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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 $|S| < \alpha$ .

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  • $\begingroup$ Is $\mu(N+rm)$ a typo? Unless $rm\in N$, $N+rm$ isn’t a module. $\endgroup$
    – Jeremy Rickard
    Commented Mar 18, 2018 at 19:05
  • $\begingroup$ @JeremyRickard: Sorry, I have edited. $\endgroup$
    – user111524
    Commented Mar 18, 2018 at 19:11
  • $\begingroup$ Are there known examples of $N$ and $M$, where $\mu(M) < \alpha$ and $N \subset M$ but $\mu(N)\ge \alpha$? $\endgroup$
    – Not Mike
    Commented Mar 19, 2018 at 0:34
  • $\begingroup$ @NotMike: Definitely ... take any non-Noetherian ring $R$. $R$ is an $R$-module satisfying $\mu(R)=1 < \aleph_0$ but $\mu(I) \ge \aleph_0$ for some ideal $I$ since $R$ is not Noetherian. $\endgroup$
    – user111524
    Commented Mar 19, 2018 at 9:37

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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 $xS\cap yS=I\oplus0<S\oplus S/I$.

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  • $\begingroup$ Sorry about my previous comment, it was pretty stupid ... do you think there would be any counterexample if $R$ was a domain ? $\endgroup$
    – user111524
    Commented Mar 19, 2018 at 10:30
  • $\begingroup$ To be more specific, if $R$ was a domain, then for which infinite cardinal $\alpha$ does my claim hold ? $\endgroup$
    – user111524
    Commented Mar 19, 2018 at 10:44
  • $\begingroup$ @users I don’t know, sorry. $\endgroup$
    – Jeremy Rickard
    Commented Mar 19, 2018 at 13:50
  • $\begingroup$ With the $x,y$ you have taken in the last line , $xy$ is non-zero ... $\endgroup$
    – user111524
    Commented May 23, 2018 at 17:43
  • $\begingroup$ @users No. $R$ is the ring $S\oplus A$ where $A$ is a square zero ideal, and $x,y\in A$, so $xy=0$. I'm afraid my notation is a bit confusing, as $A$ is itself defined to be a direct sum. But I'm describing $x$ and $y$ as elements of the ideal $A=S\oplus S/I$, not of the ring $R=S\oplus A$; as elements of $R=S\oplus(S\oplus S/I)$ they would be $(0,(1,1))$ and $(0,(1,0))$. $\endgroup$
    – Jeremy Rickard
    Commented May 23, 2018 at 20:01

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