Slight generalization of the main theorem of .

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
$$\label{more-algebra-equation-define-M} 0 \to M \to M' \oplus M_1 \to (M')_ f \to 0.$$

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
$$\label{more-algebra-equation-exact-mod-torsion} 0 \to M/M[f^\infty ] \to M_1 \to (M')_ f/M' \to 0.$$

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
$$\label{more-algebra-equation-mod-torsion-sequence} 0 \to (M/M[f^\infty ]) \otimes _ R R' \to (M')_ f \to (M')_ f/M' \to 0$$

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. $\square$

Comment #5480 by Takagi Benseki（高城 辨積） on

In the second paragraph of the proof, should $a_{i}^{-1}$ be $a_{1}^{-1}$?

Comment #8685 by Anne on

It doesn’t seem to be clear that $H^0 \circ \mathrm{Can} = id$ suffices to prove that $\mathrm{Can}$ 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 $H^0 \circ Can = id$ and $Can \circ H^0 = id$ according to Definition 4.2.17. The latter equality results from the proof of essential surjectivity.

Comment #9367 by on

Thanks for pointing this out! The functors Can and $H^0$ are adjoint which is why it works; I guess this was at the back of my mind when I wrote this. See changes here.

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