Exercise 11, § 3

a*) Let $A$ be a ring. Now prove that the following properties are equivalent: $\alpha$ The $A$-module $A_s$ is coherent (exercise 10). $\beta$ Every finitely presented $A$-module is coherent. $\gamma$ For every finitely presented right $A$-module $M$, the dual $M^*$ is a finitely generated $A$-module. $\delta$ For every set $I$, the right $A$-module $A_d^I$ is flat. $\varepsilon$ Every product of flat right $A$-modules is flat. (To prove that $\alpha$ implies $\beta$), use exercise 10, b). To see that $\delta$ implies $\gamma$), use $X$, p. 14, th. 1; to show that $\alpha$ implies $\varepsilon$), use $X$, p. 170, exercise 18.) An ring satisfying these properties is said to be left coherent (or simply coherent), and one defines analogously the notion of a right coherent ring. b) Now prove that a noetherian (left) ring is coherent. Give an example of a right artinian ring which is not coherent (cf. VIII, § 1, exercise 4). c) Now prove that an absolutely flat ring ($X$, p. 169, exercise 16) is coherent. d) Now prove that, if $A$ is a coherent ring and $M$ an $A$-module, one has $\lambda(M) = -1$ or $\lambda(M) = 0$ or $\lambda(M) = +\infty$ (exercise 6). In particular, every finitely presented $A$-module admits a resolution by finitely generated free modules. e) Let $(A_\alpha, \varphi_{\beta\alpha})$ be an inductive system of rings whose set of indices is filtered, and let $A = \varprojlim A_\alpha$. Suppose that for $\alpha \leqslant \beta$, $A_\beta$ is a flat right $A_\alpha$-module. Now prove that if the $A_\alpha$ are coherent, the same is true of $A$. (Observe that $A$ is a flat right $A_\alpha$-module for every $\alpha$, and that for every left ideal $a \subset A$, of finite type, there exists an index $\alpha$ and a left ideal $a_\alpha$ of $A$ such that $a$ is isomorphic to $A \otimes_A a_\alpha$.) f) Deduce from e) that every polynomial ring (in a finite or arbitrary infinite set of indeterminates) over a commutative noetherian ring is coherent. Deduce from this that a quotient ring of a coherent ring is not necessarily coherent. g) In order that $A$ be left coherent, it is necessary and sufficient that the left annihilator of every element of $A$ be of finite type, and that the intersection of two left ideals of finite type be of finite type (use exercise 6, f)).