
Lemma 10.23.5. Let $R$ be a ring. Let $f_1, \ldots , f_ n \in R$. Suppose we are given the following data:

1. For each $i$ an $R_{f_ i}$-module $M_ i$.

2. For each pair $i, j$ an $R_{f_ if_ j}$-module isomorphism $\psi _{ij} : (M_ i)_{f_ j} \to (M_ j)_{f_ i}$.

which satisfy the “cocycle condition” that all the diagrams

$\xymatrix{ (M_ i)_{f_ jf_ k} \ar[rd]_{\psi _{ij}} \ar[rr]^{\psi _{ik}} & & (M_ k)_{f_ if_ j} \\ & (M_ j)_{f_ if_ k} \ar[ru]_{\psi _{jk}} }$

commute (for all triples $i, j, k$). Given this data define

$M = \mathop{\mathrm{Ker}}\left( \bigoplus \nolimits _{1 \leq i \leq n} M_ i \longrightarrow \bigoplus \nolimits _{1 \leq i, j \leq n} (M_ i)_{f_ j} \right)$

where $(m_1, \ldots , m_ n)$ maps to the element whose $(i, j)$th entry is $m_ i/1 - \psi _{ji}(m_ j/1)$. Then the natural map $M \to M_ i$ induces an isorphism $M_{f_ i} \to M_ i$. Moreover $\psi _{ij}(m/1) = m/1$ for all $m \in M$ (with obvious notation).

Proof. To show that $M_{f_1} \to M_1$ is an isomorphism, it suffices to show that its localization at every prime $\mathfrak p'$ of $R_{f_1}$ is an isomorphism, see Lemma 10.22.1. Write $\mathfrak p' = \mathfrak p R_{f_1}$ for some prime $\mathfrak p \subset R$, $f_1 \not\in \mathfrak p$, see Lemma 10.16.6. Since localization is exact (Proposition 10.9.12), we see that

\begin{align*} (M_{f_1})_{\mathfrak p'} & = M_\mathfrak p \\ & = \mathop{\mathrm{Ker}}\left( \bigoplus \nolimits _{1 \leq i \leq n} M_{i, \mathfrak p} \longrightarrow \bigoplus \nolimits _{1 \leq i, j \leq n} ((M_ i)_{f_ j})_\mathfrak p \right) \\ & = \mathop{\mathrm{Ker}}\left( \bigoplus \nolimits _{1 \leq i \leq n} M_{i, \mathfrak p} \longrightarrow \bigoplus \nolimits _{1 \leq i, j \leq n} (M_{i, \mathfrak p})_{f_ j} \right) \end{align*}

Here we also used Proposition 10.9.11. Since $f_1$ is a unit in $R_\mathfrak p$, this reduces us to the case where $f_1 = 1$ by replacing $R$ by $R_\mathfrak p$, $f_ i$ by the image of $f_ i$ in $R_\mathfrak p$, $M$ by $M_\mathfrak p$, and $f_1$ by $1$.

Assume $f_1 = 1$. Then $\psi _{1j} : (M_1)_{f_ j} \to M_ j$ is an isomorphism for $j = 2, \ldots , n$. If we use these isomorphisms to identify $M_ j = (M_1)_{f_ j}$, then we see that $\psi _{ij} : (M_1)_{f_ if_ j} \to (M_1)_{f_ if_ j}$ is the canonical identification. Thus the complex

$0 \to M_1 \to \bigoplus \nolimits _{1 \leq i \leq n} (M_1)_{f_ i} \longrightarrow \bigoplus \nolimits _{1 \leq i, j \leq n} (M_1)_{f_ if_ j}$

is exact by Lemma 10.23.1. Thus the first map identifies $M_1$ with $M$ in this case and everything is clear. $\square$

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