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?


1 Answer 1


This is largely correct, except you've made a key misunderstanding in your application of the theorem on formal functions.

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 $i\geq 0$.

Since $R^if_*(\mathcal{F})_y^\wedge=0$ iff $R^if_*(\mathcal{F})_y=0$ and $\mathcal{F}=0$ iff $\mathcal{F}_y=0$ for all $y\in Y$, it suffices to show that $R^if_*(\mathcal{F})_y^\wedge=0$ for all $y\in Y$ in order to show that $R^if_*(\mathcal{F})=0$ (we already know that $R^if_*(\mathcal{F})_x=0$ for all $x\in X\setminus Y$ since $f$ is an isomorphism there). So that's the strategy we'll pursue.

The error in your argument is twofold: instead of choosing $y$ to be the generic point of $Y$, one should let $y\in Y$ be arbitrary; secondly, $\widetilde{X}_n$ is defined to be the preimage of the $n^{th}$ thickening of the fiber over whatever $y$ you pick, which is not the $n^{th}$ thickening of $E$ - this isn't true at any point, much less the generic point (think about blowing up $\operatorname{Spec} k[x]\subset \Bbb A^3$: the fiber over the generic point is a copy of $\Bbb P^1_{k(x)}$ which is definitely not the same as $\Bbb P^1_k \times \Bbb A^1_k$, so it's not even true for $E_1$).

Once you fix this, you should be able to see that the space $X_n$ over any point $y\in Y$ is a projective space over the ring $\mathcal{O}_Y/\mathfrak{m}_y^n$ and correctly conclude that the higher cohomology over this space is zero, which implies the result you're after as per the discussion immediately following the theorem.

Edit: The old concluding paragraph was wrong, as pointed out by Remy in the comments. Here's a version that's correct, taking after Hartshorne proposition V.3.4.

Pick $y\in Y\subset X$. Let $E_n:= \widetilde{X} \times_X \operatorname{Spec} \mathcal{O}_{X,y}/\mathfrak{m}_y^n$. We see that $E_1$ is a projective space cut out by a sheaf of ideals $\mathcal{I}$, and that we have natural exact sequences $$ 0\to \mathcal{I}^n/\mathcal{I}^{n+1} \to \mathcal{O}_{E_{n+1}} \to \mathcal{O}_{E_n} \to 0$$ for all $n$. Noting that $\mathcal{I}^{d}/\mathcal{I}^{d+1}=\mathcal{O}_E(d)$, we see that $H^i(E,\mathcal{O}_E(d))=0$ for $i>0$ and $d\geq 0$, which implies $R^if_*\mathcal{O}_E(d)=0$ and $R^if_*\mathcal{O}_E=0$ for all $i>0$ and $d>0$.

Taking the long exact sequence $$ 0\to R^0f_*(\mathcal{I}^n/\mathcal{I}^{n+1}) \to R^0f_*\mathcal{O}_{E_{n+1}} \to R^0f_*\mathcal{O}_{E_n} \to R^1f_*(\mathcal{I}^n/\mathcal{I}^{n+1}) \to \cdots $$ we see that we may conclude that $R^if_*\mathcal{O}_{E_n}=0$ for all $i,n>0$ by induction, which finishes the proof.

  • $\begingroup$ Generic fiber $\neq$ the domain, of course! Thank you, now I see my mistake! $\endgroup$ Aug 1, 2019 at 10:03
  • $\begingroup$ 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, $\mathscr I_E/\mathscr I_E^2\cong\mathcal O_E(-E)$). $\endgroup$
    – Remy
    Aug 20, 2019 at 21:28
  • $\begingroup$ 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. $\endgroup$
    – Remy
    Aug 20, 2019 at 21:35
  • $\begingroup$ 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. $\endgroup$
    – Remy
    Aug 20, 2019 at 22:03
  • $\begingroup$ 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.) $\endgroup$
    – Remy
    Aug 20, 2019 at 22:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.