Section 10.23 (00EN): Glueing properties

10.23 Glueing properties

In this section we put a number of standard results of the form: if something is true for all members of a standard open covering then it is true. In fact, it often suffices to check things on the level of local rings as in the following lemma.

Lemma 10.23.1. Let $R$ be a ring.

For an element $x$ of an $R$ -module $M$ the following are equivalent
1. $x = 0$ ,
2. $x$ maps to zero in $M_\mathfrak p$ for all $\mathfrak p \in \mathop{\mathrm{Spec}}(R)$ ,
3. $x$ maps to zero in $M_{\mathfrak m}$ for all maximal ideals $\mathfrak m$ of $R$ .
In other words, the map $M \to \prod _{\mathfrak m} M_{\mathfrak m}$ is injective.
Given an $R$ -module $M$ the following are equivalent
1. $M$ is zero,
2. $M_{\mathfrak p}$ is zero for all $\mathfrak p \in \mathop{\mathrm{Spec}}(R)$ ,
3. $M_{\mathfrak m}$ is zero for all maximal ideals $\mathfrak m$ of $R$ .
Given a complex $M_1 \to M_2 \to M_3$ of $R$ -modules the following are equivalent
1. $M_1 \to M_2 \to M_3$ is exact,
2. for every prime $\mathfrak p$ of $R$ the localization $M_{1, \mathfrak p} \to M_{2, \mathfrak p} \to M_{3, \mathfrak p}$ is exact,
3. for every maximal ideal $\mathfrak m$ of $R$ the localization $M_{1, \mathfrak m} \to M_{2, \mathfrak m} \to M_{3, \mathfrak m}$ is exact.
Given a map $f : M \to M'$ of $R$ -modules the following are equivalent
1. $f$ is injective,
2. $f_{\mathfrak p} : M_\mathfrak p \to M'_\mathfrak p$ is injective for all primes $\mathfrak p$ of $R$ ,
3. $f_{\mathfrak m} : M_\mathfrak m \to M'_\mathfrak m$ is injective for all maximal ideals $\mathfrak m$ of $R$ .
Given a map $f : M \to M'$ of $R$ -modules the following are equivalent
1. $f$ is surjective,
2. $f_{\mathfrak p} : M_\mathfrak p \to M'_\mathfrak p$ is surjective for all primes $\mathfrak p$ of $R$ ,
3. $f_{\mathfrak m} : M_\mathfrak m \to M'_\mathfrak m$ is surjective for all maximal ideals $\mathfrak m$ of $R$ .
Given a map $f : M \to M'$ of $R$ -modules the following are equivalent
1. $f$ is bijective,
2. $f_{\mathfrak p} : M_\mathfrak p \to M'_\mathfrak p$ is bijective for all primes $\mathfrak p$ of $R$ ,
3. $f_{\mathfrak m} : M_\mathfrak m \to M'_\mathfrak m$ is bijective for all maximal ideals $\mathfrak m$ of $R$ .

Proof. Let $x \in M$ as in (1). Let $I = \{ f \in R \mid fx = 0\}$ . It is easy to see that $I$ is an ideal (it is the annihilator of $x$ ). Condition (1)(c) means that for all maximal ideals $\mathfrak m$ there exists an $f \in R \setminus \mathfrak m$ such that $fx =0$ . In other words, $V(I)$ does not contain a closed point. By Lemma 10.17.2 we see $I$ is the unit ideal. Hence $x$ is zero, i.e., (1)(a) holds. This proves (1).

Part (2) follows by applying (1) to all elements of $M$ simultaneously.

Proof of (3). Let $H$ be the homology of the sequence, i.e., $H = \mathop{\mathrm{Ker}}(M_2 \to M_3)/\mathop{\mathrm{Im}}(M_1 \to M_2)$ . By Proposition 10.9.12 we have that $H_\mathfrak p$ is the homology of the sequence $M_{1, \mathfrak p} \to M_{2, \mathfrak p} \to M_{3, \mathfrak p}$ . Hence (3) is a consequence of (2).

Parts (4) and (5) are special cases of (3). Part (6) follows formally on combining (4) and (5). $\square$

Lemma 10.23.2.slogan Let $R$ be a ring. Let $M$ be an $R$ -module. Let $S$ be an $R$ -algebra. Suppose that $f_1, \ldots , f_ n$ is a finite list of elements of $R$ such that $\bigcup D(f_ i) = \mathop{\mathrm{Spec}}(R)$ , in other words $(f_1, \ldots , f_ n) = R$ .

If each $M_{f_ i} = 0$ then $M = 0$ .
If each $M_{f_ i}$ is a finite $R_{f_ i}$ -module, then $M$ is a finite $R$ -module.
If each $M_{f_ i}$ is a finitely presented $R_{f_ i}$ -module, then $M$ is a finitely presented $R$ -module.
Let $M \to N$ be a map of $R$ -modules. If $M_{f_ i} \to N_{f_ i}$ is an isomorphism for each $i$ then $M \to N$ is an isomorphism.
Let $0 \to M'' \to M \to M' \to 0$ be a complex of $R$ -modules. If $0 \to M''_{f_ i} \to M_{f_ i} \to M'_{f_ i} \to 0$ is exact for each $i$ , then $0 \to M'' \to M \to M' \to 0$ is exact.
If each $R_{f_ i}$ is Noetherian, then $R$ is Noetherian.
If each $S_{f_ i}$ is a finite type $R$ -algebra, so is $S$ .
If each $S_{f_ i}$ is of finite presentation over $R$ , so is $S$ .

Proof. We prove each of the parts in turn.

$\square$

Lemma 10.23.3. Let $R \to S$ be a ring map. Suppose that $g_1, \ldots , g_ n$ is a finite list of elements of $S$ such that $\bigcup D(g_ i) = \mathop{\mathrm{Spec}}(S)$ in other words $(g_1, \ldots , g_ n) = S$ .

If each $S_{g_ i}$ is of finite type over $R$ , then $S$ is of finite type over $R$ .
If each $S_{g_ i}$ is of finite presentation over $R$ , then $S$ is of finite presentation over $R$ .

Proof. Choose $h_1, \ldots , h_ n \in S$ such that $\sum h_ i g_ i = 1$ .

Proof of (1). For each $i$ choose a finite list of elements $x_{i, j} \in S_{g_ i}$ , $j = 1, \ldots , m_ i$ which generate $S_{g_ i}$ as an $R$ -algebra. Write $x_{i, j} = y_{i, j}/g_ i^{n_{i, j}}$ for some $y_{i, j} \in S$ and some $n_{i, j} \ge 0$ . Consider the $R$ -subalgebra $S' \subset S$ generated by $g_1, \ldots , g_ n$ , $h_1, \ldots , h_ n$ and $y_{i, j}$ , $i = 1, \ldots , n$ , $j = 1, \ldots , m_ i$ . Since localization is exact (Proposition 10.9.12), we see that $S'_{g_ i} \to S_{g_ i}$ is injective. On the other hand, it is surjective by our choice of $y_{i, j}$ . The elements $g_1, \ldots , g_ n$ generate the unit ideal in $S'$ as $h_1, \ldots , h_ n \in S'$ . Thus $S' \to S$ viewed as an $S'$ -module map is an isomorphism by Lemma 10.23.2.

Proof of (2). We already know that $S$ is of finite type. Write $S = R[x_1, \ldots , x_ m]/J$ for some ideal $J$ . For each $i$ choose a lift $g'_ i \in R[x_1, \ldots , x_ m]$ of $g_ i$ and we choose a lift $h'_ i \in R[x_1, \ldots , x_ m]$ of $h_ i$ . Then we see that

$S_{g_ i} = R[x_1, \ldots , x_ m, y_ i]/(J_ i + (1 - y_ ig'_ i))$

where $J_ i$ is the ideal of $R[x_1, \ldots , x_ m, y_ i]$ generated by $J$ . Small detail omitted. By Lemma 10.6.3 we may choose a finite list of elements $f_{i, j} \in J$ , $j = 1, \ldots , m_ i$ such that the images of $f_{i, j}$ in $J_ i$ and $1 - y_ ig'_ i$ generate the ideal $J_ i + (1 - y_ ig'_ i)$ . Set

$S' = R[x_1, \ldots , x_ m]/\left(\sum h'_ ig'_ i - 1, f_{i, j}; i = 1, \ldots , n, j = 1, \ldots , m_ i\right)$

There is a surjective $R$ -algebra map $S' \to S$ . The classes of the elements $g'_1, \ldots , g'_ n$ in $S'$ generate the unit ideal and by construction the maps $S'_{g'_ i} \to S_{g_ i}$ are injective. Thus we conclude as in part (1). $\square$

Comments (4)

Comment #3234 by Jiachang Xu on March 26, 2018 at 17:11

For Lemma 10.23.4., I think we could remove the condition
$f_1, \ldots, f_n \in R$ be elements which generate the unit ideal. Or add $M$ is the unique module have this property

Comment #3333 by Johan on May 19, 2018 at 14:53

Dear Jiachang Xu, yes I think you are right. OK, I decided to actually go ahead and provide a proof as well as add some earlier missing proofs. See here for the changes.

Comment #6672 by Ivan on October 28, 2021 at 19:47

Is it true that (7) only need it to be finite type $R_{f_i}$ -algebra?

Comment #6886 by Johan on January 18, 2022 at 14:11

@#6672: OK, if we have $A \to A_f \to B$ then $B$ is of finite type over $A$ if and only if $B$ is of finite type over $A_f$ , see Lemma 10.6.2.

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10.23 Glueing properties

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