The Stacks project

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10.89 Coherent rings

We use the discussion on interchanging $\prod $ and $\otimes $ to determine for which rings products of flat modules are flat. It turns out that these are the so-called coherent rings. You may be more familiar with the notion of a coherent $\mathcal{O}_ X$-module on a ringed space, see Modules, Section 17.12.

Definition 10.89.1. Let $R$ be a ring. Let $M$ be an $R$-module.

  1. We say $M$ is a coherent module if it is finitely generated and every finitely generated submodule of $M$ is finitely presented over $R$.

  2. We say $R$ is a coherent ring if it is coherent as a module over itself.

Thus a ring is coherent if and only if every finitely generated ideal is finitely presented as a module.

Example 10.89.2. A valuation ring is a coherent ring. Namely, every nonzero finitely generated ideal is principal (Lemma 10.49.15), hence free as a valuation ring is a domain, hence finitely presented.

The category of coherent modules is abelian.

Lemma 10.89.3. Let $R$ be a ring.

  1. A finite submodule of a coherent module is coherent.

  2. Let $\varphi : N \to M$ be a homomorphism from a finite module to a coherent module. Then $\mathop{\mathrm{Ker}}(\varphi )$ is finite.

  3. Let $\varphi : N \to M$ be a homomorphism of coherent modules. Then $\mathop{\mathrm{Ker}}(\varphi )$ and $\mathop{\mathrm{Coker}}(\varphi )$ are coherent modules.

  4. Given a short exact sequence of $R$-modules $0 \to M_1 \to M_2 \to M_3 \to 0$ if two out of three are coherent so is the third.

Proof. The first statement is immediate from the definition. During the rest of the proof we will use the results of Lemma 10.5.3 without further mention.

Let $\varphi : N \to M$ satisfy the assumptions of (2). Suppose that $N$ is generated by $x_1, \ldots , x_ n$. By Definition 10.89.1 the kernel $K$ of the induced map $R^{\oplus n} \to M$, $e_ i \mapsto \varphi (x_ i)$ is of finite type. Hence $\mathop{\mathrm{Ker}}(\varphi )$ which is the image of the composition $K \to R^{\oplus n} \to N$ is of finite type. This proves (2).

Let $\varphi : N \to M$ satisfy the assumptions of (3). By (2) the kernel of $\varphi $ is of finite type and hence by (1) it is coherent.

With the same hypotheses let us show that $\mathop{\mathrm{Coker}}(\varphi )$ is coherent. Since $M$ is finite so is $\mathop{\mathrm{Coker}}(\varphi )$. Let $\overline{x}_ i \in \mathop{\mathrm{Coker}}(\varphi )$. We have to show that the kernel of the associated morphism $\overline{\Psi } : R^{\oplus n} \to \mathop{\mathrm{Coker}}(\varphi )$ is finite. Choose $x_ i \in M$ lifting $\overline{x}_ i$. Choose additionally generators $y_1, \ldots , y_ m$ of $\mathop{\mathrm{Im}}(\varphi )$. Let $\Phi : R^{\oplus m} \to \mathop{\mathrm{Im}}(\varphi )$ using $y_ j$ and $\Psi : R^{\oplus m} \oplus R^{\oplus n} \to M$ using $y_ j$ and $x_ i$ be the corresponding maps. Consider the following commutative diagram

\[ \xymatrix{ 0 \ar[r] & R^{\oplus m} \ar[d]_\Phi \ar[r] & R^{\oplus m} \oplus R^{\oplus n} \ar[d]_\Psi \ar[r] & R^{\oplus n} \ar[d]_{\overline{\Psi }} \ar[r] & 0 \\ 0 \ar[r] & \mathop{\mathrm{Im}}(\varphi ) \ar[r] & M \ar[r] & \mathop{\mathrm{Coker}}(\varphi ) \ar[r] & 0 } \]

with exact rows. By Lemma 10.4.1 we get an exact sequence $\mathop{\mathrm{Ker}}(\Psi ) \to \mathop{\mathrm{Ker}}(\overline{\Psi }) \to 0$. Since $\mathop{\mathrm{Ker}}(\Psi )$ is a finite $R$-module, we see that $\mathop{\mathrm{Ker}}(\overline{\Psi })$ is finite.

Statement (4) follows from (3).

Let $0 \to M_1 \to M_2 \to M_3 \to 0$ be a short exact sequence of $R$-modules. It suffices to prove that if $M_1$ and $M_3$ are coherent so is $M_2$. By Lemma 10.5.3 we see that $M_2$ is finite. Let $x_1, \ldots , x_ n$ be finitely many elements of $M_2$. We have to show that the module of relations $K$ between them is finite. Consider the following commutative diagram

\[ \xymatrix{ 0 \ar[r] & 0 \ar[r] \ar[d] & \bigoplus _{i = 1}^{n} R \ar[r] \ar[d] & \bigoplus _{i = 1}^{n} R \ar[r] \ar[d] & 0 \\ 0 \ar[r] & M_1 \ar[r] & M_2 \ar[r] & M_3 \ar[r] & 0 } \]

with obvious notation. By the snake lemma we get an exact sequence $0 \to K \to K_3 \to M_1$ where $K_3$ is the module of relations among the images of the $x_ i$ in $M_3$. Since $M_3$ is coherent we see that $K_3$ is a finite module. Since $M_1$ is coherent we see that the image $I$ of $K_3 \to M_1$ is coherent. Hence $K$ is the kernel of the map $K_3 \to I$ between a finite module and a coherent module and hence finite by (2). $\square$

Lemma 10.89.4. Let $R$ be a ring. If $R$ is coherent, then a module is coherent if and only if it is finitely presented.

Proof. It is clear that a coherent module is finitely presented (over any ring). Conversely, if $R$ is coherent, then $R^{\oplus n}$ is coherent and so is the cokernel of any map $R^{\oplus m} \to R^{\oplus n}$, see Lemma 10.89.3. $\square$

Proof. By Lemma 10.30.4 any finite $R$-module is finitely presented. In particular any ideal of $R$ is finitely presented. $\square$

reference

Proposition 10.89.6. Let $R$ be a ring. The following are equivalent

  1. $R$ is coherent,

  2. any product of flat $R$-modules is flat, and

  3. for every set $A$ the module $R^ A$ is flat.

Proof. Assume $R$ coherent, and let $Q_\alpha $, $\alpha \in A$ be a set of flat $R$-modules. We have to show that $I \otimes _ R \prod _\alpha Q_\alpha \to \prod Q_\alpha $ is injective for every finitely generated ideal $I$ of $R$, see Lemma 10.38.5. Since $R$ is coherent $I$ is an $R$-module of finite presentation. Hence $I \otimes _ R \prod _\alpha Q_\alpha = \prod I \otimes _ R Q_\alpha $ by Proposition 10.88.3. The desired injectivity follows as $I \otimes _ R Q_\alpha \to Q_\alpha $ is injective by flatness of $Q_\alpha $.

The implication (2) $\Rightarrow $ (3) is trivial.

Assume that the $R$-module $R^ A$ is flat for every set $A$. Let $I$ be a finitely generated ideal in $R$. Then $I \otimes _ R R^ A \to R^ A$ is injective by assumption. By Proposition 10.88.2 and the finiteness of $I$ the image is equal to $I^ A$. Hence $I \otimes _ R R^ A = I^ A$ for every set $A$ and we conclude that $I$ is finitely presented by Proposition 10.88.3. $\square$


Comments (2)

Comment #818 by Charles on

What is a "finite type" module, as in 10.85.2 and its proof?

Comment #819 by on

This should be a 'finite submodule', i.e., a submodule which is a finite module, i.e., a finitely generated submodule. Thanks for pointing this out. Fixed here. Next time leave your full name and I will add you to the contributors


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