Modules over $\mathbb{C}[x]$ Suppose I have a module $M$ over $\mathbb{C}[x]$ such that $M\otimes_{\mathbb{C}[x]}\mathbb{C}[x]/(x-\lambda)\cong\mathbb{C}$ for all $\lambda \in \mathbb{C}$.
What kind of general hypotheses could I add to conclude that $M=\mathbb{C}[x]$ (e.g. an infinite sum of $\mathbb{C}[x]/(x-\lambda)$ is currently a counterexample)?
 A: More generally let $R$ be a PID and let $M$ be an $R$-module such that $M \otimes_R R/p \cong R/p$ as $R$-modules for every prime element $p \in R$. If $M$ is finitely generated, then we have $M \cong R$.
Proof. Since $M$ is finitely genetated, we may apply the structure theorem: Write $M = F \oplus \bigoplus_{\ell} M_{\ell}$, where $F$ is a finitely generated free $R$-module of some rank $d$, in the direct sum $\ell$ runs through the primes and $M_{\ell}$ is the $\ell^{\infty}$-torsion part of $M$, which we may write as $M_{\ell} \cong R/\ell^{k_{s(\ell)}} \oplus \dotsc \oplus R/\ell^{k_1}$ for some $k_{s(\ell)} \geq \dotsc \geq k_1 \geq 1$ (of course depending on $\ell$). For $p \neq \ell$ we have $M_{\ell} = p M_{\ell}$, i.e. $M_{\ell} \otimes_R R/p = 0$, whereas for $p=\ell$ we have $M_p \otimes_R R/p = (R/p)^{\oplus s(p)}$. Hence, $R/p \cong M \otimes_R R/p \cong (R/p)^{\oplus d} \oplus (R/p)^{\oplus s(p)}$. By comparing the $R/p$-dimensions, this is equivalent to $$1=d+s(p).$$ Thus, $s(p)$ doesn't depend on $p$. But since $M_p=0$ and hence $s(p)=0$ for almost all $p$, it follows that $s(p)=0$ for all $p$, and then $d=1$, i.e. $M \cong R$.
