Exercise 6, § 2

Let A be a ring and M a left A-module. An exact sequence

where $L_i$ is a free left A-module $(0 \leq i \leq n)$, is called a presentation of M of length n or an n-presentation of M. The presentation is called finite if all the $L_i$ are finitely generated free modules.

If M is a finitely generated left A-module, we denote by $\lambda(M)$ the least upper bound (finite or $+\infty$) of the integers $n \geq 0$ such that M has a finite $n$-presentation. If M is not finitely generated, we set $\lambda(M) = -1$.

(a) Let $0 \to P \to N \to M \to 0$ be an exact sequence of left A-modules. Then $\lambda(N) \geq \inf(\lambda(P), \lambda(M))$. (Starting with two n-presentations of P and M respectively, derive one for N using Exercise 5 of § 1.)

(b) Let $M, \xrightarrow{u_n} M_{n-1} \to \cdots \to M_0 \xrightarrow{u_0} M \to 0$ be a finite n-presentation of M; show that, if $\lambda(M) > n$, $\operatorname{Ker}(u_n)$ is a finitely generated A-module. (Let

be a finite $(n+1)$-presentation of M and let $P = \operatorname{Ker}(v_0)$ be such that there is an n-presentation of P: Applying the method of (a) to the exact sequence $0 \to P \to L_0 \to M \to 0$ we obtain an exact sequence and exact sequences $0 \to \operatorname{Ker}(v_{i+1}) \to \operatorname{Ker}(w_i) \to \operatorname{Ker}(u_i) \to 0$ ($§ 1$, no. 4, Proposition 2). Observe finally that $\operatorname{Ker}(w_i)$ is a direct factor of $M_i \oplus L_{i+1}$.

(c) Show that with the hypotheses of (a)

(If $n \leq \inf(\lambda(N), \lambda(P) + 1)$, show by induction on $n$ that $\lambda(M) \geq n$, arguing as in (a) and using (b).)

(d) Show that with the hypotheses of (a)

(Same method as in (c).) Deduce that, if $\lambda(N) = +\infty$, then $\lambda(M) = \lambda(P) + 1$.

(e) Deduce from (a), (c) and (d) that, if $N = M \oplus P$, then

In particular, for N to admit a finite presentation it is necessary and sufficient that M and P do so too.

(f) Let $N_1, N_2$ be two submodules of an A-module M. Suppose that $N_1$ and $N_2$ admit finite presentations. For $N_1 + N_2$ to admit a finite presentation it is necessary and sufficient that $N_1 \cap N_2$ be finitely generated.