The Stacks project

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. $\square$

Proof. Special case of Lemma 10.24.1. $\square$

Proof. By Lemma 10.21.3 we have $U = D(e)$ and $V = D(1-e)$ for some idempotent $e$. By Lemma 10.24.2 we see that $R \cong R_ e \times R_{1 - e}$ (since clearly $R_{e(1-e)} = 0$ so the glueing condition is trivial; of course it is trivial to prove the product decomposition directly in this case). The lemma follows. $\square$

Proof. The map $M \to \bigoplus M_{f_ i}$ is injective if and only if for all $m \in M$ and $e_1, \ldots , e_ n \geq 1$ such that $f_ i^{e_ i}m = 0$, $i = 1, \ldots , n$ we have $m = 0$. This clearly implies the displayed map is injective. Conversely, suppose the displayed map is injective and $m \in M$ and $e_1, \ldots , e_ n \geq 1$ are such that $f_ i^{e_ i}m = 0$, $i = 1, \ldots , n$. If $e_ i = 1$ for all $i$, then we immediately conclude that $m = 0$ from the injectivity of the displayed map. Next, we prove this holds for any such data by induction on $e = \sum e_ i$. The base case is $e = n$, and we have just dealt with this. If some $e_ i > 1$, then set $m' = f_ im$. By induction we see that $m' = 0$. Hence we see that $f_ i m = 0$, i.e., we may take $e_ i = 1$ which decreases $e$ and we win. $\square$

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 10.23.1. Write $\mathfrak p' = \mathfrak p R_{f_1}$ for some prime $\mathfrak p \subset R$, $f_1 \not\in \mathfrak p$, see Lemma 10.17.6. Since localization is exact (Proposition 10.9.12), we see that

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

is exact by Lemma 10.24.1. Thus the first map identifies $M_1$ with $M$ in this case and everything is clear. $\square$