$f^{-1}\mathcal I \cdot \mathcal O_X$ vs $f^\ast \mathcal I$

Let $X$ ad $Y$ be (noetherian) schemes and let $\mathcal I \subseteq \mathcal O_Y$ be a sheaf of ideals on $Y$. Let $f \colon X \to Y$ be a morphism of schemes. In general the sheaf $f^\ast \mathcal I$ is not a sheaf of ideals on $X$ (since it can be not contained in $\mathcal O_X$). Nevertheless, there is a natural morphism $$f^\ast \mathcal I \to \mathcal O_X$$ and we can set $f^{-1}\mathcal I \cdot \mathcal O_X$ as the image of this morphism. In this wat $f^{-1}\mathcal I \cdot \mathcal O_X$ is a sheaf of ideals on $X$.

So, my question is the following:

Is there any reasonable condition under which we have $f^{-1}\mathcal I \cdot \mathcal O_X = f^\ast \mathcal I$?

I think this is an interesting question in general, but the real reason why I ask this question is that have a problem with the book of Faltings and Chai "Degeneration of abelian varieties". In chapter V, Section 5 of that book, they prove that the toroidal compactification of the Siegel modular variety is isomorphic to the normalization of the blowup of the minimal compactification with respect to some sheaf of ideals $\mathcal I$. The proof is quite complicated, but at page 178 they use the universal property of blow-ups to define the morphism. The point is that the show that $f^\ast \mathcal I$ is invertible, while the universal property is about $f^{-1}\mathcal I \cdot \mathcal O_X$. I suspect I am missing something quite obvious.

Any help is appreciated!

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For what it's worth, many people write $f^* \mathcal{I}$ when they mean $f^{-1} \mathcal{I} \cdot \mathcal{O}_X$ (this is especially prevalent in resolution of singularities texts). I haven't looked at your source, but are you sure this isn't what's meant there? – Karl Schwede May 23 '13 at 14:11

Let $Z\subset Y$ be the subscheme given by $I$. There is an exact sequence $$0 \to I \to O_Y \to O_Z\to 0.$$ Pulling it back one gets an exact sequence $$0 \to L_1f^*O_Z \to f^*I \to O_X \to f^*O_Z \to 0$$ which shows that $f^{-1}I\cdot O_X = f^*I$ if and only if $L_1f^*O_Z = 0$. For example this holds if $f$ is flat.
A birational morphism is flat only if it is an open embedding. So, I doubt very much that the morphism you are interested in is flat. But of course, the flatness of the morphism is much stronger than what you want. For example, if $f$ is flat only over $Z$ everything is OK. – Sasha May 23 '13 at 12:41