## 52.12 Algebraization of formal sections, II

It is a bit difficult to succintly state all possible consequences of the results in Sections 52.8 and 52.10 for cohomology of coherent sheaves on quasi-affine schemes and their completion with respect to an ideal. This section gives a nonexhaustive list of applications to $H^0$. The next section contains applications to higher cohomology.

The following lemma will be superceded by Proposition 52.12.2.

Lemma 52.12.1. Let $I \subset \mathfrak a$ be ideals of a Noetherian ring $A$. Let $\mathcal{F}$ be a coherent module on $U = \mathop{\mathrm{Spec}}(A) \setminus V(\mathfrak a)$. Assume

$A$ is $I$-adically complete and has a dualizing complex,

if $x \in \text{Ass}(\mathcal{F})$, $x \not\in V(I)$, $\overline{\{ x\} } \cap V(I) \not\subset V(\mathfrak a)$ and $z \in \overline{\{ x\} } \cap V(\mathfrak a)$, then $\dim (\mathcal{O}_{\overline{\{ x\} }, z}) > \text{cd}(A, I) + 1$,

one of the following holds:

the restriction of $\mathcal{F}$ to $U \setminus V(I)$ is $(S_1)$

the dimension of $V(\mathfrak a)$ is at most $2$^{1}.

Then we obtain an isomorphism

\[ \mathop{\mathrm{colim}}\nolimits H^0(V, \mathcal{F}) \longrightarrow \mathop{\mathrm{lim}}\nolimits H^0(U, \mathcal{F}/I^ n\mathcal{F}) \]

where the colimit is over opens $V \subset U$ containing $U \cap V(I)$.

**Proof.**
Choose a finite $A$-module $M$ such that $\mathcal{F}$ is the restriction to $U$ of the coherent module associated to $M$, see Local Cohomology, Lemma 51.8.2. Set $d = \text{cd}(A, I)$. Let $\mathfrak p$ be a prime of $A$ not contained in $V(I)$ and let $\mathfrak q \in V(\mathfrak p) \cap V(\mathfrak a)$. Then either $\mathfrak p$ is not an associated prime of $M$ and hence $\text{depth}(M_\mathfrak p) \geq 1$ or we have $\dim ((A/\mathfrak p)_\mathfrak q) > d + 1$ by (2). Thus the hypotheses of Lemma 52.8.5 are satisfied for $s = 1$ and $d$; here we use condition (3). Thus we find there exists an ideal $J_0 \subset \mathfrak a$ with $V(J_0) \cap V(I) = V(\mathfrak a)$ such that for any $J \subset J_0$ with $V(J) \cap V(I) = V(\mathfrak a)$ the maps

\[ H^ i_ J(M) \longrightarrow H^ i(R\Gamma _\mathfrak a(M)^\wedge ) \]

are isomorphisms for $i = 0, 1$. Consider the morphisms of exact triangles

\[ \xymatrix{ R\Gamma _ J(M) \ar[d] \ar[r] & M \ar[r] \ar[d] & R\Gamma (V, \mathcal{F}) \ar[d] \ar[r] & R\Gamma _ J(M)[1] \ar[d] \\ R\Gamma _ J(M)^\wedge \ar[r] & M \ar[r] & R\Gamma (V, \mathcal{F})^\wedge \ar[r] & R\Gamma _ J(M)^\wedge [1] \\ R\Gamma _\mathfrak a(M)^\wedge \ar[r] \ar[u] & M \ar[r] \ar[u] & R\Gamma (U, \mathcal{F})^\wedge \ar[r] \ar[u] & R\Gamma _\mathfrak a(M)^\wedge [1] \ar[u] } \]

where $V = \mathop{\mathrm{Spec}}(A) \setminus V(J)$. Recall that $R\Gamma _\mathfrak a(M)^\wedge \to R\Gamma _ J(M)^\wedge $ is an isomorphism (because $\mathfrak a$, $\mathfrak a + I$, and $J + I$ cut out the same closed subscheme, for example see proof of Lemma 52.8.5). Hence $R\Gamma (U, \mathcal{F})^\wedge = R\Gamma (V, \mathcal{F})^\wedge $. This produces a commutative diagram

\[ \xymatrix{ 0 \ar[r] & H^0_ J(M) \ar[r] \ar[d] & M \ar[r] \ar[d] \ar[r] & \Gamma (V, \mathcal{F}) \ar[d] \ar[r] & H^1_ J(M) \ar[d] \ar[r] & 0 \\ 0 \ar[r] & H^0(R\Gamma _ J(M)^\wedge ) \ar[r] & M \ar[r] & H^0(R\Gamma (V, \mathcal{F})^\wedge ) \ar[r] & H^1(R\Gamma _ J(M)^\wedge ) \ar[r] & 0 \\ 0 \ar[r] & H^0(R\Gamma _\mathfrak a(M)^\wedge ) \ar[r] \ar[u] & M \ar[r] \ar[u] & H^0(R\Gamma (U, \mathcal{F})^\wedge ) \ar[r] \ar[u] & H^1(R\Gamma _\mathfrak a(M)^\wedge ) \ar[r] \ar[u] & 0 } \]

with exact rows and isomorphisms for the lower vertical arrows. Hence we obtain an isomorphism $\Gamma (V, \mathcal{F}) \to H^0(R\Gamma (U, \mathcal{F})^\wedge )$. By Lemmas 52.6.20 and 52.7.2 we have

\[ R\Gamma (U, \mathcal{F})^\wedge = R\Gamma (U, \mathcal{F}^\wedge ) = R\Gamma (U, R\mathop{\mathrm{lim}}\nolimits \mathcal{F}/I^ n\mathcal{F}) \]

and we find $H^0(R\Gamma (U, \mathcal{F})^\wedge ) = \mathop{\mathrm{lim}}\nolimits H^0(U, \mathcal{F}/I^ n\mathcal{F})$ by Cohomology, Lemma 20.35.1.
$\square$

Now we bootstrap the preceding lemma to get rid of condition (3).

Proposition 52.12.2. Let $I \subset \mathfrak a$ be ideals of a Noetherian ring $A$. Let $\mathcal{F}$ be a coherent module on $U = \mathop{\mathrm{Spec}}(A) \setminus V(\mathfrak a)$. Assume

$A$ is $I$-adically complete and has a dualizing complex,

if $x \in \text{Ass}(\mathcal{F})$, $x \not\in V(I)$, $\overline{\{ x\} } \cap V(I) \not\subset V(\mathfrak a)$ and $z \in \overline{\{ x\} } \cap V(\mathfrak a)$, then $\dim (\mathcal{O}_{\overline{\{ x\} }, z}) > \text{cd}(A, I) + 1$.

Then we obtain an isomorphism

\[ \mathop{\mathrm{colim}}\nolimits H^0(V, \mathcal{F}) \longrightarrow \mathop{\mathrm{lim}}\nolimits H^0(U, \mathcal{F}/I^ n\mathcal{F}) \]

where the colimit is over opens $V \subset U$ containing $U \cap V(I)$.

**Proof.**
Let $T \subset U$ be the set of points $x$ with $\overline{\{ x\} } \cap V(I) \subset V(\mathfrak a)$. Let $\mathcal{F} \to \mathcal{F}'$ be the surjection of coherent modules on $U$ constructed in Local Cohomology, Lemma 51.15.1. Since $\mathcal{F} \to \mathcal{F}'$ is an isomorphism over an open $V \subset U$ containing $U \cap V(I)$ it suffices to prove the lemma with $\mathcal{F}$ replaced by $\mathcal{F}'$. Hence we may and do assume for $x \in U$ with $\overline{\{ x\} } \cap V(I) \subset V(\mathfrak a)$ we have $\text{depth}(\mathcal{F}_ x) \geq 1$.

Let $\mathcal{V}$ be the set of open subschemes $V \subset U$ containing $U \cap V(I)$ ordered by reverse inclusion. This is a directed set. We first claim that

\[ \mathcal{F}(V) \longrightarrow \mathop{\mathrm{lim}}\nolimits H^0(U, \mathcal{F}/I^ n\mathcal{F}) \]

is injective for any $V \in \mathcal{F}$ (and in particular the map of the lemma is injective). Namely, an associated point $x$ of $\mathcal{F}$ must have $\overline{\{ x\} } \cap U \cap Y \not= \emptyset $ by the previous paragraph. If $y \in \overline{\{ x\} } \cap U \cap Y$ then $\mathcal{F}_ x$ is a localization of $\mathcal{F}_ y$ and $\mathcal{F}_ y \subset \mathop{\mathrm{lim}}\nolimits \mathcal{F}_ y/I^ n \mathcal{F}_ y$ by Krull's intersection theorem (Algebra, Lemma 10.51.4). This proves the claim as a section $s \in \mathcal{F}(V)$ in the kernel would have to have empty support, hence would have to be zero.

Choose a finite $A$-module $M$ such that $\mathcal{F}$ is the restriction of $\widetilde{M}$ to $U$, see Local Cohomology, Lemma 51.8.2. We may and do assume that $H^0_\mathfrak a(M) = 0$. Let $\text{Ass}(M) \setminus V(I) = \{ \mathfrak p_1, \ldots , \mathfrak p_ n\} $. We will prove the lemma by induction on $n$. After reordering we may assume that $\mathfrak p_ n$ is a minimal element of the set $\{ \mathfrak p_1, \ldots , \mathfrak p_ n\} $ with respect to inclusion, i.e, $\mathfrak p_ n$ is a generic point of the support of $M$. Set

\[ M' = H^0_{\mathfrak p_1 \ldots \mathfrak p_{n - 1} I}(M) \]

and $M'' = M/M'$. Let $\mathcal{F}'$ and $\mathcal{F}''$ be the coherent $\mathcal{O}_ U$-modules corresponding to $M'$ and $M''$. Dualizing Complexes, Lemma 47.11.6 implies that $M''$ has only one associated prime, namely $\mathfrak p_ n$. On the other hand, since $\mathfrak p_ n \not\in V(\mathfrak p_1 \ldots \mathfrak p_{n - 1} I)$ we see that $\mathfrak p_ n$ is not an associated prime of $M'$. Hence the induction hypothesis applies to $M'$; note that since $\mathcal{F}' \subset \mathcal{F}$ the condition $\text{depth}(\mathcal{F}'_ x) \geq 1$ at points $x$ with $\overline{\{ x\} } \cap V(I) \subset V(\mathfrak a)$ holds, see Algebra, Lemma 10.72.6.

Let $\hat s$ be an element of $\mathop{\mathrm{lim}}\nolimits H^0(U, \mathcal{F}/I^ n\mathcal{F})$. Let $\hat s''$ be the image in $\mathop{\mathrm{lim}}\nolimits H^0(U, \mathcal{F}''/I^ n\mathcal{F}'')$. Since $\mathcal{F}''$ has only one associated point, namely the point corresponding to $\mathfrak p_ n$, we see that Lemma 52.12.1 applies and we find an open $U \cap V(I) \subset V \subset U$ and a section $s'' \in \mathcal{F}''(V)$ mapping to $\hat s''$. Let $J \subset A$ be an ideal such that $V(J) = \mathop{\mathrm{Spec}}(A) \setminus V$. By Cohomology of Schemes, Lemma 30.10.5 after replacing $J$ by a power, we may assume there is an $A$-linear map $\varphi : J \to M''$ corresponding to $s''$. Since $M \to M''$ is surjective, for each $g \in J$ we can choose $m_ g \in M$ mapping to $\varphi (g) \in M''$. Then $\hat s'_ g = g \hat s - m_ g$ is in $\mathop{\mathrm{lim}}\nolimits H^0(U, \mathcal{F}'/I^ n\mathcal{F}')$. By induction hypothesis there is a $V' \geq V$ section $s'_ g \in \mathcal{F}'(V')$ mapping to $\hat s'_ g$. All in all we conclude that $g \hat s$ is in the image of $\mathcal{F}(V') \to \mathop{\mathrm{lim}}\nolimits H^0(U, \mathcal{F}/I^ n\mathcal{F})$ for some $V' \subset V$ possibly depending on $g$. However, since $J$ is finitely generated we can find a single $V' \in \mathcal{V}$ which works for each of the generators and it follows that $V'$ works for all $g$.

Combining the previous paragraph with the injectivity shown in the second paragraph we find there exists a $V' \geq V$ and an $A$-module map $\psi : J \to \mathcal{F}(V')$ such that $\psi (g)$ maps to $g\hat s$. This determines a map $\widetilde{J} \to (V' \to \mathop{\mathrm{Spec}}(A))_*\mathcal{F}|_{V'}$ whose restriction to $V'$ provides an element $s \in \mathcal{F}(V')$ mapping to $\hat s$. This finishes the proof.
$\square$

Lemma 52.12.3. Let $I \subset \mathfrak a$ be ideals of a Noetherian ring $A$. Let $\mathcal{F}$ be a coherent module on $U = \mathop{\mathrm{Spec}}(A) \setminus V(\mathfrak a)$. Assume

$A$ is $I$-adically complete and has a dualizing complex,

if $x \in \text{Ass}(\mathcal{F})$, $x \not\in V(I)$, $z \in V(\mathfrak a) \cap \overline{\{ x\} }$, then $\dim (\mathcal{O}_{\overline{\{ x\} }, z}) > \text{cd}(A, I) + 1$,

for $x \in U$ with $\overline{\{ x\} } \cap V(I) \subset V(\mathfrak a)$ we have $\text{depth}(\mathcal{F}_ x) \geq 2$,

Then we obtain an isomorphism

\[ H^0(U, \mathcal{F}) \longrightarrow \mathop{\mathrm{lim}}\nolimits H^0(U, \mathcal{F}/I^ n\mathcal{F}) \]

**Proof.**
Let $\hat s \in \mathop{\mathrm{lim}}\nolimits H^0(U, \mathcal{F}/I^ n\mathcal{F})$. By Proposition 52.12.2 we find that $\hat s$ is the image of an element $s \in \mathcal{F}(V)$ for some $V \subset U$ open containing $U \cap V(I)$. However, condition (3) shows that $\text{depth}(\mathcal{F}_ x) \geq 2$ for all $x \in U \setminus V$ and hence we find that $\mathcal{F}(V) = \mathcal{F}(U)$ by Divisors, Lemma 31.5.11 and the proof is complete.
$\square$

Example 52.12.4. Let $A$ be a Noetherian domain which has a dualizing complex and which is complete with respect to a nonzero $f \in A$. Let $f \in \mathfrak a \subset A$ be an ideal. Assume every irreducible component of $Z = V(\mathfrak a)$ has codimension $> 2$ in $X = \mathop{\mathrm{Spec}}(A)$. Equivalently, assume every irreducible component of $Z$ has codimension $> 1$ in $Y = V(f)$. Then with $U = X \setminus Z$ every element of

\[ \mathop{\mathrm{lim}}\nolimits _ n \Gamma (U, \mathcal{O}_ U/f^ n \mathcal{O}_ U) \]

is the restriction of a section of $\mathcal{O}_ U$ defined on an open neighbourhood of

\[ V(f) \setminus Z = V(f) \cap U = Y \setminus Z = U \cap Y \]

In particular we see that $Y \setminus Z$ is connected. See Lemma 52.14.2 below.

Lemma 52.12.5. Let $A$ be a Noetherian ring. Let $f \in \mathfrak a \subset A$ be an element of an ideal of $A$. Let $M$ be a finite $A$-module. Assume

$A$ is $f$-adically complete,

$f$ is a nonzerodivisor on $M$,

$H^1_\mathfrak a(M/fM)$ is a finite $A$-module.

Then with $U = \mathop{\mathrm{Spec}}(A) \setminus V(\mathfrak a)$ the map

\[ \mathop{\mathrm{colim}}\nolimits _ V \Gamma (V, \widetilde{M}) \longrightarrow \mathop{\mathrm{lim}}\nolimits \Gamma (U, \widetilde{M/f^ nM}) \]

is an isomorphism where the colimit is over opens $V \subset U$ containing $U \cap V(f)$.

**Proof.**
Set $\mathcal{F} = \widetilde{M}|_ U$. The finiteness of $H^1_\mathfrak a(M/fM)$ implies that $H^0(U, \mathcal{F}/f\mathcal{F})$ is finite, see Local Cohomology, Lemma 51.8.2. By Lemma 52.3.3 (which applies as $f$ is a nonzerodivisor on $\mathcal{F}$) we see that $N = \mathop{\mathrm{lim}}\nolimits H^0(U, \mathcal{F}/f^ n\mathcal{F})$ is a finite $A$-module, is $f$-torsion free, and $N/fN \subset H^0(U, \mathcal{F}/f\mathcal{F})$. On the other hand, we have $M \to N$ and the map

\[ M/fM \longrightarrow H^0(U, \mathcal{F}/f\mathcal{F}) \]

is an isomorphism upon localization at any prime $\mathfrak q$ in $U_0 = V(f) \setminus \{ \mathfrak m\} $ (details omitted). Thus $M_\mathfrak q \to N_\mathfrak q$ induces an isomorphism

\[ M_\mathfrak q/fM_\mathfrak q = (M/fM)_\mathfrak q \to (N/fN)_\mathfrak q = N_\mathfrak q/fN_\mathfrak q \]

Since $f$ is a nonzerodivisor on both $N$ and $M$ we conclude that $M_\mathfrak q \to N_\mathfrak q$ is an isomorphism (use Nakayama to see surjectivity). We conclude that $M$ and $N$ determine isomorphic coherent modules over an open $V$ as in the statement of the lemma. This finishes the proof.
$\square$

Lemma 52.12.6. Let $A$ be a Noetherian ring. Let $f \in \mathfrak a \subset A$ be an element of an ideal of $A$. Let $M$ be a finite $A$-module. Assume

$A$ is $f$-adically complete,

$H^1_\mathfrak a(M)$ and $H^2_\mathfrak a(M)$ are annihilated by a power of $f$.

Then with $U = \mathop{\mathrm{Spec}}(A) \setminus V(\mathfrak a)$ the map

\[ \Gamma (U, \widetilde{M}) \longrightarrow \mathop{\mathrm{lim}}\nolimits \Gamma (U, \widetilde{M/f^ nM}) \]

is an isomorphism.

**Proof.**
We may apply Lemma 52.3.6 to $U$ and $\mathcal{F} = \widetilde{M}|_ U$ because $\mathcal{F}$ is a Noetherian object in the category of coherent $\mathcal{O}_ U$-modules. Since $H^1(U, \mathcal{F}) = H^2_\mathfrak a(M)$ (Local Cohomology, Lemma 51.8.2) is annihilated by a power of $f$, we see that its $f$-adic Tate module is zero. Hence the lemma shows $\mathop{\mathrm{lim}}\nolimits H^0(U, \mathcal{F}/f^ n \mathcal{F})$ is the $0$th cohomology group of the derived $f$-adic completion of $H^0(U, \mathcal{F})$. Consider the exact sequence

\[ 0 \to H^0_\mathfrak a(M) \to M \to \Gamma (U, \mathcal{F}) \to H^1_\mathfrak a(M) \to 0 \]

of Local Cohomology, Lemma 51.8.2. Since $H^1_\mathfrak a(M)$ is annihilated by a power of $f$ it is derived complete with respect to $(f)$. Since $M$ and $H^0_\mathfrak a(M)$ are finite $A$-modules they are complete (Algebra, Lemma 10.97.1) hence derived complete (More on Algebra, Proposition 15.91.5). By More on Algebra, Lemma 15.91.6 we conclude that $\Gamma (U, \mathcal{F})$ is derived complete as desired.
$\square$

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