Lemma 15.90.1. Let $R$ be a ring and let $f \in R$. For every positive integer $n$ the map $R/f^ nR \to R^\wedge /f^ n R^\wedge $ is an isomorphism.
15.90 The Beauville-Laszlo theorem
Let $R$ be a ring and let $f$ be an element of $R$. Denote $R^\wedge = \mathop{\mathrm{lim}}\nolimits R/f^ n R$ the $f$-adic completion of $R$. In this section we discuss and slightly generalize a theorem of Beauville and Laszlo, see [Beauville-Laszlo]. The theorem asserts that under suitable conditions, a module over $R$ can be constructed by “glueing together” modules over $R^\wedge $ and $R_ f$ along an isomorphism between the base extensions to $(R^\wedge )_ f$.
In [Beauville-Laszlo] it is assumed that $f$ is a nonzerodivisor on both $R$ and $M$. In fact, one only needs to assume that
\[ R[f^\infty ] \longrightarrow R^\wedge [f^\infty ] \]
is bijective and that
\[ M[f^\infty ] \longrightarrow M \otimes _ R R^\wedge \]
is injective. This optimization was partly inspired by an alternate approach to glueing introduced in [§1.3, Kedlaya-Liu-I] for use in the theory of nonarchimedean analytic spaces.
In fact, we will establish the Beauville-Laszlo theorem in the more general setting of a ring map
\[ R \longrightarrow R' \]
which induces isomorphisms $R/f^ nR \to R'/f^ nR'$ for every $n > 0$ and an isomorphism $R[f^\infty ] \to R'[f^\infty ]$. This is better suited for globalizing and does not formally follow from the case when $R'$ is the completion of $R$ because, for instance, the condition that $R[f^\infty ] \to R'[f^\infty ]$ is a bijection does not imply that $R[f^\infty ] \to R^\wedge [f^\infty ]$ is a bijection.
The theorem of Beauville and Laszlo as proved in this section can be viewed as a non-flat version of Theorem 15.89.18 and in the case where $R' = R^\wedge $ can be viewed as a non-Noetherian version of Proposition 15.89.19. For a comparison with flat descent, please see Remark 15.90.5.
One can establish even stronger results (without imposing restrictions on $M$ for example) but for this one must work at the level of derived categories. See [§5, Bhatt-Algebraize] for more details.
Proof. This is a special case of Algebra, Lemma 10.96.3. $\square$
We will use the notation introduced in Section 15.88. Thus for an $R$-module $M$, we denote $M[f^ n]$ the submodule of $M$ annihilated by $f^ n$ and we put
\[ M[f^\infty ] = \bigcup \nolimits _{n = 1}^\infty M[f^ n] = \mathop{\mathrm{Ker}}(M \to M_ f). \]
If $M = M[f^\infty ]$, we say that $M$ is an $f$-power torsion module.
Lemma 15.90.2. Let $R$ be a ring, let $f \in R$, and let $R \to R'$ be a ring map which induces isomorphisms $R/f^ nR \to R'/f^ nR'$ for $n > 0$. The $R$-module $R' \oplus R_ f$ is faithful: for every nonzero $R$-module $M$, the module $M \otimes _ R (R' \oplus R_ f)$ is also nonzero. For example, if $M$ is nonzero, then $M \otimes _ R (R^\wedge \oplus R_ f)$ is nonzero.
However, the map $M \to M \otimes _ R (R' \oplus R_ f)$ need not be injective; see Example 15.90.9.
Proof. If $M \neq 0$ but $M \otimes _ R R_ f = 0$, then $M$ is $f$-power torsion. By Lemma 15.88.8 we find that $M \otimes _ R R' \cong M \neq 0$. The last statement is a special case of the first statement by Lemma 15.90.1. $\square$
Lemma 15.90.3. Let $R$ be a ring, let $f \in R$, and let $R \to R'$ be a ring map which induces an isomorphism $R/fR \to R'/fR'$. The map $\mathop{\mathrm{Spec}}(R') \amalg \mathop{\mathrm{Spec}}(R_ f) \to \mathop{\mathrm{Spec}}(R)$ is surjective. For example, the map $\mathop{\mathrm{Spec}}(R^\wedge ) \amalg \mathop{\mathrm{Spec}}(R_ f) \to \mathop{\mathrm{Spec}}(R)$ is surjective.
Proof. Recall that $\mathop{\mathrm{Spec}}(R) = V(f) \amalg D(f)$ where $V(f) = \mathop{\mathrm{Spec}}(R/fR)$ and $D(f) = \mathop{\mathrm{Spec}}(R_ f)$, see Algebra, Section 10.17 and especially Lemmas 10.17.7 and 10.17.6. Thus the lemma follows as the map $R \to R/fR$ factors through $R'$. The last statement is a special case of the first statement by Lemma 15.90.1. $\square$
Lemma 15.90.4. Let $R$ be a ring, let $f \in R$, and let $R \to R'$ be a ring map which induces isomorphisms $R/f^ nR \to R'/f^ nR'$ for $n > 0$. An $R$-module $M$ is finitely generated if and only if the ($R' \oplus R_ f$)-module $M \otimes _ R (R' \oplus R_ f)$ is finitely generated. For example, if $M \otimes _ R (R^\wedge \oplus R_ f)$ is finitely generated as a module over $R^\wedge \oplus R_ f$, then $M$ is a finitely generated $R$-module.
Proof. The ‘only if' is clear, so we assume that $M \otimes _ R (R' \oplus R_ f)$ is finitely generated. In this case, by writing each generator as a sum of simple tensors, $M \otimes _ R (R' \oplus R_ f)$ admits a finite generating set consisting of elements of $M$. That is, there exists a morphism from a finite free $R$-module to $M$ whose cokernel is killed by tensoring with $R' \oplus R_ f$; we may thus deduce $M$ is finite generated by applying Lemma 15.90.2 to this cokernel. The last statement is a special case of the first statement by Lemma 15.90.1. $\square$
Remark 15.90.5. While $R \to R_ f$ is always flat, $R \to R^\wedge $ is typically not flat unless $R$ is Noetherian (see Algebra, Lemma 10.97.2 and the discussion in Examples, Section 110.12). Consequently, we cannot in general apply faithfully flat descent as discussed in Descent, Section 35.3 to the morphism $R \to R^\wedge \oplus R_ f$. Moreover, even in the Noetherian case, the usual definition of a descent datum for this morphism refers to the ring $R^\wedge \otimes _ R R^\wedge $, which we will avoid considering in this section.
Glueing pairs. Let $R \to R'$ be a ring map that induces isomorphisms $R/f^ nR \to R'/f^ nR'$ for $n > 0$. Consider the sequence
15.90.5.1
\begin{equation} \label{more-algebra-equation-BL-cech-re} 0 \to R \to R' \oplus R_ f \to R'_ f \to 0, \end{equation}
in which the map on the right is the difference between the two canonical homomorphisms. If this sequence is exact, then we say that $(R \to R', f)$ is a glueing pair. We will say that $(R, f)$ is a glueing pair if $(R \to R^\wedge , f)$ is a glueing pair; this makes sense by Lemma 15.90.1. Thus $(R, f)$ is a glueing pair if and only if the sequence
15.90.5.2
\begin{equation} \label{more-algebra-equation-BL-cech} 0 \to R \to R^\wedge \oplus R_ f \to (R^\wedge )_ f \to 0, \end{equation}
is exact.
Lemma 15.90.6. Let $R$ be a ring, let $f \in R$, and let $R \to R'$ be a ring map which induces isomorphisms $R/f^ nR \to R'/f^ nR'$ for $n > 0$. The sequence (15.90.5.1) is
exact on the right,
exact on the left if and only if $R[f^\infty ] \to R'[f^\infty ]$ is injective, and
exact in the middle if and only if $R[f^\infty ] \to R'[f^\infty ]$ is surjective.
In particular, $(R \to R', f)$ is a glueing pair if and only if $R[f^\infty ] \to R'[f^\infty ]$ is bijective. For example, $(R, f)$ is a glueing pair if and only if $R[f^\infty ] \to R^\wedge [f^\infty ]$ is bijective.
Proof. Let $x \in R'_ f$. Write $x = x'/f^ n$ with $x' \in R'$. Write $x' = x'' + f^ n y$ with $x'' \in R$ and $y \in R'$. Then we see that $(y, -x''/f^ n)$ maps to $x$. Thus (1) holds.
Part (2) follows from the fact that $\mathop{\mathrm{Ker}}(R \to R_ f) = R[f^\infty ]$.
If the sequence is exact in the middle, then elements of the form $(x, 0)$ with $x \in R'[f^\infty ]$ are in the image of the first arrow. This implies that $R[f^\infty ] \to R'[f^\infty ]$ is surjective. Conversely, assume that $R[f^\infty ] \to R'[f^\infty ]$ is surjective. Let $(x, y)$ be an element in the middle which maps to zero on the right. Write $y = y'/f^ n$ for some $y' \in R$. Then we see that $f^ n x - y'$ is annihilated by some power of $f$ in $R'$. By assumption we can write $f^ nx - y' = z$ for some $z \in R[f^\infty ]$. Then $y = y''/f^ n$ where $y'' = y' + z$ is in the kernel of $R \to R/f^ nR$. Hence we see that $y$ can be represented as $y'''/1$ for some $y''' \in R$. Then $x - y'''$ is in $R'[f^\infty ]$. Thus $x - y''' = z' \in R[f^\infty ]$. Then $(x, y'''/1) = (y''' + z', (y''' + z')/1)$ as desired.
The last statement of the lemma is a special case of the penultimate statement by Lemma 15.90.1. $\square$
Remark 15.90.7. Suppose that $f$ is a nonzerodivisor. Then Algebra, Lemma 10.96.4 shows that $f$ is a nonzerodivisor in $R^\wedge $. Hence $(R, f)$ is a glueing pair.
Remark 15.90.8. If $R \to R^\wedge $ is flat, then for each positive integer $n$ tensoring the sequence $0 \to R[f^ n] \to R \to R$ with $R^\wedge $ gives the sequence $0 \to R[f^ n] \otimes _ R R^\wedge \to R^\wedge \to R^\wedge $. Combined with Lemma 15.88.8 we conclude that $R[f^ n] \to R^\wedge [f^ n]$ is an isomorphism. Thus $(R, f)$ is a glueing pair. This holds in particular if $R$ is Noetherian, see Algebra, Lemma 10.97.2.
Example 15.90.9. Let $k$ be a field and put Then $(R, f)$ is not a glueing pair because the map $R[f^\infty ] \to R^\wedge [f^\infty ]$ is not injective as the image of $T_1$ is $f$-divisible in $R^\wedge $. For the map $R[f^\infty ] \to R^\wedge [f^\infty ]$ is not surjective as the element $T_1 + fT_2 + f^2 T_3 + \ldots $ is not in the image. In particular, by Remark 15.90.8, these are both examples where $R \to R^\wedge $ is not flat.
\[ R = k[f, T_1, T_2, \ldots ]/(fT_1, fT_2 - T_1, fT_3 - T_2, \ldots ). \]
\[ R = k[f, T_1, T_2, \ldots ]/(fT_1, f^2T_2, \ldots ), \]
Glueable modules. Let $R \to R'$ be a ring map which induces isomorphisms $R/f^ nR \to R'/f^ nR'$ for $n > 0$. For any $R$-module $M$, we may tensor (15.90.5.1) with $M$ to obtain a sequence
15.90.9.1
\begin{equation} \label{more-algebra-equation-BL-cech-mod-re} 0 \to M \to (M \otimes _ R R') \oplus (M \otimes _ R R_ f) \to M \otimes _ R R'_ f \to 0 \end{equation}
Observe that $M \otimes _ R R_ f = M_ f$ and that $M \otimes _ R R'_ f = (M \otimes _ R R')_ f$. If this sequence is exact, we say that $M$ is glueable for $(R \to R', f)$. If $R$ is a ring and $f \in R$, then we say an $R$-module is glueable if $M$ is glueable for $(R \to R^\wedge , f)$. Thus $M$ is glueable if and only if the sequence
15.90.9.2
\begin{equation} \label{more-algebra-equation-BL-cech-mod} 0 \to M \to (M \otimes _ R R^\wedge ) \oplus (M \otimes _ R R_ f) \to M \otimes _ R (R^\wedge )_ f \to 0 \end{equation}
is exact.
Lemma 15.90.10. Let $R$ be a ring, let $f \in R$, and let $R \to R'$ be a ring map which induces isomorphisms $R/f^ nR \to R'/f^ nR'$ for $n > 0$. The sequence (15.90.9.1) is
exact on the right,
exact on the left if and only if $M[f^\infty ] \to (M \otimes _ R R')[f^\infty ]$ is injective, and
exact in the middle if and only if $M[f^\infty ] \to (M \otimes _ R R')[f^\infty ]$ is surjective.
Thus $M$ is glueable for $(R \to R', f)$ if and only if $M[f^\infty ] \to (M \otimes _ R R')[f^\infty ]$ is bijective. If $(R \to R', f)$ is a glueing pair, then $M$ is glueable for $(R \to R', f)$ if and only if $M[f^\infty ] \to (M \otimes _ R R')[f^\infty ]$ is injective. For example, if $(R, f)$ is a glueing pair, then $M$ is glueable if and only if $M[f^\infty ] \to (M \otimes _ R R^\wedge )[f^\infty ]$ is injective.
Proof. We will use the results of Lemma 15.90.6 without further mention. The functor $M \otimes _ R -$ is right exact (Algebra, Lemma 10.12.10) hence we get (1).
The kernel of $M \to M \otimes _ R R_ f = M_ f$ is $M[f^\infty ]$. Thus (2) follows.
If the sequence is exact in the middle, then elements of the form $(x, 0)$ with $x \in (M \otimes _ R R')[f^\infty ]$ are in the image of the first arrow. This implies that $M[f^\infty ] \to (M \otimes _ R R')[f^\infty ]$ is surjective. Conversely, assume that $M[f^\infty ] \to (M \otimes _ R R')[f^\infty ]$ is surjective. Let $(x, y)$ be an element in the middle which maps to zero on the right. Write $y = y'/f^ n$ for some $y' \in M$. Then we see that $f^ n x - y'$ is annihilated by some power of $f$ in $M \otimes _ R R'$. By assumption we can write $f^ nx - y' = z$ for some $z \in M[f^\infty ]$. Then $y = y''/f^ n$ where $y'' = y' + z$ is in the kernel of $M \to M/f^ nM$. Hence we see that $y$ can be represented as $y'''/1$ for some $y''' \in M$. Then $x - y'''$ is in $(M \otimes _ R R')[f^\infty ]$. Thus $x - y''' = z' \in M[f^\infty ]$. Then $(x, y'''/1) = (y''' + z', (y''' + z')/1)$ as desired.
If $(R \to R', f)$ is a glueing pair, then (15.90.9.1) is exact in the middle for any $M$ by Algebra, Lemma 10.12.10. This gives the penultimate statement of the lemma. The final statement of the lemma follows from this and the fact that $(R, f)$ is a glueing pair if and only if $(R \to R^\wedge , f)$ is a glueing pair. $\square$
Remark 15.90.11. Let $(R \to R', f)$ be a glueing pair and let $M$ be an $R$-module. Here are some observations which can be used to determine whether $M$ is glueable for $(R \to R', f)$.
By Lemma 15.90.10 we see that $M$ is glueable for $(R \to R^\wedge , f)$ if and only if $M[f^\infty ] \to M \otimes _ R R^\wedge $ is injective. This holds if $M[f] \to M^\wedge $ is injective, i.e., when $M[f] \cap \bigcap _{n = 1}^\infty f^ n M = 0$.
If $\text{Tor}_1^ R(M, R'_ f) = 0$, then $M$ is glueable for $(R \to R', f)$ (use Algebra, Lemma 10.75.2). This is equivalent to saying that $\text{Tor}_1^ R(M, R')$ is $f$-power torsion. In particular, any flat $R$-module is glueable for $(R \to R', f)$.
If $R \to R'$ is flat, then $\text{Tor}_1^ R(M, R') = 0$ for every $R$-module so every $R$-module is glueable for $(R \to R', f)$. This holds in particular when $R$ is Noetherian and $R' = R^\wedge $, see Algebra, Lemma 10.97.2
Example 15.90.12 (Non glueable module). Let $R$ be the ring of germs at $0$ of $C^\infty $ functions on $\mathbf{R}$. Let $f \in R$ be the function $f(x) = x$. Then $f$ is a nonzerodivisor in $R$, so $(R, f)$ is a glueing pair and $R^\wedge \cong \mathbf{R}[[x]]$. Let $\varphi \in R$ be the function $\varphi (x) = \text{exp}(-1/x^2)$. Then $\varphi $ has zero Taylor series, so $\varphi \in \mathop{\mathrm{Ker}}(R \to R^\wedge )$. Since $\varphi (x) \neq 0$ for $x \neq 0$, we see that $\varphi $ is a nonzerodivisor in $R$. The function $\varphi /f$ also has zero Taylor series, so its image in $M = R/\varphi R$ is a nonzero element of $M[f]$ which maps to zero in $M \otimes _ R R^\wedge = R^\wedge /\varphi R^\wedge = R^\wedge $. Hence $M$ is not glueable.
We next make some calculations of Tor groups.
Lemma 15.90.13. Let $(R \to R', f)$ be a glueing pair. Then $\text{Tor}^ R_1(R', f^ n R) = 0$ for each $n > 0$.
Proof. From the exact sequence $0 \to R[f^ n] \to R \to f^ n R \to 0$ we see that it suffices to check that $R[f^ n] \otimes _ R R' \to R'$ is injective. By Lemma 15.88.8 we have $R[f^ n] \otimes _ R R' = R[f^ n]$ and by Lemma 15.90.6 we see that $R[f^ n] \to R'$ is injective as $(R \to R', f)$ is a glueing pair. $\square$
Lemma 15.90.14. Let $(R \to R',f)$ be a glueing pair. Then $\text{Tor}^ R_1(R', R/R[f^\infty ]) = 0$.
Proof. We have $R/R[f^\infty ] = \mathop{\mathrm{colim}}\nolimits R/R[f^ n] = \mathop{\mathrm{colim}}\nolimits f^ nR$. As formation of Tor groups commutes with filtered colimits (Algebra, Lemma 10.76.2) we may apply Lemma 15.90.13. $\square$
Lemma 15.90.15. Let $(R \to R', f)$ be a glueing pair. For every $R$-module $M$, we have $\text{Tor}^ R_1(R', \mathop{\mathrm{Coker}}(M \to M_ f)) = 0$.
Proof. Set $\overline{M} = M/M[f^\infty ]$. Then $\mathop{\mathrm{Coker}}(M \to M_ f) \cong \mathop{\mathrm{Coker}}(\overline{M} \to \overline{M}_ f)$ hence we may and do assume that $f$ is a nonzerodivisor on $M$. In this case $M \subset M_ f$ and $M_ f/M = \mathop{\mathrm{colim}}\nolimits M/f^ nM$ where the transition maps are given by multiplication by $f$. Since formation of Tor groups commutes with colimits (Algebra, Lemma 10.76.2) it suffices to show that $\text{Tor}^ R_1(R', M/f^ n M) = 0$.
We first treat the case $M = R/R[f^\infty ]$. By Lemma 15.90.6 we have $M \otimes _ R R' = R'/R'[f^\infty ]$. From the short exact sequence $0 \to M \to M \to M/f^ nM \to 0$ we obtain the exact sequence
\[ \xymatrix{ \text{Tor}_1^ R(R', R/R[f^\infty ]) \ar[r] & \text{Tor}_1^ R(R', M/f^ n M) \ar[r] & R'/R'[f^\infty ] \ar[dll]_{f^ n} \\ R'/R'[f^\infty ] \ar[r] & (R'/R'[f^\infty ])/(f^ n (R'/R'[f^\infty ])) \ar[r] & 0 } \]
by Algebra, Lemma 10.75.2. Here the diagonal arrow is injective. Since the first group $\text{Tor}_1^ R(R', R/R[f^\infty ])$ is zero by Lemma 15.90.14, we deduce that $\text{Tor}_1^ R(R', M/f^ nM) = 0$ as desired.
To treat the general case, choose a surjection $F \to M$ with $F$ a free $R/R[f^\infty ]$-module, and form an exact sequence
\[ 0 \to N \to F/f^ n F \to M/f^ n M \to 0. \]
By Lemma 15.88.8 this sequence remains unchanged, and hence exact, upon tensoring with $R'$. Since $\text{Tor}^ R_1(R', F/f^ n F) = 0$ by the previous paragraph, we deduce that $\text{Tor}^ R_1(R', M/f^ n M) = 0$ as desired. $\square$
Let $(R \to R', f)$ be a glueing pair. This means that $R/f^ nR \to R'/f^ nR'$ is an isomorphism for $n > 0$ and the sequence
\[ 0 \to R \to R' \oplus R_ f \to R_ f' \to 0 \]
is exact. Consider the category $\text{Glue}(R \to R', f)$ introduced in Remark 15.89.10. We will call an object $(M', M_1, \alpha _1)$ of $\text{Glue}(R \to R', f)$ a glueing datum. It consists of an $R'$-module $M'$, an $R_ f$-module $M_1$, and an isomorphism $\alpha _1 : (M')_ f \to M_1 \otimes _ R R'$. There is an obvious functor
\[ \text{Can} : \text{Mod}_ R \longrightarrow \text{Glue}(R \to R', f),\quad M \longmapsto (M \otimes _ R R', M_ f, \text{can}), \]
and there is a functor
\[ H^0 : \text{Glue}(R \to R', f) \longrightarrow \text{Mod}_ R,\quad (M', M_1, \alpha _1) \longmapsto \mathop{\mathrm{Ker}}(M' \oplus M_1 \to (M')_ f) \]
in the reverse direction, see Remark 15.89.10 for the precise definition.
Theorem 15.90.16. Let $(R \to R',f)$ be a glueing pair. The functor $\text{Can} : \text{Mod}_ R \longrightarrow \text{Glue}(R \to R', f)$ determines an equivalence of the category of $R$-modules glueable for $(R \to R', f)$ and the category $\text{Glue}(R \to R', f)$ of glueing data.
Proof. Let $(M', M_1, \alpha _1)$ be a glueing datum. We will show that $M = H^0((M', M_1, \alpha _1))$ is a glueable for $(R \to R', f)$ and that $(M', M_1, \alpha _1) \cong \text{Can}(M)$.
We first check that the map $\text{d} : M' \oplus M_1 \to (M')_ f$ used in the definition of the functor $H^0$ is surjective. Observe that $(x, y) \in M' \oplus M_1$ maps to $\text{d}(x, y) = x/1 - \alpha _1^{-1}(y \otimes 1)$ in $(M')_ f$. If $z \in (M')_ f$, then we can write $\alpha _1(z) = \sum y_ i \otimes g_ i$ with $g_ i \in R'$ and $y_ i \in M_1$. Write $\alpha _1^{-1}(y_ i \otimes 1) = y_ i'/f^ n$ for some $y'_ i \in M'$ and $n \geq 0$ (we can pick the same $n$ for all $i$). Write $g_ i = a_ i + f^ n b_ i$ with $a_ i \in R$ and $b_ i \in R'$. Then with $y = \sum a_ i y_ i \in M_1$ and $x = \sum b_ i y'_ i \in M'$ we have $\text{d}(x, -y) = z$ as desired.
Since $M = H^0((M', M_1, \alpha _1)) = \mathop{\mathrm{Ker}}(\text{d})$ we obtain an exact sequence of $R$-modules
15.90.16.1
\begin{equation} \label{more-algebra-equation-define-M} 0 \to M \to M' \oplus M_1 \to (M')_ f \to 0. \end{equation}
We will prove that the maps $M \to M'$ and $M \to M_1$ induce isomorphisms $M \otimes _ R R' \to M'$ and $M \otimes _ R R_ f \to M_1$. This will imply that $M$ is glueable for $(R \to R', f)$ and $\text{Can}(M) \cong (M', M_1, \alpha _1)$ as desired.
Since $f$ is a nonzerodivisor on $M_1$, we have $M[f^\infty ] \cong M'[f^\infty ]$. This yields an exact sequence
15.90.16.2
\begin{equation} \label{more-algebra-equation-exact-mod-torsion} 0 \to M/M[f^\infty ] \to M_1 \to (M')_ f/M' \to 0. \end{equation}
Since $R \to R_ f$ is flat, we may tensor this exact sequence with $R_ f$ to deduce that $M \otimes _ R R_ f = (M/M[f^\infty ]) \otimes _ R R_ f \to M_1$ is an isomorphism.
By Lemma 15.90.15 we have $\text{Tor}_1^ R(R', \mathop{\mathrm{Coker}}(M' \to (M')_ f)) = 0$. The sequence (15.90.16.2) thus remains exact upon tensoring over $R$ with $R'$. Using $\alpha _1$ and Lemma 15.88.8 the resulting exact sequence can be written as
15.90.16.3
\begin{equation} \label{more-algebra-equation-mod-torsion-sequence} 0 \to (M/M[f^\infty ]) \otimes _ R R' \to (M')_ f \to (M')_ f/M' \to 0 \end{equation}
This yields an isomorphism $(M/M[f^\infty ]) \otimes _ R R' \cong M'/M'[f^\infty ]$. This implies that in the diagram
\[ \xymatrix{ & M[f^\infty ] \otimes _ R R' \ar[r] \ar[d] & M \otimes _ R R' \ar[r] \ar[d] & (M/M[f^\infty ]) \otimes _ R R' \ar[r] \ar[d] & 0 \\ 0 \ar[r] & M'[f^\infty ] \ar[r] & M' \ar[r] & M'/M'[f^\infty ] \ar[r] & 0, } \]
the third vertical arrow is an isomorphism. Since the rows are exact and the first vertical arrow is an isomorphism by Lemma 15.88.8 and $M[f^\infty ] = M'[f^\infty ]$, the five lemma implies that $M \otimes _ R R' \to M'$ is an isomorphism.
The above shows that $\text{Can}$ is essentially surjective and that the functor $H^0$ maps into the category of glueable modules. Due to the exactness of (15.90.9.1) for glueable modules we have $H^0 \circ \text{Can} = \text{id}$ on the category of glueable modules. This implies $\text{Can}$ is fully faithful by Lemma 15.89.11 combined with Categories, Lemma 4.24.4. This finishes the proof. $\square$
Remark 15.90.17. Let $(R \to R', f)$ be a glueing pair. Let $M$ be an $R$-module that is not necessarily glueable for $(R \to R', f)$. Setting $M' = M \otimes _ R R'$ and $M_1 = M_ f$ we obtain the glueing datum $\text{Can}(M) = (M', M_1, \text{can})$. Then $\tilde M = H^0(M', M_1, \text{can})$ is an $R$-module that is glueable for $(R \to R', f)$ and the canonical map $M \to \tilde M$ gives isomorphisms $M \otimes _ R R' \to \tilde M \otimes _ R R'$ and $M_ f \to \tilde M_ f$, see Theorem 15.90.16. From the exactness of the sequences and we conclude that the map $M \to \tilde M$ is surjective.
\[ M \to (M \otimes _ R R' )\oplus M_ f \to M \otimes _ R (R')_ f \to 0 \]
\[ 0 \to \tilde M \to (\tilde M \otimes _ R R') \oplus \tilde M_ f \to \tilde M \otimes _ R (R')_ f \to 0 \]
Recall that flat $R$-modules over a glueing pair $(R \to R', f)$ are glueable (Remark 15.90.11). Hence the following lemma shows that Theorem 15.90.16 determines an equivalence between the category of flat $R$-modules and the category of glueing data $(M', M_1, \alpha _1)$ where $M'$ and $M_1$ are flat over $R'$ and $R_ f$.
Lemma 15.90.18. Let $(R \to R', f)$ be a glueing pair. Let $M$ be an $R$-module which is not necessarily glueable for $(R \to R', f)$. Then $M$ is flat over $R$ if and only if $M \otimes _ R R'$ is flat over $R'$ and $M_ f$ is flat over $R_ f$.
Proof. One direction of the lemma follows from Algebra, Lemma 10.39.7. For the other direction, assume $M \otimes _ R R'$ is flat over $R'$ and $M_ f$ is flat over $R_ f$. Let $\tilde M$ be as in Remark 15.90.17. If $\tilde M$ is flat over $R$, then applying Algebra, Lemma 10.39.12 to the short exact sequence $0 \to \mathop{\mathrm{Ker}}(M \to \tilde M) \to M \to \tilde M \to 0$ we find that $\mathop{\mathrm{Ker}}(M \to \tilde M) \otimes _ R (R' \oplus R_ f)$ is zero. Hence $M = \tilde M$ by Lemma 15.90.2 and we conclude. In other words, we may replace $M$ by $\tilde M$ and assume $M$ is glueable for $(R \to R', f)$. Let $N$ be a second $R$-module. It suffices to prove that $\text{Tor}_1^ R(M, N) = 0$, see Algebra, Lemma 10.75.8.
The long the exact sequence of Tors associated to the short exact sequence $0 \to R \to R' \oplus R_ f \to (R')_ f \to 0$ and $N$ gives an exact sequence
\[ 0 \to \text{Tor}_1^ R(R', N) \to \text{Tor}_1^ R((R')_ f, N) \]
and isomorphisms $\text{Tor}_ i^ R(R', N) = \text{Tor}_ i^ R((R')_ f, N)$ for $i \geq 2$. Since $\text{Tor}_ i^ R((R')_ f, N) = \text{Tor}_ i^ R(R', N)_ f$ we conclude that $f$ is a nonzerodivisor on $\text{Tor}_1^ R(R', N)$ and invertible on $\text{Tor}_ i^ R(R', N)$ for $i \geq 2$. Since $M \otimes _ R R'$ is flat over $R'$ we have
\[ \text{Tor}_ i^ R(M \otimes _ R R', N) = (M \otimes _ R R') \otimes _{R'} \text{Tor}_ i^ R(R', N) \]
by the spectral sequence of Example 15.62.2. Writing $M \otimes _ R R'$ as a filtered colimit of finite free $R'$-modules (Algebra, Theorem 10.81.4) we conclude that $f$ is a nonzerodivisor on $\text{Tor}_1^ R(M \otimes _ R R', N)$ and invertible on $\text{Tor}_ i^ R(M \otimes _ R R', N)$. Next, we consider the exact sequence $0 \to M \to M \otimes _ R R' \oplus M_ f \to M \otimes _ R (R')_ f \to 0$ coming from the fact that $M$ is glueable and the associated long exact sequence of $\text{Tor}$. The relevant part is
\[ \xymatrix{ \text{Tor}_1^ R(M, N) \ar[r] & \text{Tor}_1^ R(M \otimes _ R R', N) \ar[r] & \text{Tor}_1^ R(M \otimes _ R (R')_ f, N) \\ & \text{Tor}_2^ R(M \otimes _ R R', N) \ar[r] & \text{Tor}_2^ R(M \otimes _ R (R')_ f, N) \ar[llu] } \]
We conclude that $\text{Tor}_1^ R(M, N) = 0$ by our remarks above on the action on $f$ on $\text{Tor}_ i^ R(M \otimes _ R R', N)$. $\square$
Observe that we have seen the result of the following lemma for “finitely generated” in Lemma 15.90.4.
Lemma 15.90.19. Let $(R \to R', f)$ be a glueing pair. Let $M$ be an $R$-module which is not necessarily glueable for $(R \to R', f)$. Then $M$ is a finite projective $R$-module if and only if $M \otimes _ R R'$ is finite projective over $R'$ and $M_ f$ is finite projective over $R_ f$.
Proof. Assume that $M \otimes _ R R'$ is a finite projective module over $R'$ and that $M_ f$ is a finite projective module over $R_ f$. Our task is to prove that $M$ is finite projective over $R$. We will use Algebra, Lemma 10.78.2 without further mention. By Lemma 15.90.18 we see that $M$ is flat. By Lemma 15.90.4 we see that $M$ is finite. Choose a short exact sequence $0 \to K \to R^{\oplus n} \to M \to 0$. Since a finite projective module is of finite presentation and since the sequence remains exact after tensoring with $R'$ (by Algebra, Lemma 10.39.12) and $R_ f$, we conclude that $K \otimes _ R R'$ and $K_ f$ are finite modules. Using the lemma above we conclude that $K$ is finitely generated. Hence $M$ is finitely presented and hence finite projective. $\square$
Remark 15.90.20. In [Beauville-Laszlo] it is assumed that $f$ is a nonzerodivisor in $R$ and $R' = R^\wedge $, which gives a glueing pair by Lemma 15.90.6. Even in this setting Theorem 15.90.16 says something new: the results of [Beauville-Laszlo] only apply to modules on which $f$ is a nonzerodivisor (and hence glueable in our sense, see Lemma 15.90.10). Lemma 15.90.19 also provides a slight extension of the results of [Beauville-Laszlo]: not only can we allow $M$ to have nonzero $f$-power torsion, we do not even require it to be glueable.
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