## Tag `00EO`

Chapter 10: Commutative Algebra > Section 10.23: More glueing results

**Zariski-local properties of modules and algebras**

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 $\bigcup D(f_i) = \mathop{\rm Spec}(R)$ 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.

- By Proposition 10.9.10 this implies $M_\mathfrak p = 0$ for all $\mathfrak p \in \mathop{\rm Spec}(R)$, so we conclude by Lemma 10.23.1.
- 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 surjective by Lemma 10.23.1 and $M$ is finitely generated.
- 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.
- 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.
- 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.
- 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).
- 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.
- By the previous item, there exists a surjective $R$-algebra homomorphism $R[X_1,...,X_n] \rightarrow S$. Let $K$ be the kernel of this map. This is an ideal in $R[X_1,..X_n]$, 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 $R[X_1,...,X_n]$, so an application of (2) finishes the proof.
$\square$

The code snippet corresponding to this tag is a part of the file `algebra.tex` and is located in lines 4145–4169 (see updates for more information).

```
\begin{lemma}
\label{lemma-cover}
\begin{slogan}
Zariski-local properties of modules and algebras
\end{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 $\bigcup D(f_i) = \Spec(R)$
in other words $(f_1, \ldots, f_n) = R$.
\begin{enumerate}
\item If each $M_{f_i} = 0$ then $M = 0$.
\item If each $M_{f_i}$ is a finite $R_{f_i}$-module,
then $M$ is a finite $R$-module.
\item If each $M_{f_i}$ is a finitely presented $R_{f_i}$-module,
then $M$ is a finitely presented $R$-module.
\item 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.
\item 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.
\item If each $R_{f_i}$ is Noetherian, then $R$ is Noetherian.
\item If each $S_{f_i}$ is a finite type $R$-algebra, so is $S$.
\item If each $S_{f_i}$ is of finite presentation over $R$, so is $S$.
\end{enumerate}
\end{lemma}
\begin{proof}
We prove each of the parts in turn.
\begin{enumerate}
\item By Proposition \ref{proposition-localize-twice}
this implies $M_\mathfrak p = 0$ for all $\mathfrak p \in \Spec(R)$,
so we conclude by Lemma \ref{lemma-characterize-zero-local}.
\item 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 surjective by
Lemma \ref{lemma-characterize-zero-local}
and $M$ is finitely generated.
\item 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 \ref{lemma-extension}.
By (2) this implies that $K$ is a finite $R$-module and therefore
$M$ is finitely presented.
\item By Proposition \ref{proposition-localize-twice}
the assumption implies that the induced morphism
on localizations at all prime ideals is an isomorphism, so we conclude
by Lemma \ref{lemma-characterize-zero-local}.
\item By Proposition \ref{proposition-localize-twice} the assumption
implies that the induced
sequence of localizations at all prime ideals is short exact, so we
conclude by Lemma \ref{lemma-characterize-zero-local}.
\item 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 \ref{proposition-localization-exact}
each $I_{f_i} \subset R_{f_i}$ is an ideal. These are all
finitely generated by assumption, so we conclude by (2).
\item 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.
\item By the previous item, there exists a surjective $R$-algebra
homomorphism $R[X_1,...,X_n] \rightarrow S$. Let $K$ be the kernel
of this map. This is an ideal in $R[X_1,..X_n]$, 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 $R[X_1,...,X_n]$, so an
application of (2) finishes the proof.
\end{enumerate}
\end{proof}
```

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