My goal is to classify all finitely generated modules over $R=Q[x]/(x^2+1)^2$. So, let $M$ be such a module. I am given a hint that such a module is also a $Q[x]$ module.
First off, in what way should I be seeing $M$ as a $Q[x]$ module? The only thing I can think to do is extend the action of $R$ on $M$ to $Q[x]$ on $M$ by letting $q(x)\cdot m=q(x)$ $(mod (x^2+1)^2)\cdot m$. It will follow easily that this is an action on $M$ and makes $M$ into a $Q[x]$ module. It is also clear that since $M$ is finitely generated over $R$, it will be finitely generated over $Q[x]$ as well. So, I can then write, using the fundamental theorem of finitely generated modules over PIDs, $$M\cong Q[x]^k\oplus Q[x]/(a_1^{\alpha_1}(x))\oplus\cdots\oplus Q[x]/(a_t^{\alpha_t}(x)) $$ as $Q[x]$ modules, and where each $a_i(x)$ is prime. Is this the right idea?
My question now is how do I turn this into an isomorphism of $R$ modules? My (naive) gut reaction is to reduce everything modulo $(x^2+1)^2$. However, how do I deal with these quotients? What are the images of these direct summands in $R$. Does it even make sense to "reduce modulo $(x^2+1)^2$"?
I feel like any $R$ module will look like $$\left[Q[x]/(x^2+1)^2\right]^{n_1}\oplus \left[Q[x]/(x^2+1)\right]^{n_2}$$ where possibly $n_1,n_2=0$, but I'm not sure at all how to justify this. Any help at all is appreciated.
That's all there is to it: using that "lifted" action of $Q[x]$ on the $Q[x]/(x^2+1)$ module, the $Q[x]/(x^2+1)$ modules are exactly the $Q[x]$ modules which are annihilated by $(x^2+1)$.
The finitely generated ones correspond with each other, and you draw conclusions using the structure theorem for f.g. modules over $Q[x]$.