## 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.

1. 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.

2. 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 $R$.

3. 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.

4. 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 $R$.

5. 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 $R$.

6. 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. 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 $(f_1, \ldots , f_ n) = R$.

1. If each $M_{f_ i} = 0$ then $M = 0$.

2. If each $M_{f_ i}$ is a finite $R_{f_ i}$-module, then $M$ is a finite $R$-module.

3. If each $M_{f_ i}$ is a finitely presented $R_{f_ i}$-module, then $M$ is a finitely presented $R$-module.

4. 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.

5. 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.

6. If each $R_{f_ i}$ is Noetherian, then $R$ is Noetherian.

7. If each $S_{f_ i}$ is a finite type $R$-algebra, so is $S$.

8. If each $S_{f_ i}$ is of finite presentation over $R$, so is $S$.

Proof. We prove each of the parts in turn.

1. By Proposition 10.9.10 this implies $M_\mathfrak p = 0$ for all $\mathfrak p \in \mathop{\mathrm{Spec}}(R)$, so we conclude by Lemma 10.23.1.

2. For each $i$ take a finite generating set $X_ i$ of $M_{f_ i}$. Without loss of generality, we may assume that the elements of $X_ i$ are in the image of the localization map $M \rightarrow M_{f_ i}$, so we take a finite set $Y_ i$ of preimages of the elements of $X_ i$ in $M$. Let $Y$ be the union of these sets. This is still a finite set. Consider the obvious $R$-linear map $R^ Y \rightarrow M$ sending the basis element $e_ y$ to $y$. By assumption this map is surjective after localizing at an arbitrary prime ideal $\mathfrak p$ of $R$, so it is surjective by Lemma 10.23.1 and $M$ is finitely generated.

3. By (2) we have a short exact sequence

$0 \rightarrow K \rightarrow R^ n \rightarrow M \rightarrow 0$

Since localization is an exact functor and $M_{f_ i}$ is finitely presented we see that $K_{f_ i}$ is finitely generated for all $1 \leq i \leq n$ by Lemma 10.5.3. By (2) this implies that $K$ is a finite $R$-module and therefore $M$ is finitely presented.

4. By Proposition 10.9.10 the assumption implies that the induced morphism on localizations at all prime ideals is an isomorphism, so we conclude by Lemma 10.23.1.

5. By Proposition 10.9.10 the assumption implies that the induced sequence of localizations at all prime ideals is short exact, so we conclude by Lemma 10.23.1.

6. We will show that every ideal of $R$ has a finite generating set: For this, let $I \subset R$ be an arbitrary ideal. By Proposition 10.9.12 each $I_{f_ i} \subset R_{f_ i}$ is an ideal. These are all finitely generated by assumption, so we conclude by (2).

7. For each $i$ take a finite generating set $X_ i$ of $S_{f_ i}$. Without loss of generality, we may assume that the elements of $X_ i$ are in the image of the localization map $S \rightarrow S_{f_ i}$, so we take a finite set $Y_ i$ of preimages of the elements of $X_ i$ in $S$. Let $Y$ be the union of these sets. This is still a finite set. Consider the algebra homomorphism $R[X_ y]_{y \in Y} \rightarrow S$ induced by $Y$. Since it is an algebra homomorphism, the image $T$ is an $R$-submodule of the $R$-module $S$, so we can consider the quotient module $S/T$. By assumption, this is zero if we localize at the $f_ i$, so it is zero by (1) and therefore $S$ is an $R$-algebra of finite type.

8. By the previous item, there exists a surjective $R$-algebra homomorphism $R[X_1, \ldots , X_ n] \rightarrow S$. Let $K$ be the kernel of this map. This is an ideal in $R[X_1, \ldots , X_ n]$, finitely generated in each localization at $f_ i$. Since the $f_ i$ generate the unit ideal in $R$, they also generate the unit ideal in $R[X_1, \ldots , X_ n]$, so an application of (2) finishes the proof.

$\square$

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 $(g_1, \ldots , g_ n) = S$.

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

2. 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). $\square$

Comment #3234 by Jiachang Xu on

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

Comment #3333 by on

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

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

Comment #6886 by on

@#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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