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.

  • $\begingroup$ maybe you could somehow use that $N \otimes (M_1 \oplus M_2) \cong (N \otimes M_1 )\oplus (N \otimes M_2)$. just a random idea $\endgroup$
    – Riquelme
    Aug 30, 2019 at 14:07
  • $\begingroup$ Answered here: math.stackexchange.com/a/41671/121097 $\endgroup$
    – user26857
    Sep 1, 2019 at 23:29

2 Answers 2


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 $\mathbb{Z}$.

  • $\begingroup$ Thank you for the counter-example. It would seem that there are errors in Bourbaki? In any case, I suppose that the exercise would be true provided $R$ is local. $\endgroup$
    – Andrei.B
    Aug 30, 2019 at 16:04
  • 1
    $\begingroup$ Yes, it is true in local case by the following theorem: Theorem Let $M$ be a flat module. Then $M$ is faithfully flat if and only if $M \ne mM$ for each maximal ideal $m$ of $R$. $\endgroup$
    – E.R
    Aug 30, 2019 at 16:28
  • $\begingroup$ Too many $P$'s. Confusing! $\endgroup$
    – user26857
    Sep 1, 2019 at 23:24


It' simpler to prove that for any maximal ideal $\mathfrak m\in\operatorname{Max}A$, $$\bigoplus_{i\in I}M_i\Big/\mathfrak m\bigl(\bigoplus_{i\in I}M_i\bigr)\ne\{0\}.$$


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