# Faithfully Flatness Free Modules

A $R$-module $M$ is called flat if every exact sequence of $R$ -modules $0 \to N_1 \to N_2 \to N_3 \to 0$ stays after tensoring with $M$ exact:

$0 \to M \otimes _ R N_1 \to M \otimes _ R N_2 \to M \otimes _ R N_3 \to 0$

Futhermore $M$ is called faithfully flat iff the complex of $R$-modules $N_1→N_2→N_3$ is exact if and only if the sequence $M⊗_R N_1→M⊗_ RN_2→M⊗_R N_3$ is exact.

I have two questions concerning faithfully flatness:

1. Since tensor products commute with direct sums obviously free $R$-modules $R^I$ are flat.

Are free $R$-modules also faithfully flat? Especially if we have exact sequence $M⊗_R N_1→M⊗_ RN_2→M⊗_R N_3$ for $M \cong R^I$ for arbitrary index set $I$ and therefore the exact sequence $N_1^I→N_2^I→N_3^I$, should $N_1→N_2→N_3$ be also exact and how to prove that?

1. Let $A$ be a flat $R$ algebra and therefore a $R$ module. Is that true that in this case $A$ is faithfully flat iff the induced map on Specs $\operatorname{Spec}(A) \to \operatorname{Spec}(R)$ is surjective?

If yes, does anybody have a good reference for a proof that this characterisation of faithfully flatness coincides with the previous one?

• The answer to 2. is that a map on Spec is faithfully flat if and only if it is flat and surjective on prime ideals. I think Atiyah-MacDonald has a proof of this. I guess this answers the first question, as the induced map is surjective on prime ideals - you have already said that you know it is flat. Commented Sep 10, 2018 at 0:16

Yes free modules are faithfully flat, because tensor products commute with direct sums, and a free $R$-module is isomorphic to some $R^{(I)}$ (and not $R^I$ as you erroneously wrote).