The Stacks project

Proof. The first statement is immediate from the definition.

Let $\varphi : N \to M$ satisfy the assumptions of (2). First, $\mathop{\mathrm{Im}}(\varphi )$ is finite, hence coherent by (1). In particular $\mathop{\mathrm{Im}}(\varphi )$ is finitely presented, so applying Lemma 10.5.3 to the exact sequence $0 \to \mathop{\mathrm{Ker}}(\varphi ) \to N \to \mathop{\mathrm{Im}}(\varphi ) \to 0$ we see that $\mathop{\mathrm{Ker}}(\varphi )$ is finite. To prove that $\mathop{\mathrm{Coker}}(\varphi )$ is coherent, let $E \subset \mathop{\mathrm{Coker}}(\varphi )$ be a finite submodule, and let $E'$ be its inverse image in $M$. From the exact sequence $0 \to \mathop{\mathrm{Im}}(\varphi ) \to E' \to E \to 0$ and since $\mathop{\mathrm{Ker}}(\varphi )$ is finite we conclude by Lemma 10.5.3 that $E' \subset M$ is finite, hence finitely presented because $M$ is coherent. The same exact sequence then shows that $E$ is finitely presented, whence our claim.

Part (3) follows immediately from (1) and (2).

Let $0 \to M_1 \xrightarrow {i} M_2 \xrightarrow {p} M_3 \to 0$ be a short exact sequence of $R$-modules as in (4). It remains 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 $N_2 \subset M_2$ be a finite submodule. Put $N_3 = p(N_2) \subset M_3$ and $N_1 = i^{-1}(N_2) \subset M_1$. We have an exact sequence $0 \to N_1 \to N_2 \to N_3 \to 0$. Clearly $N_3$ is finite (as a quotient of $N_2$), hence finitely presented (as a finite submodule of $M_3$). It follows by Lemma 10.5.3 (5) that $N_1$ is finite, hence finitely presented (as a finite submodule of $M_1$). We conclude by Lemma 10.5.3 (2) that $M_2$ is finitely presented. $\square$

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 10.90.3. $\square$

Proof. By Lemma 10.31.4 any finite $R$-module is finitely presented. In particular any ideal of $R$ is finitely presented. $\square$

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 10.39.5. 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 10.89.3. 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 10.89.2 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 10.89.3. $\square$