The Stacks project

Slight generalization of the main theorem of [Beauville-Laszlo].

Theorem 15.90.17. 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 (15.90.10.1) 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 $\alpha _1^{-1}(y_ i \otimes 1) = y_ i'/f^ n$ 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.

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

15.90.17.1
\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 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

15.90.17.2
\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 15.90.16 we have $\text{Tor}_1^ R(R', \mathop{\mathrm{Coker}}(M' \to (M')_ f)) = 0$. The sequence (15.90.17.2) thus remains exact upon tensoring over $R$ with $R'$. Using $\alpha _1$ and Lemma 15.90.2 the resulting exact sequence can be written as

15.90.17.3
\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 15.90.2 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. $\square$


Comments (4)

Comment #5480 by Takagi Benseki(高城 辨積) on

In the second paragraph of the proof, should be ?

Comment #8685 by Anne on

It doesn’t seem to be clear that suffices to prove that is fully faithful. I thought about using Tag 05ER? But in general, \varphi is not flat.

Comment #8704 by Anne on

Okay, so it is indeed not clear because "fully faithful" has not been shown explicitly. Instead, the proof shows and according to Definition 4.2.17. The latter equality results from the proof of essential surjectivity.

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  • 4 comment(s) on Section 15.90: The Beauville-Laszlo theorem

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