MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

My question is if this can be generalized as well: Let $\beta:\widetilde{X}\mathrel{\mathop:}=\mathop{\mathrm{Bl}}_Z(X)\to X$ be the blow-up of a smooth algebraic variety $X$ along a irreducible generically smooth subscheme $Z$. Let $E\mathrel{\mathop:}=\beta^{-1}(Z)$ be the exceptional divisor. Now, let us assume I have a divisor $D$ on $X$, with $Z$ closed subscheme of $D$ ($D$ smooth and irreducible, if it helps). Then I ask if $\beta^\ast D \sim \widetilde{D} + \alpha E$, for some integer $\alpha$, where $\widetilde{D}$ denotes the strict transform of $D$ and "$\sim$" is linear equivalence.

share|cite|improve this question
up vote 5 down vote accepted

I think the answer is, almost.

$E$ might have more than one component. Indeed, since you only assume that $Z$ is generically smooth, $E$ might have many different components. $E = \sum E_i$.

You can always write $\beta^* D = \widetilde{D} + \sum_i \alpha_i E_i$, but you can't use the same $\alpha$ for all of them.

Let me give an example. Suppose $X = \mathbb{A}^2$, $Z = V(x \cdot (x,y) \cdot (x,y^2)) = V(x^3, x^2y^2, x^2y, xy^3)$.
Certainly $Z$ is generically smooth and irreducible (its just not reduced, in higher dimensions, I'm think I can rig reduced examples as well). Set $D = \text{Div}(x^3)$. The blow-up of $X$ has two exceptional divisors.
Note $Z$ is a closed subscheme of $D$ as well. By working in higher dimensions, one can rig examples where the support of $Z$ and $D$ are different.

Anyway, one chart of the blow-up looks like $k[x/y, y^2/x]$ with the obvious map to $\text{Spec} k[x,y]$. On that chart, the two exceptional divisors are $E_1 = \text{Div}(x/y)$ and $E_2 =\text{Div}(y^2/x)$.

On this chart, we compute $\beta^*D$. Note $$x^3 = ((x/y)^2 \cdot (y^2/x))^3$$ and so $$\beta^*D = \widetilde{D} + 6 E_1 + 3 E_2.$$

share|cite|improve this answer
Thank you for your answer. Hence what I wanted would be valid only if $E$ is irreducible. I think also that find conditions of $Z$ to ensure the irreducibility of $E$ is very complicated :(... bye – gio Jul 11 '11 at 15:43
gio, another point you can consider is what do you mean by $E = \beta^{-1}(Z)$. Do you mean scheme theoretically, or set theoretically. In particular, does $E$ have the reduced structure? That's how I interpreted your question in my answer. If not, then the thing you wanted is still not true, but there might be a little more hope in some examples. – Karl Schwede Jul 11 '11 at 17:20

Your Answer


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.