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I would like to prove the following:

Let $M$ be a non-finitely generated right module over a ring $R$, and presume axiom of choice is true. Then there exists a subset of $M$, such that it generates $M$, and it's minimal as far as subsets that generate $M$ go.

The discussion that follows in the comments in the link above didn't clear up much, as it seemed too advanced for me. Also I'm not too interested in whether full choice is needed etc., I'm fine with presuming full choice.

Here's what I've done so far. The idea is we use Zorn's lemma on subsets of linearly independent elements. We get a maximal element, a subset $A$ of independent elements. If we could show that this subset generates the whole module, we would be done, because any proper subset would not generate the whole module (if that were not true, we would get a contradiction with independence of $A$).

However, I don't know how to show that $A$ actually generates the whole module. Presuming it's not, if we take some element $c$ that's not in $A$, it doesn't seem like I can simply add $c$ to $A$ (in hopes of getting a contradiction with maximality of $A$), because I don't really know that $cR \cap A=0$.

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This isn't true without some additional hypotheses. For instance, if $R=\mathbb{Z}$ and $p$ is a prime, the module $\mathbb{Z}[1/p]/\mathbb{Z}$ has no minimal generating set (a generating set must contain elements of order $p^n$ for arbitrarily large $n$, but an element of order $p^n$ generates the entire submodule of elements of order $\leq p^n$ and thus any individual generator is redundant).

On the other hand, it is true if $R$ is commutative and artinian. Suppose $R$ is commutative and artinian and let $M$ be any $R$-module. Then $M$ is equal to the direct sum of its localizations at each of the finitely many primes of $R$, so by considering each direct summand separately we may assume $R$ is local with maximal ideal $p$. Choose a basis for $M/pM$ over $R/p$ and lift it to a set $B$ of elements of $M$. Clearly no proper subset of $B$ can generate $M$ (since it does not generate $M/pM$), so it suffices to show that $B$ generates $M$.

Let $N\subseteq M$ be the submodule generated by $B$. Since $B$ generates $M/pM$, we have $N+pM=M$. But then we get $$M=N+pM=N+p(N+pM)=N+p^2M=N+p^2(N+pM)=N+p^3M$$ and so on by induction. Thus for all $n$, $M=N+p^nM$. But since $R$ is artinian, $p^n=0$ for $n$ sufficiently large. For such $n$ we have $$M=N+p^nM=N+0=N.$$ That is, $B$ generates $M$.

While the context of the linked discussion was modules over a commutative artinian ring, I doubt this is the argument Tobias Kildetoft had in mind when he commented that "the existence of such a minimal generating set in general can be obtained from Zorn's lemma in a similar way to the one for vector spaces". I'm guessing he instead had an erroneous argument that involved taking a maximal irredundant set of generators (i.e., a set of elements such that if you remove any one, it generates a smaller submodule). The problem with this argument is that there is no reason such a maximal set has to actually generate the whole module. You might think you can always add one more generator if you don't have the whole module, but adding one more generator might make some of your previous generators redundant.

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  • $\begingroup$ You are pretty much spot on in your assessment of my error (at least as far as I recall, since that post is quite old and I should have deleted it a long time ago). $\endgroup$ – Tobias Kildetoft Nov 21 '16 at 10:35
  • $\begingroup$ Thanks for the answer and edit. I don't really understand your proof, since I'm not sure what local, localization and prime means, in this context. Just to save me some time, could I ask for a link or book, where I'll find these definitions used as in your proof? Also, presuming the ring is artinian, it should also follow from Hopkins theorem, right? $\endgroup$ – John P Nov 21 '16 at 15:17
  • $\begingroup$ I'm not quite sure whether it shouldn't say $\mathbb{Z}_{p}/\mathbb{Z}$ as opposed to $\mathbb{Z}_{(p)}/\mathbb{Z}$ - I understood your example as the Prüfer group. I also deleted the first two lines in my original post with the link to the other thread. I'm not sure whether there's something wrong with that, but it only seemed to be unnecessarily confusing at this point. I'll delete this comment later. $\endgroup$ – John P Nov 21 '16 at 15:17
  • $\begingroup$ Take a look at any textbook on commutative algebra (e.g., Matsumura's Commutative Algebra) for all the definitions and facts about commutative artinian rings I'm using here. I'm not sure what you mean by Hopkins theorem, or how that would help here. And yes, sorry, $\mathbb{Z}_{(p)}$ should have been $\mathbb{Z}[1/p]$; I've fixed that now. $\endgroup$ – Eric Wofsey Nov 21 '16 at 20:04
  • $\begingroup$ You're right. Hopkins theorem doesn't help here, I got confused. $\endgroup$ – John P Nov 22 '16 at 5:35

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