In this section we show that given an open covering
by standard opens, and given an element $h_ i \in R_{f_ i}$ for each $i$ such that $h_ i = h_ j$ as elements of $R_{f_ i f_ j}$ then there exists a unique $h \in R$ such that the image of $h$ in $R_{f_ i}$ is $h_ i$. This result can be interpreted in two ways:
The rule $D(f) \mapsto R_ f$ is a sheaf of rings on the standard opens, see Sheaves, Section 6.30.
If we think of elements of $R_ f$ as the “algebraic” or “regular” functions on $D(f)$, then these glue as would continuous, resp. differentiable functions on a topological, resp. differentiable manifold.
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 is exact, where $\alpha (m) = (m/1, \ldots , m/1)$ and $\beta (m_1/f_1^{e_1}, \ldots , m_ n/f_ n^{e_ n}) = (m_ i/f_ i^{e_ i} - m_ j/f_ j^{e_ j})_{(i, j)}$.
\[ 0 \to M \xrightarrow {\alpha } \bigoplus \nolimits _{i = 1}^ n M_{f_ i} \xrightarrow {\beta } \bigoplus \nolimits _{i, j = 1}^ n M_{f_ i f_ j} \]
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. $\square$
Lemma 10.24.2. Let $R$ be a ring, and let $f_1, f_2, \ldots f_ n\in R$ generate the unit ideal in $R$. Then the following sequence is exact: where the maps $\alpha : R \longrightarrow \bigoplus _ i R_{f_ i}$ and $\beta : \bigoplus _ i R_{f_ i} \longrightarrow \bigoplus _{i, j} R_{f_ if_ j}$ are defined as
\[ 0 \longrightarrow R \longrightarrow \bigoplus \nolimits _ i R_{f_ i} \longrightarrow \bigoplus \nolimits _{i, j}R_{f_ if_ j} \]
\[ \alpha (x) = \left(\frac{x}{1}, \ldots , \frac{x}{1}\right) \text{ and } \beta \left(\frac{x_1}{f_1^{r_1}}, \ldots , \frac{x_ n}{f_ n^{r_ n}}\right) = \left(\frac{x_ i}{f_ i^{r_ i}}-\frac{x_ j}{f_ j^{r_ j}}~ \text{in}~ R_{f_ if_ j}\right). \]
Proof. Special case of Lemma 10.24.1. $\square$
The following we have already seen above, but we state it explicitly here for convenience.
Lemma 10.24.3. Let $R$ be a ring. If $\mathop{\mathrm{Spec}}(R) = U \amalg V$ with both $U$ and $V$ open then $R \cong R_1 \times R_2$ with $U \cong \mathop{\mathrm{Spec}}(R_1)$ and $V \cong \mathop{\mathrm{Spec}}(R_2)$ via the maps in Lemma 10.21.2. Moreover, both $R_1$ and $R_2$ are localizations as well as quotients of the ring $R$.
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. $\square$
Lemma 10.24.4. Let $R$ be a ring. Let $f_1, \ldots , f_ n \in R$. Let $M$ be an $R$-module. Then $M \to \bigoplus M_{f_ i}$ is injective if and only if is injective.
\[ M \longrightarrow \bigoplus \nolimits _{i = 1, \ldots , n} M, \quad m \longmapsto (f_1m, \ldots , f_ nm) \]
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. $\square$
The following lemma is better stated and proved in the more general context of flat descent. However, it makes sense to state it here since it fits well with the above.
Lemma 10.24.5. Let $R$ be a ring. Let $f_1, \ldots , f_ n \in R$. Suppose we are given the following data:
For each $i$ an $R_{f_ i}$-module $M_ i$.
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
commute (for all triples $i, j, k$). Given this data define
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 isomorphism $M_{f_ i} \to M_ i$. 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 10.23.1. 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. $\square$
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