Let $R$ be an integral domain. Let $\alpha$ be an infinite cardinal . Let $M$ be a faithful $R$-module such that $\mu(M)< \alpha$ . Let $N$ be a submodule of $M$ and $m\in M$ and $r\in R$ be such that $rm \in N$, $\mu (N+rM) < \alpha$ and $\mu (N+Rm) < \alpha$ . Then is it true that $\mu (N) < \alpha$ ?


If this is not true in general for all infinite cardinal $\alpha$, can we atleast characterize those $\alpha$ for which it is true ? In particular, is it true for $\alpha= \aleph\_0$ ? 


The answer here [On cardinality of generating subsets of some submodules](https://mathoverflow.net/questions/295519/on-cardinality-of-generating-subsets-of-some-submodules) shows that when $R$ is not an integral domain, we can have counterexamples.


NOTE : For an $R$-module $M$, by $\mu (M)$ we denote the minimal no. of generators of $M$ . So $\mu (M) < \alpha$ means $M$ can be generated by a set $S \subseteq M$ such that $|S| < \alpha$ . 



Cardinality of generating sets of faithful modules over integral domain
Let $W$ be a well-ordered set with cofinality at least $\alpha$ (for example, $W$ could be the cardinal $2^\alpha$).


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.


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.


Let $1\in\mathbb{Z}^W$ be the constant function taking the value $1$, and let $I\_{>1}=\{x\in R\mid \nu(x)>1\}\cup\{0\}$ and $I\_{\geq1}=\{x\in R\mid \nu(x)\geq1\}\cup\{0\}$. 


Then $\mu(I\_{\geq1})=1$, since $I\_{\geq1}$ is generated as an ideal by any element $x$ with $\nu(x)=1$.


$\mu(I\_{>1})$ is equal to the cofinality of $W$, and is therefore at least $\alpha$. 


To see this, suppose $\{n\_j\mid j\in J\}\subseteq I\_{>1}$, and for each $j$, $w\_j$ is the least element of $W$ such that $\nu(n\_j)(w\_j)\neq1$. Then $\{n\_j\mid j\in J\}$ generates $I\_{>1}$ if and only if $\{w\_j\mid j\in J\}$ is cofinal in $W$, since if $w'\in W$ with $w'>w\_j$ for every $j\in J$, then choosing $f\in\mathbb{Z}^W$ with $f(w)=1$ for $w<w'$ and $f(w')=2$, and $n\in I\_{>1}$ with $\nu(n)=f$, then $n$ is not in the subideal of $I\_{>1}$ generated by $\{n\_j\mid j\in J\}$. 


Now take $M=R$, $N=I\_{>1}$ and $r=m$ with $\nu(r)=1$.


Then $N+rM=I\_{\geq1}=N+Rm$, so $\mu(R)=\mu(N+rM)=\nu(N+Rm)=1$, but $\mu(N)\geq\alpha$.


