Assume that $\beta:\tilde{X}\to X$ is the blowup 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?

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 pairwise intersection, just the total intersection. My example in the above edit doesn't help there. I think his answer below is then correct. 


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


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.
Did I miss something? Or is this correct? 

