On minimum cardinality of generating sets of modules Let $\alpha$ be an infinite cardinal. Let $M$ be an $R$-module with $\mu (M) < \alpha$ , where $R$ is a commutative ring with unity. Let $\mathcal F$ be the collection of all submodules $N$ of $M$ such that $\mu (N) \ge \alpha$ . If $\mathcal F$ is non-empty, then does that imply that $\mathcal F$ has a maximal element ? (where $\mu (M)$ denotes the minimum cardinality among generating sets of $M$ ) 
The only place where I am stuck while applying Zorn's lemma is that I can't seem to conclude that every totally ordered subset of $\mathcal F$ has an upper bound w.r.t. inclusion. If $\alpha = \aleph_0$ , then I could easily conclude the upper bound because if every member a totally ordered collection $\{N_i\}$ is not finitely generated, then $\cup_i N_i$ is also not finitely generated. But I don't know what to do if we have a general infinte cardinal $\alpha$ . 
 A: To a man with a hammer, everything looks like a nail, and following another recent question of yours on MathOverflow, I have a new hammer. Quite probably it's a bit of a sledgehammer for this question, and I'd be interested in easier examples.
I'll give an example of a cyclic module which has submodules that are not countably generated, but has no maximal such submodule, answering your question in the negative for $\alpha=\aleph_1$.
Let $W$ be a well-ordered set with the order type of $\aleph_\omega=\bigcup_{n\in\mathbb{N}}\aleph_n$, and for $n\in\mathbb{N}$ let $w_n$ be the least element of $W$ such that the initial segment $\{w\in W\mid w<w_n\}$ has cardinality $\aleph_n$. Then $\{w_n\mid n\in\mathbb{N}\}$ is a countable cofinal subset of $W$.
Let $\mathbb{Z}^W$ be the group of functions $W\to\mathbb{Z}$, ordered lexicographically using the well-order on $W$ (i.e., $f<g$ if and only if $f(w)<g(w)$ for the least $w\in W$ with $f(w)\neq g(w)$). Then $\mathbb{Z}^W$ is a totally ordered abelian group.
For $n\in\mathbb{N}$ let $f_n\in\mathbb{Z}^W$ be the function 
$$f_n(w)=\begin{cases}
1&\text{ if }w=w_n\\
0&\text{ otherwise.}
\end{cases}$$
For $0\neq f\in\mathbb{Z}^W$ let $d(f)\in W$ be the least $w\in W$ with $f(w)\neq0$. For some $n$, $w_n>d(f)$ and so if $f>0$ then $f_n\leq f$ for some $n\in\mathbb{N}$.
By a theorem of Krull, every totally ordered abelian group is the value group of some valuation domain. Let $R$ be a valuation domain with value group $\mathbb{Z}^W$, and $\nu:R\setminus\{0\}\to \mathbb{Z}^W$ the associated valuation.
Recall that in a valuation domain $R$, if $x,y\in R$ with $\nu(x)\leq \nu(y)$ then $x$ divides $y$, or equivalently the principal ideal generated by $x$ contains the principal ideal generated by $y$.
Ideals of $R$ correspond to upwards closed subsets $U$ of non-negative elements of $\mathbb{Z}^W$ (i.e., subsets such that if $u\in U$ and $u<v$ then $v\in U$), with $I_U=\{r\in R\mid \nu(r)\in U\}\cup\{0\}$ the ideal corresponding to $U$. [Certainly $I_U$ is an ideal. Conversely, if $I$ is any ideal and $U=\left\{\nu(x)\mid 0\neq x\in I\right\}$ then $I=I_U$.]
The unique maximal ideal $I_{U_{>0}}$ is the ideal corresponding to the set $U_{>0}$ of strictly positive elements of $\mathbb{Z}^W$, and is countably generated by any set $\{r_n\mid n\in\mathbb{N}\}$ of elements of $R$ with $\nu(r_n)=f_n$, since if $0<f\in\mathbb{Z}^W$ we have seen that there is some $n$ with $f_n\leq f$, and so $U_{>0}$ is the upwards closure of $\left\{f_n\mid n\in\mathbb{N}\right\}$.
We'll take $M=R$ and suppose that $I_U$ is a maximal uncountably generated ideal of $R$. Since $I_{U_{>0}}$ is countably generated, there must be some strictly positive element $g\in\mathbb{Z}^W$ that is not in $U$. Let $w=d(g)$, choose $0<n\in\mathbb{N}$ with $w<w_n$, and let $V$ be the set of positive elements $f\in\mathbb{Z}^W$ for which $d(f)<w_n$. Then $V$ is an upwards closed subset of $\mathbb{Z}^W$ that properly contains $U$, and so $I_U$ is a proper subideal of $I_V$.
But $I_V$ is not countably generated, since $\aleph_n$ does not have countable cofinality. If $\{s_k\mid k\in\mathbb{N}\}$ is a countable subset of $I_V$ then there is an element $s\in I_V$ with $d\left(\nu(s)\right)>d(\left(\nu(s_k)\right)$ for every $n\in\mathbb{N}$, and so $s$ is not in the ideal generated by $\{s_k\mid k\in\mathbb{N}\}$
