Lemma 15.81.1. 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.

## 15.81 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

is bijective and that

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

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 15.80.17 and in the case where $R' = R^\wedge $ can be viewed as a non-Noetherian version of Proposition 15.80.18. For a comparison with flat descent, please see Remark 15.81.6.

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 10.95.3.
$\square$

We will use the notation introduced in Section 15.79. Thus for an $R$-module $M$, we denote $M[f^ n]$ the submodule of $M$ annihilated by $f^ n$ and we put

If $M = M[f^\infty ]$, we say that $M$ is an $f$-power torsion module.

Lemma 15.81.2. Let $R$ be a ring, let $f \in R$ be an element, and let $R \to R'$ be a ring map which induces isomorphisms $R/f^ nR \to R'/f^ nR'$ for $n > 0$. For any $f$-power torsion $R$-module $M$ the map $M \to M \otimes _ R R'$ is an isomorphism. For example, we have $M \cong M \otimes _ R R^\wedge $.

**Proof.**
If $M$ is annihilated by $f^ n$, then

Since $M = \bigcup M[f^ n]$ and since tensor products commute with direct limits (Algebra, Lemma 10.11.9), we obtain the desired isomorphism. The last statement is a special case of the first statement by Lemma 15.81.1. $\square$

Lemma 15.81.3. 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 15.81.10.

**Proof.**
If $M \neq 0$ but $M \otimes _ R R_ f = 0$, then $M$ is $f$-power torsion. By Lemma 15.81.2 we find that $M \otimes _ R R' \cong M \neq 0$. The last statement is a special case of the first statement by Lemma 15.81.1.
$\square$

Lemma 15.81.4. 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 10.16 and especially Lemmas 10.16.7 and 10.16.6. 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 15.81.1.
$\square$

Lemma 15.81.5. 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 15.81.3 to this cokernel. The last statement is a special case of the first statement by Lemma 15.81.1.
$\square$

Remark 15.81.6. While $R \to R_ f$ is always flat, $R \to R^\wedge $ is typically not flat unless $R$ is Noetherian (see Algebra, Lemma 10.96.2 and the discussion in Examples, Section 102.11). Consequently, we cannot in general apply faithfully flat descent as discussed in Descent, Section 34.3 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

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 15.81.1. Thus $(R, f)$ is a glueing pair if and only if the sequence

is exact.

Lemma 15.81.7. 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 (15.81.6.1) 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 15.81.1. $\square$

Remark 15.81.8. Suppose that $f$ is a nonzerodivisor. Then Algebra, Lemma 10.95.4 shows that $f$ is a nonzerodivisor in $R^\wedge $. Hence $(R, f)$ is a glueing pair.

Remark 15.81.9. 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 15.81.2 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 10.96.2.

Example 15.81.10. 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 15.81.9, these are both examples where $R \to R^\wedge $ is not flat.

**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 (15.81.6.1) with $M$ to obtain a sequence

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

is exact.

Lemma 15.81.11. 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 (15.81.10.1) 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 15.81.7 without further mention. The functor $M \otimes _ R -$ is right exact (Algebra, Lemma 10.11.10) 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 (15.81.10.1) is exact in the middle for any $M$ by Algebra, Lemma 10.11.10. 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 15.81.12. 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 15.81.11 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 10.74.2). 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 10.96.2

Example 15.81.13 (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 15.81.14. 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 15.81.2 we have $R[f^ n] \otimes _ R R' = R[f^ n]$ and by Lemma 15.81.7 we see that $R[f^ n] \to R'$ is injective as $(R \to R', f)$ is a glueing pair.
$\square$

Lemma 15.81.15. 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 10.75.2) we may apply Lemma 15.81.14.
$\square$

Lemma 15.81.16. 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 10.75.2) 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 15.81.7 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

by Algebra, Lemma 10.74.2. Here the diagonal arrow is injective. Since the first group $\text{Tor}_1^ R(R', R/R[f^\infty ])$ is zero by Lemma 15.81.15, 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

By Lemma 15.81.2 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

is exact. Consider the category $\text{Glue}(R \to R', f)$ introduced in Remark 15.80.10. 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

and there is a functor

in the reverse direction, see Remark 15.80.10 for the precise definition.

Theorem 15.81.17. 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.**
The functor is fully faithful due to the exactness of (15.81.10.1) for glueable modules, which tells us exactly that $H^0 \circ \text{Can} = \text{id}$ on the full subcategory of glueable modules. Hence it suffices to check essential surjectivity. That is, we must show that an arbitrary glueing datum $(M', M_1, \alpha _1)$ arises from some glueable $R$-module.

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 $\alpha _ i^{-1}(y_ i \otimes 1) = y_ i'/f^ n$ 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.

Put $M = H^0((M', M_1, \alpha _1)) = \mathop{\mathrm{Ker}}(\text{d})$. We obtain an exact sequence of $R$-modules

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 gives rise to the original glueing datum.

Since $f$ is a nonzerodivisor on $M_1$, we have $M[f^\infty ] \cong M'[f^\infty ]$. This yields an exact sequence

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 15.81.16 we have $\text{Tor}_1^ R(R', \mathop{\mathrm{Coker}}(M' \to (M')_ f)) = 0$. The sequence (15.81.17.2) thus remains exact upon tensoring over $R$ with $R'$. Using $\alpha _1$ and Lemma 15.81.2 the resulting exact sequence can be written as

This yields an isomorphism $(M/M[f^\infty ]) \otimes _ R R' \cong M'/M'[f^\infty ]$. This implies that in the diagram

the third vertical arrow is an isomorphism. Since the rows are exact and the first vertical arrow is an isomorphism by Lemma 15.81.2 and $M[f^\infty ] = M'[f^\infty ]$, the five lemma implies that $M \otimes _ R R' \to M'$ is an isomorphism. This completes the proof. $\square$

Remark 15.81.18. 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 15.81.17. From the exactness of the sequences

and

we conclude that the map $M \to \tilde M$ is surjective.

Recall that flat $R$-modules over a glueing pair $(R \to R', f)$ are glueable (Remark 15.81.12). Hence the following lemma shows that Theorem 15.81.17 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 15.81.19. 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 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 10.38.7. 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 15.81.18. If $\tilde M$ is flat over $R$, then applying Algebra, Lemma 10.38.12 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 15.81.3 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 10.74.8.

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

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

by the spectral sequence of Example 15.60.2. Writing $M \otimes _ R R'$ as a filtered colimit of finite free $R'$-modules (Algebra, Theorem 10.80.4) 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

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 15.81.5.

Lemma 15.81.20. 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 10.77.2 without further mention. By Lemma 15.81.19 we see that $M$ is flat. By Lemma 15.81.5 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 10.38.12) 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 15.81.21. In [Beauville-Laszlo] it is assumed that $f$ is a nonzerodivisor in $R$ and $R' = R^\wedge $, which gives a glueing pair by Lemma 15.81.7. Even in this setting Theorem 15.81.17 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 15.81.11). Lemma 15.81.20 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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