Question

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?

Answer 1

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.

Answer 2

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 $f_2(0,0,1) = 0$.

Answer 3

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?