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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?

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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.

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  • $\begingroup$ 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

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