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How to characterize every finitely generated module over $\mathbb{Q}[X]/(X^2+1)^3$?

I feel somehow we should use the structure theorem for modules over PID but the ring here is not a domain. So I am lost for ideas.

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  • $\begingroup$ We just can say they're quotients of a finite number of copies of the ring (as a module over itself). $\endgroup$
    – Bernard
    May 7 '20 at 22:26
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    $\begingroup$ If $R$ is a (commutative) ring, and $I$ is an ideal of $R$, then an $R/I$-module is nothing more than an $I$-torsion $R$-module, so it suffices to classify finitely generated $\mathbb{Q}[X]$-modules which are $(X^{2}+1)^{3}$-torsion. $\endgroup$ May 7 '20 at 22:27
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Let $M$ be a finitely generated $\mathbb Q[x]/ (x^2+1)^3$- module. By definition, we have a ring hom $\mathbb Q[x]/ (x^2+1)^3 \to \mathrm{End}_{\mathrm{Ab}}(M)$.

This is the same as a ring hom $\mathbb Q[x] \to \mathrm{End}_{\mathrm{Ab}}(M)$ where $(x^2+1)^3$ is contained in the kernel of the action. Therefore, a $\mathbb Q[x]/ (x^2+1)^3$-module, $M$, is the same as a $\mathbb Q[x]$-module where $(x^2+1)^3 \cdot M = 0$.

$\mathbb Q[x]$ is a PID and since $M$ is a finitely generated over $\mathbb Q[x]/ (x^2+1)^3$, then it is finitely generated over $\mathbb Q[x]$.

Now you can use the classification theorem for finitely generated modules over PIDs.

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