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?



Blow-up removes intersections?
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.


