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 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 $|S| < \alpha$ .

  • $\begingroup$ Please see " R. Nekooei and E. Rostami, A Prime Submodule Principle,Algebra Colloq. 21, 697 (2014). doi.org/10.1142/S1005386714000649 ". > $\endgroup$ – E. Rostami Mar 19 '18 at 16:25
  • $\begingroup$ The paper only treats , as far as I can see, the Oka property for cardinaility of generating sets. Whereas I am asking for an Ako type of property . $\endgroup$ – user111524 Mar 19 '18 at 17:08

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 $\mu(N)\geq\alpha$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy