Exercise 6, § 3

Let $M$ be a left $A$-module. One calls presentation of length $n$ or $n$-presentation of $M$ an exact sequence

where $L_i$ is a free left $A$-module ($0 \leqslant i \leqslant n$). One says that the presentation is finite if all the $L_i$ are free modules of finitely generated type.

If $M$ is a finitely generated left $A$-module, one denotes by $\lambda(M)$ the supremum (finite or equal to $+\infty$) of the integers $n \geqslant 0$ such that $M$ possesses a finite $n$-presentation. If $M$ is not of finite type, one sets $\lambda(M) = -1$.

a) Let $0 \to P \to N \to M \to 0$ be an exact sequence of left $A$-modules. Then one has $\lambda(N) \geqslant \inf (\lambda(P), \lambda(M))$.

b) Let $L_n \xrightarrow{d_n} L_{n-1} \to \cdots \to L_0 \to M \to 0$ be a finite $n$-presentation of $M$; if $\lambda(M) > n$, now prove that $\operatorname{Ker}(d_n)$ is a finitely generated $A$-module (use exercise 5, b)).

c) Now prove that, under the hypotheses of a), one has

(If $n \leqslant \inf (\lambda(N), \lambda(P) + 1)$), now prove by induction on $n$ that $\lambda(M) \geqslant n$, reasoning as in a) and using b).)

d) Now prove that, under the hypotheses of a), one has

(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 $N = M \oplus P$, one has

In particular, in order that $N$ admit a finite presentation it is necessary and sufficient that this be so for $M$ and $P$.

f) Let $N_1, N_2$ be two submodules of an $A$-module $M$. Suppose that $N_1$ and $N_2$ admit a finite presentation. In order that $N_1 + N_2$ admit a finite presentation, it is necessary and sufficient that $N_1 \cap N_2$ be of finite type.