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Let $c_1,\dots,c_k$ be some non-zero complex numbers and $M$ be the abelian subgroup generated by $c_1,\dots,c_k$ (i.e. all $\mathbb{Z}$-linear combinations of $c_1\dots,c_k$). Suppose further that $\alpha$ and $\beta$ are complex numbers such that $\alpha^n=\beta^m=1$ for some positive integers $m$ and $n$ and $\alpha M=\beta M=M$. So $M$ is a $\mathbb{Z}[\alpha,\beta]$-module, where $R=\mathbb{Z}[\alpha,\beta]$ is the subring generated by $\alpha,\beta$ in the complex field $\mathbb{C}$, i.e. all $\mathbb{Z}$-linear combinations of $\alpha^i \beta ^j$ for all $i=0,1,\dots,n-1$, $j=0,1,\dots,m-1$.

Is there a non-zero $R$-homomorphism from $M$ to $R$? If the answer is negative, what if $\alpha=\beta$?

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    $\begingroup$ It seems to me that the $\alpha=\beta$ case is the general case. If $\alpha$ is a primitive $k$-th root of unity and $\beta$ is a primitive $l$-th root of unity, let $d$ be the greatest common divisor of $k$ and $l$, let $\zeta$ be a primitive $d$-th root of unity, and notice that $\mathbb Z[\alpha,\beta]=\mathbb Z[\zeta]$. $\endgroup$
    – Andreas Blass
    Commented Dec 13, 2018 at 15:21
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    $\begingroup$ @AndreasBlass ... and a finitely generated torsion-free module over a Dedekind domain (${\bf Z}[\zeta]$) is projective, so yes, there exist lots of homomorphisms to $R$. $\endgroup$
    – David Handelman
    Commented Dec 13, 2018 at 16:02
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    $\begingroup$ In my previous comment, "greatest common divisor" should have been "least common multiple". (Sometimes it would be nice to be able to edit comments.) $\endgroup$
    – Andreas Blass
    Commented Dec 15, 2018 at 4:28

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