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?
[Math] Blow-up removes intersections
ag.algebraic-geometryblow-ups
Best Answer
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.