Lemma 10.24.1. Let $R$ be a ring. Let $f_1, \ldots , f_ n$ be elements of $R$ generating the unit ideal. Let $M$ be an $R$-module. The sequence

\[ 0 \to M \xrightarrow {\alpha } \bigoplus \nolimits _{i = 1}^ n M_{f_ i} \xrightarrow {\beta } \bigoplus \nolimits _{i, j = 1}^ n M_{f_ i f_ j} \]

is exact, where $\alpha (m) = (m/1, \ldots , m/1)$ and $\beta (m_1/f_1^{e_1}, \ldots , m_ n/f_ n^{e_ n}) = (m_ i/f_ i^{e_ i} - m_ j/f_ j^{e_ j})_{(i, j)}$.

**Proof.**
It suffices to show that the localization of the sequence at any maximal ideal $\mathfrak m$ is exact, see Lemma 10.23.1. Since $f_1, \ldots , f_ n$ generate the unit ideal, there is an $i$ such that $f_ i \not\in \mathfrak m$. After renumbering we may assume $i = 1$. Note that $(M_{f_ i})_\mathfrak m = (M_\mathfrak m)_{f_ i}$ and $(M_{f_ if_ j})_\mathfrak m = (M_\mathfrak m)_{f_ if_ j}$, see Proposition 10.9.11. In particular $(M_{f_1})_\mathfrak m = M_\mathfrak m$ and $(M_{f_1 f_ i})_\mathfrak m = (M_\mathfrak m)_{f_ i}$, because $f_1$ is a unit. Note that the maps in the sequence are the canonical ones coming from Lemma 10.9.7 and the identity map on $M$. Having said all of this, after replacing $R$ by $R_\mathfrak m$, $M$ by $M_\mathfrak m$, and $f_ i$ by their image in $R_\mathfrak m$, and $f_1$ by $1 \in R_\mathfrak m$, we reduce to the case where $f_1 = 1$.

Assume $f_1 = 1$. Injectivity of $\alpha $ is now trivial. Let $m = (m_ i) \in \bigoplus _{i = 1}^ n M_{f_ i}$ be in the kernel of $\beta $. Then $m_1 \in M_{f_1} = M$. Moreover, $\beta (m) = 0$ implies that $m_1$ and $m_ i$ map to the same element of $M_{f_1f_ i} = M_{f_ i}$. Thus $\alpha (m_1) = m$ and the proof is complete.
$\square$

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