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Let $R$ be a Dedekind domain, $M$ be an additive subgroup of $(R, +)$. If $M$ is closed under multiplication by units of $R$, is $M$ one-generated as a $\mathbb{Z}[R^\times]$-module? What if $R^\times$ generates $R$ as a ring?

I'm particularly interested when $R = \mathcal{O}_K$ for some number field $K$. I know that in general $R$ is not a PID, but given that $R^\times$ generates $R$, can I retrieve "one-generated-ness" for $\mathbb{Z}[R^\times]$ sub-modules of $(R, +)$?

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    What do you mean by "$M$ absorbs products from $R^{\times}$"? – 2017-02-22
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    @QiaochuYuan for every $r \in R^\times$ and $m \in M$, $rm \in M$. – 2017-02-22
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    A more typical way to say that might be "$M$ is closed under multiplication by units $r \in R^{\times}$." "Absorbs" is strange terminology here because it's sometimes used in mathematics for an "absorbing element," meaning an element $m$ of a monoid which behaves like zero in the sense that $am = ma = m$ for all $a$. – 2017-02-23
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    Anyway, if I understand correctly what you mean by "one-generated as an $R^{\times}$-module," the answer is already no for $R = M = \mathbb{Z}$. Can you clarify what you mean by this? – 2017-02-23
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    @QiaochuYuan oh I see the problem, I would like to allow both the action of $R^\times$ and the abelian group structure on this generating set. Thus, if I say $M = \langle x \rangle$, then every element of $M$ can be written as a combination of elements of the set $\{ rx | r \in R^\times\}$. I just need to look at the group ring $\mathbb{Z}[R^\times]$. I'm editing the question now. – 2017-02-23
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    I'm slightly confused. Take $M$ to be a non principal ideal in $R$. This is naturally a $\mathbb Z[R]$ module and is not one generated. Viewing it as a $\mathbb Z[R^\times]$ module only loses information. – 2017-02-23

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