## 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.

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

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.

A finite submodule of a coherent module is coherent.

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

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

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.
$\square$

Lemma 10.89.5. A Noetherian ring is a coherent ring.

**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

$R$ is coherent,

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

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)

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