Question about $R$ modules are precisely $k[x]$ modules which are annihilated by $\langle x^p \rangle$? Let $R = k[x] / \langle x^p \rangle$. 
I'm trying to find all of the (finitely generated) indecomposable $R$-modules (up to isomorphism), and apparently it is sufficient to find the finitely generated indecomposable $R$-modules that are annihilated by $\langle x^p \rangle$.
But how do we know that the finitely generated indecomposable $R$ modules are precisely the finitely generated indecomposable $k[x]$ modules which are annihilated by $\langle x^p \rangle$?
 A: 
But how do we know that the finitely generated indecomposable $R$ modules are precisely the finitely generated indecomposable $k[x]$ modules which are annihilated by $\langle x^p \rangle$?

Uh, I guess the fact that the $R$ modules are exactly the $k[x]$ modules annihilated by $(x^p)$ isn't in question, just the indecomposability and/or finite generation part.
The whole point about considering the class of $R$ modules annihilated by $I$ as $R/I$ modules is that the "lifted action" (that is, $m\cdot r:= m\cdot (r+I)$ which always works even when $M$ isn't annihilated by $I$) can be reversed. That is, given an $R$ module annihilated by $I$, the action given by $m\cdot(r+I):= m\cdot r$ is well-defined.
So, what I'm trying to say here is that in our context here, given a $k[x]$ module $M$ annihilated by $I=(x^p)$, $m\cdot r=m\cdot (r+I)$. So to say that $M=\sum_{i=1}^nm_ik[x]/(x^p)$ means exactly the same thing as $M=\sum_{i=1}^nm_ik[x]$. Thus, the finitely generated modules correspond.
Now, if $M$ is an indecomposable $R=k[x]/(x^p)$ module, it is automatically a $k[x]$ module annihilated by $(x^p)$. By the fundamental theorem of finitely generated modules over a PID, $M$ is isomorphic to $\oplus_{i=1}^n k[x]/(x^{e_i})$ for a set of $e_i$, $1\leq e_i\leq p$. But each of these is obviously an $R$ module, so to be irreducible, $n=1$. So, $M=k[x]/(x^k)$ for some $k$, $1\leq k\leq p$.
This information suffices to answer the doubt you had about your previous question's solution, so I'll leave it at that.
