Let $R$ be a regular local ring and let $M$ be a faithfully flat $R$-module. Does there necessarily exist a surjective $R$-module homomorphism from $M$ to $R$?
For context, I am computing $\sum_{f\in\text{Hom}(M,R)}f(M)$. If $M$ is a nonzero finitely generated Cohen-Macaulay $R$-module where $R$ is regular local, it can be shown that $M$ must be free, and hence $\sum_{f\in\text{Hom}(M,R)}f(M)=R$ (clearly $\sum_{f\in\text{Hom}(M,R)}f(M)=R$ iff there exists a surjective homomorphism from $M$ to $R$ since $R$ is local). However, it is true that $\sum_{f\in\text{Hom}(M,R)}f(M)=R$ even when $M$ is not finitely generated. The justification someone gave me for this is that $M$ will have to be faithfully flat. However, I do not see why this would implies the existence of a surjective homomorphism from $M$ to $R$.
Best Answer
No. For instance, let $R=\mathbb{Z}_{(p)}$ and $M=\mathbb{Z}_p$. Then there are no nonzero homomorphisms $M\to R$. Indeed, any homomorphism $M\to R$ is continuous in the $p$-adic topology, and thus has compact image, but no nonzero submodule of $\mathbb{Z}_{(p)}$ is compact in the $p$-adic topology.