Faithfully flat direct sum

Question

The following appears as an exercise in Bourbaki, Commutative Algebra, Chapter I, Exercises §3, n°1 (and other books). Consider a family of $R$-modules $(M_i)_{i\in I}$ ($R$ and $I$ arbitrary). We wish to show that $\bigoplus_{i\in I}M_i$ is faithfully flat iff all the $M_i$ are flat and at least one of them is faithfully flat.

A direct sum of modules is flat iff their summands are flat. Thus the if part follows from the fact that tensor products commute with direct sums. The only if part is where I am stuck.

My attempt goes with reductio ad absurdum: suppose none of the $M_i$ are faithfully flat, so there are for every $i\in I$ a non-zero module $N_i$ such that $M_i\otimes_R N_i=0$. From this, I'd try and build a non-zero module $N$ st
$N\otimes_R\bigoplus_{i\in I}M_i = 0$, but I'm lacking ideas for such a candidate $N$. Even for $I$ reduced to 2 elements isn't obvious to me.

Is such an $N$ easy to find, or is there another way of going about this? Any suggestion is welcome. Thank you.

Answer 1

Note: First of all note that if $R$ is any commutative ring. For any prime ideal $P$ of $R$, the localization $R_P$ is flat $R$-module. Set $P := \bigoplus_{P\in Spec(R)} R_P$ is also flat. We claim that $P$ is faithfully flat. Indeed, if $M$ is any $R$-module such that $P \otimes M = 0$, then $$0 = (\bigoplus_{P\in Spec(R)} R_P)\otimes M = \bigoplus_{P\in Spec(R)} (R_P\otimes M) = \bigoplus_{P\in Spec(R)} M_P.$$ Then each localization $M_P = 0$. It is well known that this implies that $M = 0$, so $P$ is faithfully flat $R$-module.

(Note that some of above equalities are actualy isomorphism)

Example: Set $R:= \mathbb{Z}$ and $P := (p)$ where p is a prime, then for any prime $l\ne p$, the nonzero $\mathbb{Z}$-module $\mathbb{Z}/l \mathbb{Z}$ localizes to zero with respect to $P$. If $P = (0)$, any torsion $\mathbb{Z}$-module localizes to zero with respect to $P$. This shows that no $\mathbb{Z}_P$ is faithfully flat over $\mathbb{Z}$. But by above note $$P := \bigoplus_{P\in Spec(\mathbb{Z})} \mathbb{Z}_P$$ is faithfully flat over $\mathbb{Z}$.