Bourbaki

Exercise 22, § 4

Algebra, VII, p. 68

Bourbaki marks this one as harder than the rest.

A module M over a principal ideal domain is said to have finite rank if there exists an integer m≥0m \geq 0 with the following property: every finitely generated submodule of M is contained in a submodule generated by m elements or fewer. The smallest integer mm with this property is called the rank of M. a) Show that if M has finite rank, then every quotient module has finite rank, at most equal to that of M. b) For a torsion-free module M, the notion of rank defined above coincides with that defined in II, p. 315. c) Show that if M has finite rank then every submodule of M has finite rank, at most equal to that of M. d) Let M be a π\pi-primary module such that M(π)=MM(\pi) = M (in the notation of VII, p. 7). Show that if M has finite rank then it is finitely generated. e) Let M be a π\pi-primary module with no element of infinite height. Show that if M has finite rank then it is finitely generated, and the notion of rank coincides with that defined in

Ex. 8. (Notice that if M(π)M(\pi) is finitely generated then so is MM, by proving that the heights of elements of M(π)M(\pi) in MM are bounded ; use Ex. 5 of VII, p. 55 to show this.) f) Let MM be a τ\tau-primary module of finite rank rr, and let MsM_s denote the submodule consisting of all the elements of infinite height in MM. Show that MsM_s is a divisible module, and the direct sum of a finite number p≤rp \leq r of indecomposable submodules UπU_{\pi} (VII, p. 54, Ex. 3), and that MM is the direct sum of M0M_0 and a finitely generated submodule NN of rank r−pr - p. (Notice that M/M0M/M_0 is finitely generated, using e) ; then apply Ex. 3 of VII, p. 54.) g) Deduce from the above that an AA-module MM has finite rank rr if and only if the quotient module M/TM/T of MM by its torsion submodule TT has finite rank p≤rp \leq r, and that each π\pi-primary component of TT has rank ≤r−p\leq r - p, with at least one of them having rank r−pr - p. h) Let (pn)(p_n) be the sequence of distinct prime numbers ; let EE be the direct sum of the cyclic Z\mathbf{Z}-modules Z/pn2Z\mathbf{Z}/p_n^2\mathbf{Z}, and let ene_n denote the element whose coordinates all vanish except the n-th, which is equal to the residue class of 3 mod pn2p_n^2. Let MM be the submodule of the Z\mathbf{Z}-module E×QE \times \mathbf{Q} generated by the elements (en,1/pn)(e_n, 1/p_n). Show that MM is a Z\mathbf{Z}-module of rank 2, that the torsion submodule TT of MM is generated by the elements (pnen,0)(p_n e_n, 0), and that TT is not a direct factor of MM. (If xˉn\bar{x}_n is the class of (en,1/pn)(e_n, 1/p_n) mod TT, show that for all m≠nm \neq n there exists a class yˉnm\bar{y}_{nm} mod TT such that xˉn=pmyˉnm\bar{x}_n = p_m \bar{y}_{nm}, but that there is no element xnx_n in the class xˉn\bar{x}_n such that for all m≠nm \neq n there exists yn∈My_n \in M with xn=pmynmx_n = p_m y_{nm}.) i) If MM is a finitely generated AA-module, show that the rank of MM is equal to the smallest cardinal γ(M)\gamma(M) of any generating set for MM. j) Let MM be a torsion AA-module of finite rank. Show that if NN is a proper submodule of MM then NN cannot be isomorphic to MM (use f)).

Solution

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Transcribed from the printed volume. Not translated.