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I'm trying to prove the following statement:

Theorem A Let $X$ be a non-singular variety over a field $k$ and let $Y \subset X$ be a smooth subvariety. Consider the blow-up $f : \widetilde X = Bl_Y(X) \to X$. Then for $i > 0$: $$R^i f_* \mathcal O_{\widetilde X} = 0.$$

This is mentioned for example in Hironaka, Resolution of Singularities of an Algebraic Variety Over a Field of Characteristic Zero I, p. 153 without a reference or a proof. My attempt (following the proof of Proposition V.3.4 in Hartshorne's Algebraic Geometry):

let $\mathcal F^i := R^i f_* \mathcal O_{\widetilde X}$ and let $y$ be the generic point of $Y$. Then the support of $\mathcal F^i$ is contained in $Y$ and using the Formal Functions Theorem, we get:

$$ \mathcal F^i_y = \lim_{\leftarrow} H^i(E_n, \mathcal O_{E_n}),$$

where $E_1 = E = f^{-1}(Y)$ and $E_n$ is given by the ideal sheaf $\mathcal J^n$ (where $\mathcal J$ is the ideal sheaf of $E$ in $\widetilde X$). Thus the above statement should be equivalent to:

$$ H^i(E_n, \mathcal O_{E_n}) = 0 \qquad \text{for all } i, n \ge 1.$$

Also, we have an exact sequence:

$$ 0 \to \mathcal J^n/\mathcal J^{n + 1} = \mathcal O_E(n) \to \mathcal O_{E_{n+1}} \to \mathcal O_{E_{n}} \to 0 \qquad (*)$$

Thus, it seems to me that the Theorem A is equivalent to the statement that $$H^i(E, \mathcal O_{E}(n)) = 0 \qquad \text{for all } i, n > 0. $$

On the other hand, $E = \mathbb P(\mathcal I/\mathcal I^2)$ is a projective bundle over $Y$ (where $\mathcal I$ is the ideal sheaf of $Y$ in $X$). Thus $$ R^i g_* \mathcal O_E (d) = 0 $$ for $i, d > 0$ (where $g = f|_E : E \to Y$) - see e.g. Stacks. Therefore by Leray spectral sequence we obtain: $$ H^i(E, \mathcal O_{E}(n)) = H^i(Y, g_* \mathcal O_{E}(n)) = H^i(Y, S^n(\mathcal I/\mathcal I^2)). $$ The right hand side seems to be non-zero in general.

Question: where's the mistake? How to fix it? Alternatively, what is a reference for the proof of Theorem A?

  • There are similar questions ([https://math.stackexchange.com/questions/3253408/blowing-up-nonsingular-variety-along-nonsingular-subvariety-keeps-arithmetic-gen?rq=1] - here the user had a similar problem, [https://mathoverflow.net/questions/201842/higher-direct-images-of-oe]) – Jędrzej Garnek Jul 31 '19 at 13:24
  • ... but I didn't find them useful. – Jędrzej Garnek Jul 31 '19 at 13:39

1 Answers1

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This is largely correct, except you need a slightly better version of the theorem on formal functions. Here's how it's stated in Hartshorne:

Theorem on Formal Functions (Hartshorne III.11.1): Let $f:X\to Y$ be a projective morphism of noetherian schemes, let $\mathcal{F}$ be a coherent sheaf on $X$, let $y\in Y$, let $X_n = X\times_Y \operatorname{Spec} \mathcal{O}_y/\mathfrak{m}_y^n$, and let $\mathcal{F}_n = v_n^*\mathcal{F}$ where $v_n: X_n\to X$ is the natural map.

Then $R^if_*(\mathcal{F})_y^{\wedge} \cong \lim_{\leftarrow} H^i(X_n,\mathcal{F}_n)$ is an isomorphism for all $i\geq 0$.

A better version replaces $y\in Y$ by a closed subscheme $Z\subset Y$: upgrading to a closed subscheme instead of a closed point, Hartshorne's proof works verbatim to show that $$\varprojlim (R^if_*\mathcal{F})\otimes_{\mathcal{O}_Y} \mathcal{O}_Y/\mathcal{I}_Z^n \cong\varprojlim R^if_*(\mathcal{F}\otimes_{\mathcal{O}_X} \mathcal{O}_Y/\mathcal{I}_Z^n)$$ and then one can use the exact same argument as the proof of proposition V.3.4 to conclude that $R^\bullet f_*\mathcal{O}_{\widetilde{X}}=\mathcal{O}_X$ as complexes. (For a reference for the improved version of the theorem, see EGAIII, theorem 4.1.5.)

KReiser
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  • Generic fiber $\neq$ the domain, of course! Thank you, now I see my mistake! – Jędrzej Garnek Aug 01 '19 at 10:03
  • It's definitely not true that $X_n$ is a projective space over $\mathcal O_Y/\mathfrak m_y^n$. It's not even flat for $n>1$ when blowing up a point $y$ on a smooth surface $Y$. For example, the short exact sequence $$0\to\mathfrak m_y/\mathfrak m_y^2\to\mathcal O_{Y,y}/\mathfrak m_y^2\to\mathcal O_{Y,y}/\mathfrak m_y\to 0$$ does not give a short exact sequence $$0\to f^*(\mathfrak m_y/\mathfrak m_y^2)\to\mathcal O_{X_2}\to\mathcal O_E\to 0,$$ since it is not true that $\mathscr I_E/\mathscr I_E^2\cong\mathcal O_E^{\oplus 3}$ (in fact, $\mathscr I_E/\mathscr I_E^2\cong\mathcal O_E(-E)$). – Remy Aug 20 '19 at 21:28
  • A more conceptual reason why the $X_n \to \operatorname{Spec} \mathcal O_Y/\mathfrak m_y^n$ cannot all be flat comes from the flattening stratification, cf. FGA Explained Thm. 5.13. If all $X_n \to \operatorname{Spec} \mathcal O_Y/\mathfrak m_y^n$ were flat, then there would be an open set $U \subseteq Y$ where $f^{-1}(U) \to U$ is flat, which is absurd. – Remy Aug 20 '19 at 21:35
  • My previous two comments may have arisen from confusion on my part about your notation (your usage of $Y$ in the theorem of formal functions is different from that of the OP). In the notation of the OP, you have to say something about $\widetilde{X_n} = \widetilde{X} \times_X \operatorname{Spec} \mathcal O_{X,x}/\mathfrak m_x^n$. I agree that the base change $\widetilde{X} \times_X \operatorname{Spec} \mathcal O_{Y,x}/\mathfrak m_x^n$ is a projective space over $\operatorname{Spec} \mathcal O_{Y,x}/\mathfrak m_x^n$, but this is not the question. – Remy Aug 20 '19 at 22:03
  • It is still not true that $\widetilde{X_n}$ is a projective space over $\mathcal O_{Y,x}/\mathfrak m_x^n$ or over $\mathcal O_{X,x}/\mathfrak m_x^n$, as can be seen in the case of blowing up a point. (In the first case it would be reduced, and the second is covered by my first two comments.) – Remy Aug 20 '19 at 22:03
  • I believe that in order to fix @KReiser 's argument, one needs to note that the fiber $\widetilde X_1 = \widetilde X \times \textrm{Spec } \mathcal O_{X, x}/\mathfrak m_{X, x}$ is a projective space, and then proceed by induction, using the exact sequence: $0 \to I^n/I^{n+1} \to \mathcal O_{X_{n+1}} \to \mathcal O_{X_{n}} \to 0$ (this is what Hartshorne does in case when $Y = pt$). Does this sound right to you? – Jędrzej Garnek Aug 21 '19 at 09:41
  • @Remy you are correct, my apologies for the error. As to the OP's latest comment, I think that fix should work but I have not checked it all the way through. – KReiser Aug 25 '19 at 00:22