# On the number of irreducible components of an exceptional divisor

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.

• 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. – Karl Schwede May 26 '15 at 16:52
• @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. – Jesko Hüttenhain May 26 '15 at 17:05
• @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. – Sándor Kovács May 26 '15 at 19:21

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.
• 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. – Jesko Hüttenhain May 26 '15 at 7:51