The Stacks project

Proof. We will check hypotheses (a), (b), and (c) of Lemma 52.16.10. Before we start, let us point out that the modules $H^0(U, \mathcal{F}''_ n)$ and $H^1(U, \mathcal{F}''_ n)$ are finite $A$-modules for all $n$ by Local Cohomology, Lemma 51.12.1.

Observe that for each $p \geq 0$ the limit topology on $\mathop{\mathrm{lim}}\nolimits H^ p(U, \mathcal{F}_ n)$ is the $I$-adic topology by Lemma 52.4.5. In particular, hypothesis (b) holds.

We know that $M = \mathop{\mathrm{lim}}\nolimits H^0(U, \mathcal{F}_ n)$ is an $A$-module whose limit topology is the $I$-adic topology. Thus, given $n$, the module $M/I^ nM$ is a subquotient of $H^0(U, \mathcal{F}_ N)$ for some $N \gg n$. Since the inverse system $\{ H^0(U, \mathcal{F}_ N)\} $ is pro-isomorphic to an inverse system of finite $A$-modules, namely $\{ H^0(U, \mathcal{F}''_ N)\} $, we conclude that $M/I^ nM$ is finite. It follows that $M$ is finite, see Algebra, Lemma 10.96.12. In particular hypothesis (c) holds.

For each $n \geq 0$ let us write $Ob_ n = \mathop{\mathrm{lim}}\nolimits _ N H^1(U, I^ n\mathcal{F}_ N)$. A special case is $Ob = Ob_0 = \mathop{\mathrm{lim}}\nolimits _ N H^1(U, \mathcal{F}_ N)$. Arguing exactly as in the previous paragraph we find that $Ob$ is a finite $A$-module. (In fact, we also know that $Ob/I Ob$ is annihilated by a power of $\mathfrak a$, but it seems somewhat difficult to use this.)

We set $\mathcal{F} = \mathop{\mathrm{lim}}\nolimits \mathcal{F}_ n$, we pick generators $f_1, \ldots , f_ r$ of $I$, we pick $c \geq 1$, and we choose $\Phi _\mathcal {F}$ as in Lemma 52.4.4. We will use the results of Lemma 52.2.1 without further mention. In particular, for each $n \geq 1$ there are maps

The first comes from the short exact sequence $0 \to I^ n\mathcal{F} \to \mathcal{F} \to \mathcal{F}_ n \to 0$ and the second from $I^ n\mathcal{F} = \mathop{\mathrm{lim}}\nolimits I^ n\mathcal{F}_ N$. We will later use that if $\delta _ n(s) = 0$ for $s \in H^0(U, \mathcal{F}_ n)$ then we can for each $n' \geq n$ find $s' \in H^0(U, \mathcal{F}_{n'})$ mapping to $s$. Observe that there are commutative diagrams

We conclude that the obstruction map $H^0(U, \mathcal{F}_ n) \to Ob_ n$ sends the image of $H^0(U, \mathcal{F}_{nc}) \to H^0(U, \mathcal{F}_ n)$ into the submodule

where on the summand $Ob \cdot T_1^{e_1} \ldots T_ r^{e_ r}$ we use the map on cohomology coming from the reductions modulo powers of $I$ of the multiplication map $f_1^{e_1} \ldots f_ r^{e_ r} : \mathcal{F} \to I^ n\mathcal{F}$. By construction

is a finite graded module over the Rees algebra $\bigoplus _{n \geq 0} I^ n$. For each $n$ we set

Observe that $\{ M_ n\} $ is an inverse system and that $f_ j : \mathcal{F}_ n \to \mathcal{F}_{n + 1}$ on global sections maps $M_ n$ into $M_{n + 1}$. By exactly the same argument as in the proof of Cohomology, Lemma 20.35.1 we find that $\{ M_ n\} $ is ML. Namely, because the Rees algebra is Noetherian we can choose a finite number of homogeneous generators of the form $\delta _{n_ j}(z_ j)$ with $z_ j \in M_{n_ j}$ for the graded submodule $\bigoplus _{n \geq 0} \mathop{\mathrm{Im}}(M_ n \to Ob'_ n)$. Then if $k = \max (n_ j)$ we find that for $n \geq k$ and any $z \in M_ n$ we can find $a_ j \in I^{n - n_ j}$ such that $z - \sum a_ j z_ j$ is in the kernel of $\delta _ n$ and hence in the image of $M_{n'}$ for all $n' \geq n$ (because the vanishing of $\delta _ n$ means that we can lift $z - \sum a_ j z_ j$ to an element $z' \in H^0(U, \mathcal{F}_{n'c})$ for all $n' \ge n$ and then the image of $z'$ in $H^0(U, \mathcal{F}_{n'})$ is in $M_{n'}$ by what we proved above). Thus $\mathop{\mathrm{Im}}(M_ n \to M_{n - k}) = \mathop{\mathrm{Im}}(M_{n'} \to M_{n - k})$ for all $n' \geq n$.

Choose $n$. By the Mittag-Leffler property of $\{ M_ n\} $ we just established we can find an $n' \geq n$ such that the image of $M_{n'} \to M_ n$ is the same as the image of $M' \to M_ n$. By the above we see that the image of $M' \to M_ n$ contains the image of $H^0(U, \mathcal{F}_{n'c}) \to H^0(U, \mathcal{F}_ n)$. Thus we see that $\{ M_ n\} $ and $\{ H^0(U, \mathcal{F}_ n)\} $ are pro-isomorphic. Therefore $\{ H^0(U, \mathcal{F}_ n)\} $ has ML and we finally conclude that hypothesis (a) holds. This concludes the proof. $\square$

Proof. By Lemma 52.17.1 we may replace $(\mathcal{F}_ n)$ by the object $(\mathcal{H}_ n)$ of $\textit{Coh}(U, I\mathcal{O}_ U)$ found in Lemma 52.21.3. Thus we may assume that $(\mathcal{F}_ n)$ is pro-isomorphic to a inverse system $(\mathcal{F}_ n'')$ with the properties mentioned in Lemma 52.21.3. In Lemma 52.22.1 we proved that $(\mathcal{F}_ n)$ canonically extends to $X$. The final statement follows from Lemma 52.16.8. $\square$

Proof. Let $Y' \subset X'$ be the exceptional divisor. Let $Z' \subset Y'$ be the inverse image of $Z \subset Y$. Then $U' = X' \setminus Z'$ is the inverse image of $U$. With $\delta ^{Y'}_{Z'}$ as in (52.18.0.1) we set

These are specialization stable subsets of $U' \cap Y' = Y' \setminus Z'$. Consider the object $(b|_{U'}^*\mathcal{F}_ n)$ of $\textit{Coh}(U', I\mathcal{O}_{U'})$, see Cohomology of Schemes, Lemma 30.23.9. For $y' \in U' \cap Y'$ let us denote

the “stalk” of this pullback at $y'$. We claim that conditions (a), (b), (c), (d), and (e) of Lemma 52.21.2 hold for the object $(b|_{U'}^*\mathcal{F}_ n)$ on $U'$ with $d$ replaced by $1$ and the subsets $T' \subset T \subset U' \cap Y'$. Condition (a) holds because $Y'$ is an effective Cartier divisor and hence locally cut out by $1$ equation. Condition (e) holds by Lemma 52.18.1 parts (1) and (2). To prove (b), (c), and (d) we need some preparation.

Let $y' \in U' \cap Y'$ and let $\mathfrak p' \subset \mathcal{O}_{X', y'}^\wedge $ be a prime ideal not contained in $V(I\mathcal{O}_{X', y'}^\wedge )$. Denote $y = b(y') \in U \cap Y$. Choose $f \in I$ such that $y'$ is contained in the spectrum of the affine blowup algebra $A[\frac{I}{f}]$, see Divisors, Lemma 31.32.2. For any $A$-algebra $B$ denote $B' = B[\frac{IB}{f}]$ the corresponding affine blowup algebra. Denote $I$-adic completion by ${\ }^\wedge $. By our choice of $f$ we get a ring map $(\mathcal{O}_{X, y}^\wedge )' \to \mathcal{O}_{X', y'}^\wedge $. If we let $\mathfrak q' \subset (\mathcal{O}_{X, y}^\wedge )'$ be the inverse image of $\mathfrak m_{y'}^\wedge $, then we see that $((\mathcal{O}_{X, y}^\wedge )'_{\mathfrak q'})^\wedge = \mathcal{O}_{X', y'}^\wedge $. Let $\mathfrak p \subset \mathcal{O}_{X, y}^\wedge $ be the corresponding prime. At this point we have a commutative diagram

whose vertical arrows are surjective. By More on Algebra, Lemma 15.43.1 and the dimension formula (Algebra, Lemma 10.113.1) we have

Tracing through the definitions of pullbacks, stalks, localizations, and completions we find

Details omitted. The ring maps $\beta $ and $\gamma $ in the diagram are flat with Gorenstein (hence Cohen-Macaulay) fibres, as these are completions of rings having a dualizing complex. See Dualizing Complexes, Lemmas 47.23.1 and 47.23.2 and the discussion in More on Algebra, Section 15.51. Observe that $(\mathcal{O}_{X, y}^\wedge )_\mathfrak p = (\mathcal{O}_{X, y}^\wedge )'_{\tilde{\mathfrak p}}$ where $\tilde{\mathfrak p}$ is the kernel of $\alpha $ in the diagram. On the other hand, $(\mathcal{O}_{X, y}^\wedge )'_{\tilde{\mathfrak p}} \to (\mathcal{O}_{X', y'}^\wedge )_{\mathfrak p'}$ is flat with CM fibres by the above. Whence $(\mathcal{O}_{X, y}^\wedge )_\mathfrak p \to (\mathcal{O}_{X', y'}^\wedge )_{\mathfrak p'}$ is flat with CM fibres. Using Algebra, Lemma 10.163.1 we see that

where $F$ is the generic formal fibre of $(\mathcal{O}_{X, y}^\wedge /\mathfrak p)'_{\mathfrak q'}$ and $\mathfrak r$ is the prime corresponding to $\mathfrak p'$. Since $(\mathcal{O}_{X, y}^\wedge /\mathfrak p)'_{\mathfrak q'}$ is a universally catenary local domain, its $I$-adic completion is equidimensional and (universally) catenary by Ratliff's theorem (More on Algebra, Proposition 15.109.5). It then follows that

Combined with Lemma 52.18.2 we get

Please keep in mind that $\dim (\mathcal{O}_{X, y}^\wedge /\mathfrak p) \geq \dim (\mathcal{O}_{X', y'}^\wedge /\mathfrak p')$. Rewriting this we get

This inequality will allow us to check the remaning conditions.

Conditions (b) and (d) of Lemma 52.21.2. Assume $V(\mathfrak p') \cap V(I\mathcal{O}_{X', y'}^\wedge ) = \{ \mathfrak m_{y'}^\wedge \} $. This implies that $\dim (\mathcal{O}_{X', y'}^\wedge /\mathfrak p') = 1$ because $Z'$ is an effective Cartier divisor. The combination of (b) and (d) is equivalent with

If $(\mathcal{F}_ n)$ satisfies the inequalities in (3)(b) then we immediately conclude this is true by applying (52.22.3.6). If $(\mathcal{F}_ n)$ satisfies (3)(a), i.e., the $(d + 1, d + 2)$-inequalities, then we see that in any case

Looking at (52.22.3.1) and (52.22.3.6) above this gives what we want except possibly if $\dim (\mathcal{O}_{X, y}^\wedge /\mathfrak p) = 1$. However, if $\dim (\mathcal{O}_{X, y}^\wedge /\mathfrak p) = 1$, then we have $V(\mathfrak p) \cap V(I\mathcal{O}_{X, y}^\wedge ) = \{ \mathfrak m_ y^\wedge \} $ and we see that actually

as $(\mathcal{F}_ n)$ satisfies the $(d + 1, d + 2)$-inequalities and we conclude again.

Condition (c) of Lemma 52.21.2. Assume $V(\mathfrak p') \cap V(I\mathcal{O}_{X', y'}^\wedge ) \not= \{ \mathfrak m_{y'}^\wedge \} $. Then condition (c) is equivalent to

If $(\mathcal{F}_ n)$ satisfies the inequalities in (3)(b) then we see the second of the two displayed inequalities holds true by applying (52.22.3.6). If $(\mathcal{F}_ n)$ satisfies (3)(a), i.e., the $(d + 1, d + 2)$-inequalities, then this follows immediately from (52.22.3.1) and (52.22.3.6). This finishes the proof of our claim.

Choose $(b|_{U'}^*\mathcal{F}_ n) \to (\mathcal{F}_ n'')$ and $(\mathcal{H}_ n)$ in $\textit{Coh}(U', I\mathcal{O}_{U'})$ as in Lemma 52.21.2. For any affine open $W \subset X'$ observe that $\delta ^{W \cap Y'}_{W \cap Z'}(y') \geq \delta ^{Y'}_{Z'}(y')$ by Lemma 52.18.1 part (7). Hence we see that $(\mathcal{H}_ n|_ W)$ satisfies the assumptions of Lemma 52.22.1. Thus $(\mathcal{H}_ n|_ W)$ extends canonically to $W$. Let $(\mathcal{G}_{W, n})$ in $\textit{Coh}(W, I\mathcal{O}_ W)$ be the canonical extension as in Lemma 52.16.8. By Lemma 52.16.9 we see that for $W' \subset W$ there is a unique isomorphism

compatible with the given isomorphisms $(\mathcal{G}_{W, n}|_{W \cap U}) \cong (\mathcal{H}_ n|_{W \cap U})$. We conclude that there exists an object $(\mathcal{G}_ n)$ of $\textit{Coh}(X', I\mathcal{O}_{X'})$ whose restriction to $U$ is isomorphic to $(\mathcal{H}_ n)$.

If $A$ is $I$-adically complete we can finish the proof as follows. By Grothedieck's existence theorem (Cohomology of Schemes, Lemma 30.24.3) we see that $(\mathcal{G}_ n)$ is the completion of a coherent $\mathcal{O}_{X'}$-module. Then by Cohomology of Schemes, Lemma 30.23.6 we see that $(b|_{U'}^*\mathcal{F}_ n)$ is the completion of a coherent $\mathcal{O}_{U'}$-module $\mathcal{F}'$. By Cohomology of Schemes, Lemma 30.25.3 we see that there is a map

whose kernel and cokernel is annihilated by a power of $I$. Then finally, we win by applying Lemma 52.17.1.

If $A$ is not complete, then, before starting the proof, we may replace $A$ by its completion, see Lemma 52.16.6. After completion the assumptions still hold: this is immediate for condition (3), follows from Dualizing Complexes, Lemma 47.22.4 for condition (1), and from Divisors, Lemma 31.32.3 for condition (2). Thus the complete case implies the general case. $\square$

Proof. We may assume $I = (f_1, \ldots , f_ d)$, see Cohomology of Schemes, Lemma 30.23.11. Then we see that all fibres of the blowup of $X$ in $I$ have dimension at most $d - 1$. Thus we get the extension from Lemma 52.22.3. The final statement follows from Lemma 52.16.3. $\square$

Proof. We will show that the hypotheses (1), (2), (3)(b) of Proposition 52.22.4 are satisfied. This is clear for (1) and (2).

Let $y \in U \cap Y$ and let $\mathfrak p$ be a prime $\mathfrak p \subset \mathcal{O}_{X, y}^\wedge $ with $\mathfrak p \not\in V(I\mathcal{O}_{X, y}^\wedge )$. The last condition shows that $\text{depth}((\mathcal{F}_ y^\wedge )_\mathfrak p) = \text{depth}((\mathcal{O}_{X, y}^\wedge )_\mathfrak p)$. Since $X \setminus Y$ is Cohen-Macaulay we see that $(\mathcal{O}_{X, y}^\wedge )_\mathfrak p$ is Cohen-Macaulay. Thus we see that

The final equality because $\mathcal{O}_{X, y}$ is equidimensional by the second condition. Let $\delta (y) = \dim (\overline{\{ y\} })$. This is a dimension function as $A$ is a catenary local ring. By Lemma 52.18.1 we have $\delta ^ Y_ Z(y) \geq \delta (y) - \dim (Z)$. Since $X$ is equidimensional we get

Thus we get the desired inequality and we win. $\square$