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Let $X,Y$ be smooth, proper varieties, and $f: X \to Y$ be a proper birational morphism. Suppose $E$ is a smooth, irreducible exceptional divisor, with the image $f(E)$ also smooth. Let $I$ be the sheaf of ideal associated to $f(E)$.

Now, let $\pi: X_1 \to Y$ be the blowup of $Y$ with center $f(E)$. It is claimed that $f$ factor through $\pi$.

It is suggested to use the universal property of blowup as in Hartshorne Chapt. II Prop. 7.14. In order to use that result, one need to know that the sheaf $f^{-1}I \cdot \mathcal{O}_X$ is an invertible sheaf on $X$. However, I don't know how to show that fact.

Li Yutong
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1 Answers1

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Maybe this could work. The ideal $I$ detects local functions on $Y$ that vanish along $f(E)$. Now, since both $X$ and $Y$ are smooth, the exceptional locus for $f$ is purely divisorial. So, the exceptional locus over a neighborhood of $f(E)$ an $f$-exceptional divisor $F$ (let $F$ have reduced structure); furthermore, $E$ is an irreducible component of such divisor.

Now, we'd be home if we can show $f^{-1}I \cdot \mathcal{O}_X = \mathcal{O}_X(-F)$. Now, how can we think of $f^{-1}I \cdot \mathcal{O}_X$ as an ideal on $X$? What we do is taking local functions in $I$ and precomposing with $f$. Then, we get an inclusion $f^{-1}I \cdot \mathcal{O}_X \subset \mathcal{O}_X(-F)$, as the whole $F$ is mapped onto $f(E)$.

Now, we are left with showing the reversed inclusion. We can regard a local function on $X$ as a rational function on $Y$ (this is the inverse operation to precomposing a function on $Y$ with $f$). Pick a local function in $\mathcal{O}_X(-F)$, call it $\phi$, and let $\psi$ the rational funciton induced on $Y$. As $\psi \circ f =\phi$, $\psi$ needs to vanish along $f(E)$. But then this implies that $\phi$ is a local function in $f^{-1} I \cdot \mathcal{O}_X$.

Stefano
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