# Direct image of structure sheaf under blow-up along non-singular subvariety

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?

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.

• Generic fiber $\neq$ the domain, of course! Thank you, now I see my mistake! Aug 1, 2019 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, 2019 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, 2019 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, 2019 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, 2019 at 22:03