Let $X$ and $Y$ be smooth projective schemes with $Y \subset X$. Let $\pi : \widetilde{X} \to X$ be the blow up of $X$ along $Y$ with exceptional divisor $E$.

I have seen the statement that Zariski's Main Theorem implies that $\pi_{*}(\mathcal{O}_{\widetilde{X}} ) \to \mathcal{O}_{X}$ and $\pi_{*}(\mathcal{O}_{E}) \to \mathcal{O}_{Y} $ are isomorphisms. Why is this true?

References and suggestions will be appreciated.


1 Answer 1


This MO post is a great reference if you're ever trying to figure out or remember when for a morphism $f:X\to Y$ we will have $f_*\mathcal{O}_X=\mathcal{O}_Y$.

Here's the relevant portion of that answer for this post, in order to make this answer self-contained:

The case of an arbitrary projective morphism.

Now when $f:X\to Y$ is any projective morphism, then $f_*\mathscr O_X$ is a coherent $\mathscr O_Y$-module, hence we get a factorization of $f$ as $h\circ g:X\to Z\to Y$, where $h:Z\to Y$ is affine, and where also $h_*(\mathscr O_Z) = f_*\mathscr O_X$. Then $h$ is not only an affine map, but since $h_*(\mathscr O_Z)$ is a coherent $\mathscr O_Y$-module, $h$ is also a finite map. Moreover $g:X\to Z$ is also projective and since $g_*(\mathscr O_X) = \mathscr O_Z$, it can be shown that the fibers of $g$ are connected. Hence an arbitrary projective map $f$ factors through a projective map $g$ with connected fibers, followed by a finite map $h$. Thus in this case, the algebra $f_*\mathscr O_X$ determines exactly the finite part $h:Z\to Y$ of $f$, whose points over $y$ are precisely the connected components of the fiber $f^{-1}(y)$.

One corollary of this is "Zariski's connectedness theorem". If $f:X\to Y$ is projective and birational, and $Y$ is normal then $f_*\mathscr O_X= \mathscr O_Y$, and all fibers of $f$ are connected, since in this case $Z = Y$ in the Stein factorization described above. If we assume in addition that $f$ is quasi finite, i.e. has finite fibers, then $f$ is an isomorphism. More generally, if $Y$ is normal and $f:X\to Y$ is any birational, quasi - finite, morphism, then $f$ is an embedding onto an open subset of $Y$ ("Zariski's 'main theorem' "). More generally still, any quasi finite morphism factors through an open embedding and a finite morphism.

This applies to your situation as follows: the blowup map $\pi:\widetilde{X}\to X$ is a projective birational map with connected fibers. Since projective and connected fibers are preserved under base change, we see that the base change of this map $E\to Y$ is again projective with connected fibers, so we may also apply the result there, via the Stein factorization described in the first paragraph (even though this last morphism is not birational - $\dim E=\dim X-1\neq \dim Y$).

  • $\begingroup$ Hi, KReiser @. Thank you very much for the support given in the questions I ask here in the forum. $\endgroup$
    – Emanuell
    Jan 17, 2020 at 11:34
  • $\begingroup$ KReiser@. I asked this question because I am trying to understand your answer and another about Direct Image and Blow up. $\endgroup$
    – Emanuell
    Jan 17, 2020 at 11:38
  • $\begingroup$ Hi, @KReiser. I will post a question to answer two questions I have about your answer in Direct Image and Blow up. thank you. $\endgroup$
    – Emanuell
    Jan 17, 2020 at 13:13

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