Direct sum of completions is faithfully flat

Question

Let $R$ be a commutative Noetherian ring (with unity). For every maximal ideal $\mathfrak{p}$, denote by $R_{\mathfrak{p}}$ the completion of $R$ at $\mathfrak{p}$. I wish to show that $\bigoplus_{\mathfrak{p}} R_{\mathfrak{p}}$ is a faithfully flat $R$-module. If $R$ is Noetherian, then $R_{\mathfrak{p}}$ is flat for every $\mathfrak{p}$, and so $\bigoplus_{\mathfrak{p}}R_{\mathfrak{p}}$ is flat. Let $M$ be an $R$-module. We need to show that if $\left(\bigoplus_{\mathfrak{p}} R_{\mathfrak{p}}\right)\otimes_R M=0$, then $M=0$. We have $$0=\left(\bigoplus_{\mathfrak{p}} R_{\mathfrak{p}}\right)\otimes_R M\cong \bigoplus_{\mathfrak{p}} (R_{\mathfrak{p}}\otimes_R M).$$ We need to deduce that if $R_{\mathfrak{p}}\otimes_R M=0$ for every $\mathfrak{p}$, then $M=0$. This is clear if instead of the completions one considers the localisations. But is this still true in our setting?

Answer 1

If there is a nonzero element $a\in M$ then there is a maximal ideal $\mathfrak{p}$ containing the annihilator of $a$, giving us an injective map $R/\mathfrak{p} \to M$. By flatness, there is an injective map $R_\mathfrak{p} \otimes R/\mathfrak{p} \to R_\mathfrak{p} \otimes M$.

But $R_\mathfrak{p} \otimes R/\mathfrak{p} = R_\mathfrak{p} / \mathfrak{p}R_\mathfrak{p} \neq 0$, so $R_\mathfrak{p} \otimes M \neq 0$.

Note that this points to a more general fact: to check that a flat module is faithfully flat, it's enough to check that its tensor product with $R/\mathfrak{p}$ is nonzero for all maximal ideals $\mathfrak{p}$.