:51
  • $\begingroup$ My recollection is that even for several lines meeting transversally (pairwise), when you blow them up you can get all sorts of weird components over the center. Have you tried any examples? $\endgroup$ – Karl Schwede May 26 '15 at 17:43

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