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Assume that $\beta:\tilde{X}\to X$ is the blow-up of a nonsinular $\Bbbk$-variety $X$ along a sheaf of ideals $\mathcal{I}$. Let $Y:=Z(\mathcal{I})$. Given nonsingular, closed subvarieties $Z_1,\ldots,Z_r\subseteq X$ such that $\bigcap_i Z_i \subseteq Y$, is it true that $\bigcap_i \tilde{Z}_i=\emptyset$, where $\tilde{Z}_i$ denotes the strict transform of $Z_i$? If not, does this hold if we require $Y$ to be nonsingular and/or the $Z_i$ to intersect transversally?

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This is certainly false in the generality that you first state it. Think about the subvarieties defined by $y=0$ and $y=x^2$ in the plane. If you blow up at the origin, the strict transforms meet in a point, if I'm not mistaken (I'm thinking about the incidence correspondence style description of the blow-up). Of course, the problem with this example is the lack of transversality. My intuition suggests that it's true if the $Z_i$ meet transversely, but I'm not confident about it :) –  Ramsey Jul 5 '11 at 19:51
    
Yea, that was my feeling as well. Thanks (and +1) for confirming that. –  Jesko Hüttenhain Jul 5 '11 at 21:08
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3 Answers 3

up vote 7 down vote accepted

As Sasha and Ramsey point out, this isn't true in the generality requested. However, the following is true, see Hartshorne, Chapter II, Exercise 7.12.

Statement: Suppose that $X$ is a Noetherian scheme and let $Y, Z$ be closed subschemes, neither one containing the other. Let $\widetilde{X}$ be the blowing up of $Y \cap Z$ (defined by the sum of the ideal sheaves). Then the strict transforms of $Y$ and $Z$ do not meet.

In other words, you can't choose an arbitrary $Y$, but there always is a subscheme (supported where you want) which you can blow up which will work.

EDIT: With regards to why the sum of all the ideals can't work, consider the three coordinate hyperplanes $$H_1, H_2, H_3 \subseteq \mathbb{A}^3.$$ The sum of the ideals defining the hyperplanes is the ideal defining the origin in $\mathbb{A}^3$. Blowing up the origin cannot possibly separate $H_1$ and $H_2$ because $H_1 \cap H_2$ is a line.

EDIT2: As Jesko pointed out, the previous edit answers the wrong question. He's not interested in the pair-wise intersection, just the total intersection. My example in the above edit doesn't help there. I think his answer below is then correct.

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So this is because we blow up in $\mathcal{I}_Y+\mathcal{I}_Z$, not its radical, I suppose. Since there is no proof here, I have to ask, does this generalize in the obvious way to more than two closed subschemes? By blowing up $\mathcal{I}_{Z_1}+\cdots+\mathcal{I}_{Z_r}$? –  Jesko Hüttenhain Jul 6 '11 at 7:01
    
Yes, blowing up the radical won't work. It's an easy exercise, if you get stuck I can write down a proof. Also, it won't generalize in that way. You could blow up $(I_{Z_1} + I_{Z_2}) \cdot (I_{Z_1} + I_{Z_3}) \dots (I_{Z_{r-1}} + I_{Z_r})$ though. –  Karl Schwede Jul 6 '11 at 7:20
    
Renitently, I tried to generalize the statement in my answer below, using the sum of the ideals. Would you be so kind to tell me whether it checks out? –  Jesko Hüttenhain Jul 6 '11 at 10:47
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In general the answer is no. For example if $X$ is a plane, $Y$ is a point and $Z_1,Z_2$ are curves tangent in $Y$, then the strict transforms intersect. If however $Y$ is smooth and normal bundles $N_{Z_i/Y}$ do not intersect in $N_{X/Y}$ then the intersection is empty. Note that transverslity is another condition, the transversality just means that $N_{Z_i/Y} + N_{Z_j/Y} = N_{X/Y}$ which is not the same as emptiness of the intersection.

For example let $X = A^3$, $Z_1$ being the line $x = y = 0$ and $Z_2$ being the hypersurface $f_2 + f_3 = 0$, where $f_i$ are homogeneous polynomials of degree $i$ such that $f_2(0,0,1) = 0$ and $f_3(0,0,1) \ne 0$. Let finally $Y$ be the intersection of $Z_1$ and $Z_2$ (it consists of two points, one of those being $(0,0,0)$). Then the strict transforms of $Z_1$ and $Z_2$ intersect in the exceptional divisor over the point $(0,0,0)$ since $f_2(0,0,1) = 0$.

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This is sweet, do you have a proof or reference for the fact that the intersection is empty as long as the $N_{Z_i/Y}$ have empty intersection? –  Jesko Hüttenhain Jul 6 '11 at 6:54
    
The intersection of the strict preimage of $Z$ with the exceptional divizor $E$ is $P(N_{Z/Y}) \subset P(N_{X/Y}) = E$. This is more or less by definition of the blowup. –  Sasha Jul 6 '11 at 7:06
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I have tried to generalize the Exercise referenced by Karl, even though he told me that it shouldn't be possible this way. I think, however, it works:

Edit: I made a mistake concerning $J_i$ - it cannot be equal to $I_i\oplus\bigoplus_{d\ge 1} I_i^dT^d$ because that is not necessarily an ideal - it might not be closed under multiplication by elements from the ring $S$. The version below looks better.

Proposition. Let $Z_0,\ldots,Z_r$ be closed subschemes of a Noetherian scheme $X$ such that $Z_i\not\subset Z_j$ for $i\ne j$. Let $I_i:=I(Z_i)$ and denote by $\tilde{Z}_i$ the respective strict transform of $Z_i$ under the blow-up $\beta:\tilde{X}\to X$ of $X$ along $I:=\sum_{i=0}^rI_i$. Then, $\bigcap_{i=0}^r\tilde{Z}_i=\emptyset$.

Proof. The statement can be checked locally, so we may assume that $X=\mathrm{Spec}(A)$ is affine. Let $f_i:Z_i\hookrightarrow X$ be the respective closed immersion, so $Z_i=\mathrm{Spec}(A/I_i)$ and $f_i^\sharp:A\twoheadrightarrow A/I_i$. Then, the inverse image ideal sheaf of $I$ under $f_i$ is $I\cdot A/I_i$ and hence,

$\displaystyle\tilde{Z}_i=\mathrm{Proj}\left(\bigoplus_{d\ge 0} \left(I\cdot A/I_i\right)^d\cdot T^d\right)$

With $S=\bigoplus_{d\ge 0} I^d\cdot T^d$, the homogeneous ideal defining $\tilde{Z}_i$ inside $\tilde{X}=\mathrm{Proj}(S)$ is equal to

$\displaystyle J_i = \bigoplus_{d\ge 0} (I^d\cap I_i)$

In particular, $J_0+\cdots+J_r\supseteq S_+$, so any point $P\in\tilde{Z}_0\cap\cdots\cap\tilde{Z}_r$ would correspond to a homogeneous prime ideal containing each of the $J_i$ and hence, the irrelevant ideal. There is no such point.

Did I miss something? Or is this correct?

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I don't think this can be correct. Consider the coordinate hyperplances $H_1, \dots, H_4 \in \mathbb{A}^4$. Now, $I_1 + \dots I_4$ is just the ideal of the maximal ideal of the origin. Blowing up that maximal ideal will certainly not separate the $H_i$ though, since they intersect at points besides only the origin. I'll try to read through the proof now. –  Karl Schwede Jul 6 '11 at 14:58
    
Perhaps I've totally missed the problem, but I don't see why $S_+ \subseteq J_0 + \dots + J_r$. –  Karl Schwede Jul 6 '11 at 15:06
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Well, shouldn't $S_1=(J_0+\cdots+J_r)_1$ verify this, since everything is generated in degree one? Also, I do not want to remove any pairwise intersection, just their mutual intersecion $\bigcap_i\tilde{Z}_i$ should be empty. –  Jesko Hüttenhain Jul 6 '11 at 16:11
    
Ah, I misunderstood. I'll get back to you. –  Karl Schwede Jul 6 '11 at 19:33
    
I think I confused you with the blunt mistake I made concerning $J_i$, check my edit. This looks quite believable to me now, but any comments are still welcome. I'll also accept your answer because it was extremely helpful. Thanks for all the effort! –  Jesko Hüttenhain Jul 7 '11 at 16:59
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