# Surjective homomorphism from a faithfully flat module to a regular local ring.

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 $$R$$.

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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.