Does a module (over a commutative ring) always possess a minimal generating set? When the module is not finitely generated, the typical Zorn's lemma type argument doesn't seem to work. More precisely, if $(S_{\alpha})_{\alpha}$ is a chain of generating subsets of a module $M$, $M=\cap _ {\alpha}(S _ {\alpha})\supset (\cap_{\alpha}S_{\alpha})$ is not in general (I think) equality (the parentheses in the last equation indicate submodule generated by).
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$\mathbb{Q}$, as a $\mathbb{Z}$-module, has no minimal generating set.
By the way, in the paper "A characterization of left perfect rings" by Yiqiang Zhou, it is proven that a ring $R$ is left perfect if and only if every generating set of some $R$-module contains a minimal generating set.
answered Jul 27, 2010 at 16:27
Martin Brandenburg
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$\begingroup$ Wouldn't {$\frac{1}{p} \ $|$\ p \mbox{ a positive prime integer }$ } be a minimal generating set of $\mathbb{Q}$? $\endgroup$ashpool– ashpool2010-07-27 16:49:36 +00:00Commented Jul 27, 2010 at 16:49
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4$\begingroup$ @kwan: No; e.g., $\frac{1}{4}$ is not in the subgroup generated by your system of elements. Note that a somewhat easier example to think about is $\mathbb{Q}_p$ as a $\mathbb{Z}_p$-module: in this case, a generating set is a collection of elements of arbitrarily small $p$-adic valuation, but given any such set, any finite number of elements may be removed while retaining a generating set. $\endgroup$Pete L. Clark– Pete L. Clark2010-07-27 17:08:34 +00:00Commented Jul 27, 2010 at 17:08
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$\begingroup$ Your remark about perfect rings is interesting! Thanks. $\endgroup$user717– user7172010-07-27 17:33:39 +00:00Commented Jul 27, 2010 at 17:33
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$\begingroup$ Or take $\mathbb{Q}_p/\mathbb{Z}_p$ as a $\mathbb{Z}$-module: every proper submodule is cyclic, but the module itself is not finitely generated. $\endgroup$Victor Protsak– Victor Protsak2010-07-27 18:19:28 +00:00Commented Jul 27, 2010 at 18:19
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There are also simple examples of ideals of commutative rings that have no minimal generating sets, e.g., any nonzero ideal $I$ with $I = J(R)I$. (Adapt the proof of Nakayama's Lemma.) For example, let $K$ be a field, $M$ be the maximal ideal of $K[\{x^s \mid s \in \mathbb{Q}^+\}]$ consisting of the elements with zero constant term, and note that $M_M = (M_M)^2$.
Since I don't have enough points to comment yet, I would also like to point out here that there is an erratum for the paper Martin referenced where the author points out that his proof is flawed. As far as I know, it is known that that condition implies perfect, but whether the converse is true is still open.
