Assume I have a nonsingular, irreducible, algebraic variety $X$ and irreducible, nonsingular subvarieties $Z_1,\ldots,Z_k\subseteq X$. Let $\mathcal{I}_i$ be the ideal sheaf of $Z_i$ and $\mathcal{I}:=\mathcal{I}_1\cdots\mathcal{I}_k$ the product. My question is whether $\mathcal{I}$ is the ideal sheaf of the union $Z_1\cup\ldots\cup Z_k$. You may assume that $Z_1\cap\ldots\cap Z_k=\emptyset$ or, equivalently, $\mathcal{I}_1+\ldots+\mathcal{I}_k=\mathcal{O}_X$. You may also assume that the $Z_i$ intersect transversally, if that helps.

For each $i$ we have a short exact sequence $$ 0 \to I_i \to O_X \to O_{Z_i} \to 0. $$ Let us think about it as about a resolution of $I_i$. Then it follows that $$ I_1 \otimes^L I_2\otimes^L \dots \otimes^L I_k = (O_X \to O_{Z_1}) \otimes^L (O_X \to O_{Z_2}) \otimes^L \dots \otimes^L (O_X \to O_{Z_k}). $$ The transversality condition ensures that $$ O_{Z_{i_1}} \otimes^L O_{Z_{i_2}}\otimes^L \dots \otimes^L O_{Z_{i_p}} \cong O_{Z_{i_1} \cap Z_{i_2} \cap \dots \cap Z_{i_p}}. $$ Therefore we conclude that $I_1 \otimes^L I_2\otimes^L \dots \otimes^L I_k$ is quasiisomorphic to the complex $$ O_X \to \oplus O_{Z_i} \to \oplus O_{Z_i \cap Z_j} \to \dots \to O_{Z_1 \cap Z_2 \cap \dots Z_k}. $$ It follows that $$ I_1\cdot I_2\cdot \dots \cdot I_k = Im(I_1 \otimes^L I_2\otimes^L \dots \otimes^L I_k \to O_X) $$ is equal to the kernel of the map $O_X \to \oplus O_{Z_i}$ that is to $I_{Z_1 \cup Z_2 \cup \dots \cup Z_k}$. EDIT. The importance of the transversality condition can be seen at the following simple example. Let $X = A^3$ and $Z_1,Z_2$ be two lines intersecting at a point $P$ (the simplest nontransversal intersection). Then one has $O_{Z_1} \otimes O_{Z_2} = O_P$, $Tor_1(O_{Z_1},O_{Z_2}) = O_P$. Therefore one has a spectral sequence with the first term being $$ \begin{array}{ccccc} O_X & \to & O_{Z_1} \oplus O_{Z_2} & \to & O_P \cr &&&& O_P \end{array} $$ which converges to $I_{Z_1} \otimes^L I_{Z_2}$. The second term then gives exact sequence $$ 0 \to I_{Z_1}\otimes I_{Z_2} \to I_{Z_1 \cup Z_2} \to O_P \to 0, $$ which shows that $I_{Z_1}\cdot I_{Z_2}$ is the ideal of the union of lines with an additional nilpotent at the point of intersection. 


Here is a simple proof for $k=2$. Karl's examples show that the statement is false in general (without more careful assumptions). Claim If $\mathscr I, \mathscr J\subseteq \mathscr O_X$ are two ideal sheaves such that $\mathscr I +\mathscr J =\mathscr O_X$, then $\mathscr I\mathscr J=\mathscr I\cap \mathscr J$. Proof $$ \qquad \qquad\qquad \mathscr{(I+J)(I\cap J)}\subseteq \mathscr{I J}\subseteq \mathscr{I\cap J}. \qquad\qquad\qquad \square $$ 


As Sandor and Martin pointed out above, it is ok if the subschemes intersect in the emptyset pairwise. I'm going to provide 3 examples, I think the third one gives a counterexample to even the transversality statements. Example 1: Here's an example where it's false without that hypotheses, notice that the varieties are smooth and they intersect pairwise with normal crossings. EDIT: as t3suji pointed out in the comments, these varieties don't intersect transversally in the ambient space, just in the ambient $z = 0$ plane EndOfEdit Consider $X = \mathbb{A}^3$ and set $Z_1 = V(y,z)$, $Z_2 = V(x,z)$, $Z_3 = V(x, z1)$. Notice that $Z_3$ doesn't intersect any of the other subschemes. Then, $I_1 \cap I_2 = (z, xy)$. However, $I_1 \cdot I_2 = (xy, yz, xz, z^2)$. These ideals are not equal clearly. Now, we can immediately see that multiplying/intersecting by $I_3$ won't change the behavior at the origin at all since the ideal doesn't vanish there, so they are not equal. However, just to be sure, I also did the following computation (with Macaulay2): $$I_1 \cdot I_2 \cdot I_3 = (xz, yz, z^3  z^2, yz^2  yz, xz^2  xz).$$ $$I_1 \cap I_2 \cap I_3 = (z^2 z, xz, xy).$$ Macaulay2 also confirmed that the ideals were not equal. Example 2: Here's a different example. a 4th variety that doesn't intersect the others at all. $X = \mathbb{A}^3$. $I_1 = (x,y)$, $I_2 = (x,z)$, $I_3 = (y,z)$ and $I_4 = (x1,y1,z1)$. Certainly again the $I_4$ doesn't matter, it's just included so that the sum of the ideals is equal to $R = k[x,y,z]$. I wonder if it might be reasonable to say that $Z_1$, $Z_2$ and $Z_3$, as a triple, have transverse intersection at the origin. Anyways: $I_1 \cap I_2 \cap I_3 = (xy, xz, yz)$ but, $I_1 \cdot I_2 \cdot I_3 = (yz^2, xz^2, y^2z, xyz, x^2z, xy^2, x^2y)$. Example 3: Ok, now I'm just going to give three subvarieties which intersect at the origin, pairwise transversally, and such that the product of the ideals is not equal to the intersection. You may add the ideal sheaf of some other variety that doesn't intersect them at all to make the sum of ideals equal the whole structure sheaf. $X = \mathbb{A}^4 = \text{Spec}k[x,y,u,v]$. $I_1 = (x,y)$, $I_2 = (u,v)$, $I_3 = (x+u, y+v)$. I believe these have pairwise transverse intersection at the origin. Now then, it is true that $I_1 \cdot I_2 = I_1 \cap I_2$, and likewise with any pair. However, Macaulay2 can be used to verify that $I_1 \cdot I_2 \cdot I_3 \neq I_1 \cap I_2 \cap I_3$. Roughly speaking the problem is that $Z_1 \cup Z_2$ has funny intersection with $Z_3$. Ok, let me now give a proof of a correct statement showing that sometimes they are equal. Lemma: Suppose that subschemes $Z_1, \dots, Z_k$ have pairwise trivial intersection in some ambient Noetherian scheme $X$. Then $I_{Z_1} \dots I_{Z_K} = I_{Z_1} \cap I_{Z_k}$. Proof: The statement is local so we may assume that $X$ is the spectrum of a local ring $(R, \mathfrak{m})$. Now, since $I_{Z_1} + I_{Z_2} = R$, at least one of those ideals must equal $R$ (if not, both would be in the maximal ideal $\mathfrak{m}$, and so would their sum). Likewise with all pairs. Therefore, at most one of the ideals $I_{Z_i}$ is not equal to $R$. But now the statement is obvious. $R \cdot R \dots I_{Z_i} \dots R = I_{Z_i} = R \cap R \cap \dots I_{Z_i} \cap \dots R$. 

