Let $R$ be a commutative ring with unity. Let $\alpha$ be an infinite cardinal . Let $M$ be an $R$-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 $\alpha= \aleph\_0$ ? 


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 $|S| < \alpha$ . 



On cardinality of generating subsets of some submodules
This is not true for any cardinal.


For any cardinal $\alpha$, there is a commutative ring $R$ with elements $x,y\in R$ such that $xy=0$ and $\mu(Rx\cap Ry)=\alpha$. So you can take $M=R$, $m=x$, $r=y$ and $N=Rx\cap Ry$.


To construct such a ring, it's sufficient to find a commutative ring $S$ and an $S$-module $A$ with elements $x,y\in A$ such that $\mu(xS\cap yS)=\alpha$, as then you can take $R$ to be the ring $S\oplus A$ with multiplication $(s,a)(s',a')=(ss',as'+a's)$.


Then, for example, you could take $S$ to be any commutative ring which has an ideal $I$ with $\mu(I)=\alpha$ (for example, take $S$ to be a ring of polynomials in $\alpha$ variables, and $I$ the ideal of polynomials with zero constant term), and take $A=S\oplus S/I$, $x=(1,1)$, $y=(1,0)$, so $xS\cap yS=I\oplus0<S\oplus S/I$.


