Fix a Noetherian scheme $X$, and a closed subscheme $Z$ with complement $U$. Our goal is to explain how coherent sheaves on $X$ can be constructed (uniquely) from coherent sheaves on the formal completion of $X$ along $Z$, and those on $U$ with a suitable compatibility on the overlap. We first do this using only commutative algebra (this section) and later we explain this in the setting of algebraic spaces (Pushouts of Spaces, Section 81.10).
Here are some references treating some of the material in this section: [Section 2, ArtinII], [Appendix, Ferrand-Raynaud], [Beauville-Laszlo], [MB], and [Section 4.6, dJ-crystalline].
Lemma 15.89.1. Let $\varphi : R \to S$ be a ring map. Let $I \subset R$ be an ideal. The following are equivalent
$\varphi $ is flat and $R/I \to S/IS$ is faithfully flat,
$\varphi $ is flat, and the map $\mathop{\mathrm{Spec}}(S/IS) \to \mathop{\mathrm{Spec}}(R/I)$ is surjective.
$\varphi $ is flat, and the base change functor $M \mapsto M \otimes _ R S$ is faithful on modules annihilated by $I$, and
$\varphi $ is flat, and the base change functor $M \mapsto M \otimes _ R S$ is faithful on $I$-power torsion modules.
Proof.
If $R \to S$ is flat, then $R/I^ n \to S/I^ nS$ is flat for every $n$, see Algebra, Lemma 10.39.7. Hence (1) and (2) are equivalent by Algebra, Lemma 10.39.16. The equivalence of (1) with (3) follows by identifying $I$-torsion $R$-modules with $R/I$-modules, using that
\[ M \otimes _ R S = M \otimes _{R/I} S/IS \]
for $R$-modules $M$ annihilated by $I$, and Algebra, Lemma 10.39.14. The implication (4) $\Rightarrow $ (3) is immediate. Assume (3). We have seen above that $R/I^ n \to S/I^ nS$ is flat, and by assumption it induces a surjection on spectra, as $\mathop{\mathrm{Spec}}(R/I^ n) = \mathop{\mathrm{Spec}}(R/I)$ and similarly for $S$. Hence the base change functor is faithful on modules annihilated by $I^ n$. Since any $I$-power torsion module $M$ is the union $M = \bigcup M_ n$ where $M_ n$ is annihilated by $I^ n$ we see that the base change functor is faithful on the category of all $I$-power torsion modules (as tensor product commutes with colimits).
$\square$
Lemma 15.89.2. Assume $(\varphi : R \to S, I)$ satisfies the equivalent conditions of Lemma 15.89.1. The following are equivalent
for any $I$-power torsion module $M$, the natural map $M \to M \otimes _ R S$ is an isomorphism, and
$R/I \to S/IS$ is an isomorphism.
Proof.
The implication (1) $\Rightarrow $ (2) is immediate. Assume (2). First assume that $M$ is annihilated by $I$. In this case, $M$ is an $R/I$-module. Hence, we have an isomorphism
\[ M \otimes _ R S = M \otimes _{R/I} S/IS = M \otimes _{R/I} R/I = M \]
proving the claim. Next we prove by induction that $M \to M \otimes _ R S$ is an isomorphism for any module $M$ is annihilated by $I^ n$. Assume the induction hypothesis holds for $n$ and assume $M$ is annihilated by $I^{n + 1}$. Then we have a short exact sequence
\[ 0 \to I^ nM \to M \to M/I^ nM \to 0 \]
and as $R \to S$ is flat this gives rise to a short exact sequence
\[ 0 \to I^ nM \otimes _ R S \to M \otimes _ R S \to M/I^ nM \otimes _ R S \to 0 \]
Using that the canonical map is an isomorphism for $M' = I^ nM$ and $M'' = M/I^ nM$ (by induction hypothesis) we conclude the same thing is true for $M$. Finally, suppose that $M$ is a general $I$-power torsion module. Then $M = \bigcup M_ n$ where $M_ n$ is annihilated by $I^ n$ and we conclude using that tensor products commute with colimits.
$\square$
Lemma 15.89.3. Assume $\varphi : R \to S$ is a flat ring map and $I \subset R$ is a finitely generated ideal such that $R/I \to S/IS$ is an isomorphism. Then
for any $R$-module $M$ the map $M \to M \otimes _ R S$ induces an isomorphism $M[I^\infty ] \to (M \otimes _ R S)[(IS)^\infty ]$ of $I$-power torsion submodules,
the natural map
\[ \mathop{\mathrm{Hom}}\nolimits _ R(M, N) \longrightarrow \mathop{\mathrm{Hom}}\nolimits _ S(M \otimes _ R S, N \otimes _ R S) \]
is an isomorphism if either $M$ or $N$ is $I$-power torsion, and
the base change functor $M \mapsto M \otimes _ R S$ defines an equivalence of categories between $I$-power torsion modules and $IS$-power torsion modules.
Proof.
Note that the equivalent conditions of both Lemma 15.89.1 and Lemma 15.89.2 are satisfied. We will use these without further mention. We first prove (1). Let $M$ be any $R$-module. Set $M' = M/M[I^\infty ]$ and consider the exact sequence
\[ 0 \to M[I^\infty ] \to M \to M' \to 0 \]
As $M[I^\infty ] = M[I^\infty ] \otimes _ R S$ we see that it suffices to show that $(M' \otimes _ R S)[(IS)^\infty ] = 0$. Write $I = (f_1, \ldots , f_ t)$. By Lemma 15.88.4 we see that $M'[I^\infty ] = 0$. Hence for every $n > 0$ the map
\[ M' \longrightarrow \bigoplus \nolimits _{i = 1, \ldots t} M', \quad x \longmapsto (f_1^ n x, \ldots , f_ t^ n x) \]
is injective. As $S$ is flat over $R$ also the corresponding map $M' \otimes _ R S \to \bigoplus _{i = 1, \ldots t} M' \otimes _ R S$ is injective. This means that $(M' \otimes _ R S)[I^ n] = 0$ as desired.
Next we prove (2). If $N$ is $I$-power torsion, then $N \otimes _ R S = N$ and the displayed map of (2) is an isomorphism by Algebra, Lemma 10.14.3. If $M$ is $I$-power torsion, then the image of any map $M \to N$ factors through $M[I^\infty ]$ and the image of any map $M \otimes _ R S \to N \otimes _ R S$ factors through $(N \otimes _ R S)[(IS)^\infty ]$. Hence in this case part (1) guarantees that we may replace $N$ by $N[I^\infty ]$ and the result follows from the case where $N$ is $I$-power torsion we just discussed.
Next we prove (3). The functor is fully faithful by (2). For essential surjectivity, we simply note that for any $IS$-power torsion $S$-module $N$, the natural map $N \otimes _ R S \to N$ is an isomorphism.
$\square$
Lemma 15.89.4. Assume $\varphi : R \to S$ is a flat ring map and $I \subset R$ is a finitely generated ideal such that $R/I \to S/IS$ is an isomorphism. For any $f_1, \ldots , f_ r \in R$ such that $V(f_1, \ldots , f_ r) = V(I)$
the map of Koszul complexes $K(R, f_1, \ldots , f_ r) \to K(S, f_1, \ldots , f_ r)$ is a quasi-isomorphism, and
The map of extended alternating Čech complexes
\[ \xymatrix{ R \to \prod _{i_0} R_{f_{i_0}} \to \prod _{i_0 < i_1} R_{f_{i_0}f_{i_1}} \to \ldots \to R_{f_1\ldots f_ r} \ar[d] \\ S \to \prod _{i_0} S_{f_{i_0}} \to \prod _{i_0 < i_1} S_{f_{i_0}f_{i_1}} \to \ldots \to S_{f_1\ldots f_ r} } \]
is a quasi-isomorphism.
Proof.
In both cases we have a complex $K_\bullet $ of $R$ modules and we want to show that $K_\bullet \to K_\bullet \otimes _ R S$ is a quasi-isomorphism. By Lemma 15.89.2 and the flatness of $R \to S$ this will hold as soon as all homology groups of $K$ are $I$-power torsion. This is true for the Koszul complex by Lemma 15.28.6 and for the extended alternating Čech complex by Lemma 15.29.5.
$\square$
Lemma 15.89.5. Let $R$ be a ring. Let $I = (f_1, \ldots , f_ n)$ be a finitely generated ideal of $R$. Let $M$ be the $R$-module generated by elements $e_1, \ldots , e_ n$ subject to the relations $f_ i e_ j - f_ j e_ i = 0$. There exists a short exact sequence
\[ 0 \to K \to M \to I \to 0 \]
such that $K$ is annihilated by $I$.
Proof.
This is just a truncation of the Koszul complex. The map $M \to I$ is determined by the rule $e_ i \mapsto f_ i$. If $m = \sum a_ i e_ i$ is in the kernel of $M \to I$, i.e., $\sum a_ i f_ i = 0$, then $f_ j m = \sum f_ j a_ i e_ i = (\sum f_ i a_ i) e_ j = 0$.
$\square$
Lemma 15.89.6. Let $R$ be a ring. Let $I = (f_1, \ldots , f_ n)$ be a finitely generated ideal of $R$. For any $R$-module $N$ set
\[ H_1(N, f_\bullet ) = \frac{\{ (x_1, \ldots , x_ n) \in N^{\oplus n} \mid f_ i x_ j = f_ j x_ i \} }{\{ f_1x, \ldots , f_ nx) \mid x \in N\} } \]
For any $R$-module $N$ there exists a canonical short exact sequence
\[ 0 \to \mathop{\mathrm{Ext}}\nolimits _ R(R/I, N) \to H_1(N, f_\bullet ) \to \mathop{\mathrm{Hom}}\nolimits _ R(K, N) \]
where $K$ is as in Lemma 15.89.5.
Proof.
The notation above indicates the $\mathop{\mathrm{Ext}}\nolimits $-groups in $\text{Mod}_ R$ as defined in Homology, Section 12.6. These are denoted $\mathop{\mathrm{Ext}}\nolimits _ R(M, N)$. Using the long exact sequence of Homology, Lemma 12.6.4 associated to the short exact sequence $0 \to I \to R \to R/I \to 0$ and the fact that $\mathop{\mathrm{Ext}}\nolimits _ R(R, N) = 0$ we see that
\[ \mathop{\mathrm{Ext}}\nolimits _ R(R/I, N) = \mathop{\mathrm{Coker}}(N \longrightarrow \mathop{\mathrm{Hom}}\nolimits (I, N)) \]
Using the short exact sequence of Lemma 15.89.5 we see that we get a complex
\[ N \to \mathop{\mathrm{Hom}}\nolimits (M, N) \to \mathop{\mathrm{Hom}}\nolimits _ R(K, N) \]
whose homology in the middle is canonically isomorphic to $\mathop{\mathrm{Ext}}\nolimits _ R(R/I, N)$. The proof of the lemma is now complete as the cokernel of the first map is canonically isomorphic to $H_1(N, f_\bullet )$.
$\square$
Lemma 15.89.7. Let $R$ be a ring. Let $I = (f_1, \ldots , f_ n)$ be a finitely generated ideal of $R$. For any $R$-module $N$ the Koszul homology group $H_1(N, f_\bullet )$ defined in Lemma 15.89.6 is annihilated by $I$.
Proof.
Let $(x_1, \ldots , x_ n) \in N^{\oplus n}$ with $f_ i x_ j = f_ j x_ i$. Then we have $f_ i(x_1, \ldots , x_ n) = (f_ i x_ i, \ldots , f_ i x_ n)$. In other words $f_ i$ annihilates $H_1(N, f_\bullet )$.
$\square$
We can improve on the full faithfulness of Lemma 15.89.3 by showing that $\mathop{\mathrm{Ext}}\nolimits $-groups whose source is $I$-power torsion are insensitive to passing to $S$ as well. See Dualizing Complexes, Lemma 47.9.8 for a derived version of the following lemma.
Lemma 15.89.8. Assume $\varphi : R \to S$ is a flat ring map and $I \subset R$ is a finitely generated ideal such that $R/I \to S/IS$ is an isomorphism. Let $M$, $N$ be $R$-modules. Assume $M$ is $I$-power torsion. Given an short exact sequence
\[ 0 \to N \otimes _ R S \to \tilde E \to M \otimes _ R S \to 0 \]
there exists a commutative diagram
\[ \xymatrix{ 0 \ar[r] & N \ar[r] \ar[d] & E \ar[r] \ar[d] & M \ar[r] \ar[d] & 0 \\ 0 \ar[r] & N \otimes _ R S \ar[r] & \tilde E \ar[r] & M \otimes _ R S \ar[r] & 0 } \]
with exact rows.
Proof.
As $M$ is $I$-power torsion we see that $M \otimes _ R S = M$, see Lemma 15.89.2. We will use this identification without further mention. As $R \to S$ is flat, the base change functor is exact and we obtain a functorial map of $\mathop{\mathrm{Ext}}\nolimits $-groups
\[ \mathop{\mathrm{Ext}}\nolimits _ R(M, N) \longrightarrow \mathop{\mathrm{Ext}}\nolimits _ S(M \otimes _ R S, N \otimes _ R S), \]
see Homology, Lemma 12.7.3. The claim of the lemma is that this map is surjective when $M$ is $I$-power torsion. In fact we will show that it is an isomorphism. By Lemma 15.88.2 we can find a surjection $M' \to M$ with $M'$ a direct sum of modules of the form $R/I^ n$. Using the long exact sequence of Homology, Lemma 12.6.4 we see that it suffices to prove the lemma for $M'$. Using compatibility of $\mathop{\mathrm{Ext}}\nolimits $ with direct sums (details omitted) we reduce to the case where $M = R/I^ n$ for some $n$.
Let $f_1, \ldots , f_ t$ be generators for $I^ n$. By Lemma 15.89.6 we have a commutative diagram
\[ \xymatrix{ 0 \ar[r] & \mathop{\mathrm{Ext}}\nolimits _ R(R/I^ n, N) \ar[r] \ar[d] & H_1(N, f_\bullet ) \ar[r] \ar[d] & \mathop{\mathrm{Hom}}\nolimits _ R(K, N) \ar[d] \\ 0 \ar[r] & \mathop{\mathrm{Ext}}\nolimits _ S(S/I^ nS, N \otimes S) \ar[r] & H_1(N \otimes S, f_\bullet ) \ar[r] & \mathop{\mathrm{Hom}}\nolimits _ S(K \otimes S, N \otimes S) } \]
with exact rows where $K$ is as in Lemma 15.89.5. Hence it suffices to prove that the two right vertical arrows are isomorphisms. Since $K$ is annihilated by $I^ n$ we see that $\mathop{\mathrm{Hom}}\nolimits _ R(K, N) = \mathop{\mathrm{Hom}}\nolimits _ S(K \otimes _ R S, N \otimes _ R S)$ by Lemma 15.89.3. As $R \to S$ is flat we have $H_1(N, f_\bullet ) \otimes _ R S = H_1(N \otimes _ R S, f_\bullet )$. As $H_1(N, f_\bullet )$ is annihilated by $I^ n$, see Lemma 15.89.7 we have $H_1(N, f_\bullet ) \otimes _ R S = H_1(N, f_\bullet )$ by Lemma 15.89.2.
$\square$
Let $R \to S$ be a ring map. Let $f_1, \ldots , f_ t \in R$ and $I = (f_1, \ldots , f_ t)$. Then for any $R$-module $M$ we can define a complex
15.89.8.1
\begin{equation} \label{more-algebra-equation-glueing-complex} 0 \to M \xrightarrow {\alpha } M \otimes _ R S \times \prod M_{f_ i} \xrightarrow {\beta } \prod (M \otimes _ R S)_{f_ i} \times \prod M_{f_ if_ j} \end{equation}
where $\alpha (m) = (m \otimes 1, m/1, \ldots , m/1)$ and
\[ \beta (m', m_1, \ldots , m_ t) = ((m'/1 - m_1 \otimes 1, \ldots , m'/1 - m_ t \otimes 1), (m_1 - m_2, \ldots , m_{t - 1} - m_ t). \]
We would like to know when this complex is exact.
Lemma 15.89.9. Assume $\varphi : R \to S$ is a flat ring map and $I = (f_1, \ldots , f_ t) \subset R$ is an ideal such that $R/I \to S/IS$ is an isomorphism. Let $M$ be an $R$-module. Then the complex (15.89.8.1) is exact.
Proof.
First proof. Denote $\check{\mathcal{C}}_ R \to \check{\mathcal{C}}_ S$ the quasi-isomorphism of extended alternating Čech complexes of Lemma 15.89.4. Since these complexes are bounded with flat terms, we see that $M \otimes _ R \check{\mathcal{C}}_ R \to M \otimes _ R \check{\mathcal{C}}_ S$ is a quasi-isomorphism too (Lemmas 15.59.7 and 15.59.12). Now the complex (15.89.8.1) is a truncation of the cone of the map $M \otimes _ R \check{\mathcal{C}}_ R \to M \otimes _ R \check{\mathcal{C}}_ S$ and we win.
Second computational proof. Let $m \in M$. If $\alpha (m) = 0$, then $m \in M[I^\infty ]$, see Lemma 15.88.3. Pick $n$ such that $I^ n m = 0$ and consider the map $\varphi : R/I^ n \to M$. If $m \otimes 1 = 0$, then $\varphi \otimes 1_ S = 0$, hence $\varphi = 0$ (see Lemma 15.89.3) hence $m = 0$. In this way we see that $\alpha $ is injective.
Let $(m', m'_1, \ldots , m'_ t) \in \mathop{\mathrm{Ker}}(\beta )$. Write $m'_ i = m_ i/f_ i^ n$ for some $n > 0$ and $m_ i \in M$. We may, after possibly enlarging $n$ assume that $f_ i^ n m' = m_ i \otimes 1$ in $M \otimes _ R S$ and $f_ j^ nm_ i - f_ i^ nm_ j = 0$ in $M$. In particular we see that $(m_1, \ldots , m_ t)$ defines an element $\xi $ of $H_1(M, (f_1^ n, \ldots , f_ t^ n))$. Since $H_1(M, (f_1^ n, \ldots , f_ t^ n))$ is annihilated by $I^{tn + 1}$ (see Lemma 15.89.7) and since $R \to S$ is flat we see that
\[ H_1(M, (f_1^ n, \ldots , f_ t^ n)) = H_1(M, (f_1^ n, \ldots , f_ t^ n)) \otimes _ R S = H_1(M \otimes _ R S, (f_1^ n, \ldots , f_ t^ n)) \]
by Lemma 15.89.2 The existence of $m'$ implies that $\xi $ maps to zero in the last group, i.e., the element $\xi $ is zero. Thus there exists an $m \in M$ such that $m_ i = f_ i^ n m$. Then $(m', m'_1, \ldots , m'_ t) - \alpha (m) = (m'', 0, \ldots , 0)$ for some $m'' \in (M \otimes _ R S)[(IS)^\infty ]$. By Lemma 15.89.3 we conclude that $m'' \in M[I^\infty ]$ and we win.
$\square$
There is a canonical functor
\[ \text{Can} : \text{Mod}_ R \longrightarrow \text{Glue}(R \to S, f_1, \ldots , f_ t), \quad M \longmapsto (M \otimes _ R S, M_{f_ i}, \text{can}_ i, \text{can}_{ij}) \]
where $\text{can}_ i : (M \otimes _ R S)_{f_ i} \to M_{f_ i} \otimes _ R S$ and $\text{can}_{ij} : (M_{f_ i})_{f_ j} \to (M_{f_ j})_{f_ i}$ are the canonical isomorphisms. For any object $\mathbf{M} = (M', M_ i, \alpha _ i, \alpha _{ij})$ of the category $\text{Glue}(R \to S, f_1, \ldots , f_ t)$ we define
\[ H^0(\mathbf{M}) = \{ (m', m_ i) \mid \alpha _ i(m') = m_ i \otimes 1, \alpha _{ij}(m_ i) = m_ j\} \]
in other words defined by the exact sequence
\[ 0 \to H^0(\mathbf{M}) \to M' \times \prod M_ i \to \prod M'_{f_ i} \times \prod (M_ i)_{f_ j} \]
similar to (15.89.8.1). We think of $H^0(\mathbf{M})$ as an $R$-module. Thus we also get a functor
\[ H^0 : \text{Glue}(R \to S, f_1, \ldots , f_ t) \longrightarrow \text{Mod}_ R \]
Our next goal is to show that the functors $\text{Can}$ and $H^0$ are sometimes quasi-inverse to each other.
Lemma 15.89.11. Assume $\varphi : R \to S$ is a flat ring map and $I = (f_1, \ldots , f_ t) \subset R$ is an ideal such that $R/I \to S/IS$ is an isomorphism. Then the functor $H^0$ is a left quasi-inverse to the functor $\text{Can}$ of Remark 15.89.10.
Proof.
This is a reformulation of Lemma 15.89.9.
$\square$
Lemma 15.89.12. Assume $\varphi : R \to S$ is a flat ring map and let $I = (f_1, \ldots , f_ t) \subset R$ be an ideal. Then $\text{Glue}(R \to S, f_1, \ldots , f_ t)$ is an abelian category, and the functor $\text{Can}$ is exact and commutes with arbitrary colimits.
Proof.
Given a morphism $(\varphi ', \varphi _ i) : (M', M_ i, \alpha _ i, \alpha _{ij}) \to (N', N_ i, \beta _ i, \beta _{ij})$ of the category $\text{Glue}(R \to S, f_1, \ldots , f_ t)$ we see that its kernel exists and is equal to the object $(\mathop{\mathrm{Ker}}(\varphi '), \mathop{\mathrm{Ker}}(\varphi _ i), \alpha _ i, \alpha _{ij})$ and its cokernel exists and is equal to the object $(\mathop{\mathrm{Coker}}(\varphi '), \mathop{\mathrm{Coker}}(\varphi _ i), \beta _ i, \beta _{ij})$. This works because $R \to S$ is flat, hence taking kernels/cokernels commutes with $- \otimes _ R S$. Details omitted. The exactness follows from the $R$-flatness of $R_{f_ i}$ and $S$, while commuting with colimits follows as tensor products commute with colimits.
$\square$
Lemma 15.89.13. Let $\varphi : R \to S$ be a flat ring map and $(f_1, \ldots , f_ t) = R$. Then $\text{Can}$ and $H^0$ are quasi-inverse equivalences of categories
\[ \text{Mod}_ R = \text{Glue}(R \to S, f_1, \ldots , f_ t) \]
Proof.
Consider an object $\mathbf{M} = (M', M_ i, \alpha _ i, \alpha _{ij})$ of $\text{Glue}(R \to S, f_1, \ldots , f_ t)$. By Algebra, Lemma 10.24.5 there exists a unique module $M$ and isomorphisms $M_{f_ i} \to M_ i$ which recover the glueing data $\alpha _{ij}$. Then both $M'$ and $M \otimes _ R S$ are $S$-modules which recover the modules $M_ i \otimes _ R S$ upon localizing at $f_ i$. Whence there is a canonical isomorphism $M \otimes _ R S \to M'$. This shows that $\mathbf{M}$ is in the essential image of $\text{Can}$. Combined with Lemma 15.89.11 the lemma follows.
$\square$
Lemma 15.89.14. Let $\varphi : R \to S$ be a flat ring map and $I = (f_1, \ldots , f_ t)$ and ideal. Let $R \to R'$ be a flat ring map, and set $S' = S \otimes _ R R'$. Then we obtain a commutative diagram of categories and functors
\[ \xymatrix{ \text{Mod}_ R \ar[r]_-{\text{Can}} \ar[d]_{-\otimes _ R R'} & \text{Glue}(R \to S, f_1, \ldots , f_ t) \ar[r]_-{H^0} \ar[d]^{-\otimes _ R R'} & \text{Mod}_ R \ar[d]^{-\otimes _ R R'} \\ \text{Mod}_{R'} \ar[r]^-{\text{Can}} & \text{Glue}(R' \to S', f_1, \ldots , f_ t) \ar[r]^-{H^0} & \text{Mod}_{R'} } \]
Proof.
Omitted.
$\square$
Proposition 15.89.15. Assume $\varphi : R \to S$ is a flat ring map and $I = (f_1, \ldots , f_ t) \subset R$ is an ideal such that $R/I \to S/IS$ is an isomorphism. Then $\text{Can}$ and $H^0$ are quasi-inverse equivalences of categories
\[ \text{Mod}_ R = \text{Glue}(R \to S, f_1, \ldots , f_ t) \]
Proof.
We have already seen that $H^0 \circ \text{Can}$ is isomorphic to the identity functor, see Lemma 15.89.11. Consider an object $\mathbf{M} = (M', M_ i, \alpha _ i, \alpha _{ij})$ of $\text{Glue}(R \to S, f_1, \ldots , f_ t)$. We get a natural morphism
\[ \Psi : (H^0(\mathbf{M}) \otimes _ R S, H^0(\mathbf{M})_{f_ i}, \text{can}_ i, \text{can}_{ij}) \longrightarrow (M', M_ i, \alpha _ i, \alpha _{ij}). \]
Namely, by definition $H^0(\mathbf{M})$ comes equipped with compatible $R$-module maps $H^0(\mathbf{M}) \to M'$ and $H^0(\mathbf{M}) \to M_ i$. We have to show that this map is an isomorphism.
Pick an index $i$ and set $R' = R_{f_ i}$. Combining Lemmas 15.89.14 and 15.89.13 we see that $\Psi \otimes _ R R'$ is an isomorphism. Hence the kernel, resp. cokernel of $\Psi $ is a system of the form $(K, 0, 0, 0)$, resp. $(Q, 0, 0, 0)$. Note that $H^0((K, 0, 0, 0)) = K$, that $H^0$ is left exact, and that by construction $H^0(\Psi )$ is bijective. Hence we see $K = 0$, i.e., the kernel of $\Psi $ is zero.
The conclusion of the above is that we obtain a short exact sequence
\[ 0 \to H^0(\mathbf{M}) \otimes _ R S \to M' \to Q \to 0 \]
and that $M_ i = H^0(\mathbf{M})_{f_ i}$. Note that we may think of $Q$ as an $R$-module which is $I$-power torsion so that $Q = Q \otimes _ R S$. By Lemma 15.89.8 we see that there exists a commutative diagram
\[ \xymatrix{ 0 \ar[r] & H^0(\mathbf{M}) \ar[r] \ar[d] & E \ar[r] \ar[d] & Q \ar[r] \ar[d] & 0 \\ 0 \ar[r] & H^0(\mathbf{M}) \otimes _ R S \ar[r] & M' \ar[r] & Q \ar[r] & 0 } \]
with exact rows. This clearly determines an isomorphism $\text{Can}(E) \to (M', M_ i, \alpha _ i, \alpha _{ij})$ in the category $\text{Glue}(R \to S, f_1, \ldots , f_ t)$ and we win. (Of course, a posteriori we have $Q = 0$.)
$\square$
Lemma 15.89.16. Let $\varphi : R \to S$ be a flat ring map and let $I \subset R$ be a finitely generated ideal such that $R/I \to S/IS$ is an isomorphism.
Given an $R$-module $N$, an $S$-module $M'$ and an $S$-module map $\varphi : M' \to N \otimes _ R S$ whose kernel and cokernel are $I$-power torsion, there exists an $R$-module map $\psi : M \to N$ and an isomorphism $M \otimes _ R S = M'$ compatible with $\varphi $ and $\psi $.
Given an $R$-module $M$, an $S$-module $N'$ and an $S$-module map $\varphi : M \otimes _ R S \to N'$ whose kernel and cokernel are $I$-power torsion, there exists an $R$-module map $\psi : M \to N$ and an isomorphism $N \otimes _ R S = N'$ compatible with $\varphi $ and $\psi $.
In both cases we have $\mathop{\mathrm{Ker}}(\varphi ) \cong \mathop{\mathrm{Ker}}(\psi )$ and $\mathop{\mathrm{Coker}}(\varphi ) \cong \mathop{\mathrm{Coker}}(\psi )$.
Proof.
Proof of (1). Say $I = (f_1, \ldots , f_ t)$. It is clear that the localization $\varphi _{f_ i}$ is an isomorphism. Thus we see that $(M', N_{f_ i}, \varphi _{f_ i}, can_{ij})$ is an object of $\text{Glue}(R \to S, f_1, \ldots , f_ t)$, see Remark 15.89.10. By Proposition 15.89.15 we conclude that there exists an $R$-module $M$ such that $M' = M \otimes _ R S$ and $N_{f_ i} = M_{f_ i}$ compatibly with the isomorphisms $\varphi _{f_ i}$ and $can_{ij}$. There is a morphism
\[ (M \otimes _ R S, M_{f_ i}, can_ i, can_{ij}) = (M', N_{f_ i}, \varphi _{f_ i}, can_{ij}) \to (N \otimes _ R S, N_{f_ i}, can_ i, can_{ij}) \]
of $\text{Glue}(R \to S, f_1, \ldots , f_ t)$ which uses $\varphi $ in the first component. This corresponds to an $R$-module map $\psi : M \to N$ (by the equivalence of categories of Proposition 15.89.15). The composition of the base change of $M \to N$ with the isomorphism $M' \cong M \otimes _ R S$ is $\varphi $, in other words $M \to N$ is compatible with $\varphi $.
Proof of (2). This is just the dual of the argument above. Namely, the localization $\varphi _{f_ i}$ is an isomorphism. Thus we see that $(N', M_{f_ i}, \varphi _{f_ i}^{-1}, can_{ij})$ is an object of $\text{Glue}(R \to S, f_1, \ldots , f_ t)$, see Remark 15.89.10. By Proposition 15.89.15 we conclude that there exists an $R$-module $N$ such that $N' = N \otimes _ R S$ and $N_{f_ i} = M_{f_ i}$ compatibly with the isomorphisms $\varphi _{f_ i}^{-1}$ and $can_{ij}$. There is a morphism
\[ (M \otimes _ R S, M_{f_ i}, can_ i, can_{ij}) \to (N', M_{f_ i}, \varphi _{f_ i}, can_{ij}) = (N \otimes _ R S, N_{f_ i}, can_ i, can_{ij}) \]
of $\text{Glue}(R \to S, f_1, \ldots , f_ t)$ which uses $\varphi $ in the first component. This corresponds to an $R$-module map $\psi : M \to N$ (by the equivalence of categories of Proposition 15.89.15). The composition of the base change of $M \to N$ with the isomorphism $N' \cong N \otimes _ R S$ is $\varphi $, in other words $M \to N$ is compatible with $\varphi $.
The final statement follows for example from Lemma 15.89.3.
$\square$
Next, we specialize Proposition 15.89.15 to get something more useable. Namely, if $I = (f)$ is a principal ideal then the objects of $\text{Glue}(R \to S, f)$ are simply triples $(M', M_1, \alpha _1)$ and there is no cocycle condition to check!
Theorem 15.89.17. Let $R$ be a ring, and let $f \in R$. Let $\varphi : R \to S$ be a flat ring map inducing an isomorphism $R/fR \to S/fS$. Then the functor
\[ \text{Mod}_ R \longrightarrow \text{Mod}_ S \times _{\text{Mod}_{S_ f}} \text{Mod}_{R_ f}, \quad M \longmapsto (M \otimes _ R S, M_ f, \text{can}) \]
is an equivalence.
Proof.
The category appearing on the right side of the arrow is the category of triples $(M', M_1, \alpha _1)$ where $M'$ is an $S$-module, $M_1$ is a $R_ f$-module, and $\alpha _1 : M'_ f \to M_1 \otimes _ R S$ is a $S_ f$-isomorphism, see Categories, Example 4.31.3. Hence this theorem is a special case of Proposition 15.89.15.
$\square$
A useful special case of Theorem 15.89.17 is when $R$ is Noetherian, and $S$ is a completion of $R$ at an element $f$. The completion $R \to S$ is flat, and the functor $M \mapsto M \otimes _ R S$ can be identified with the $f$-adic completion functor when $M$ is finitely generated. To state this more precisely, let $\text{Mod}^{fg}_ R$ denote the category of finitely generated $R$-modules.
Proposition 15.89.18. Let $R$ be a Noetherian ring. Let $f \in R$ be an element. Let $R^\wedge $ be the $f$-adic completion of $R$. Then the functor $M \mapsto (M^\wedge , M_ f, \text{can})$ defines an equivalence
\[ \text{Mod}^{fg}_ R \longrightarrow \text{Mod}^{fg}_{R^\wedge } \times _{\text{Mod}^{fg}_{(R^\wedge )_ f}} \text{Mod}^{fg}_{R_ f} \]
Proof.
The ring map $R \to R^\wedge $ is flat by Algebra, Lemma 10.97.2. It is clear that $R/fR = R^\wedge /fR^\wedge $. By Algebra, Lemma 10.97.1 the completion of a finite $R$-module $M$ is equal to $M \otimes _ R R^\wedge $. Hence the displayed functor of the proposition is equal to the functor occurring in Theorem 15.89.17. In particular it is fully faithful. Let $(M_1, M_2, \psi )$ be an object of the right hand side. By Theorem 15.89.17 there exists an $R$-module $M$ such that $M_1 = M \otimes _ R R^\wedge $ and $M_2 = M_ f$. As $R \to R^\wedge \times R_ f$ is faithfully flat we conclude from Algebra, Lemma 10.23.2 that $M$ is finitely generated, i.e., $M \in \text{Mod}^{fg}_ R$. This proves the proposition.
$\square$
Comments (2)
Comment #4623 by Anonymous on
Comment #4624 by Anonymous on