# Faithfully flat module and exact sequences

I have the following problem.

Let $$M$$ be a faithfully flat module. I need to show that if sequence

$$0\longrightarrow N'\otimes M \stackrel{\varphi\otimes{id_{M}}}\longrightarrow N\otimes M\stackrel{\psi\otimes id_{M}}\longrightarrow N''\otimes M\longrightarrow 0$$

is exact, then sequence $$0\longrightarrow N'\stackrel{\varphi}\longrightarrow N\stackrel{\psi}\longrightarrow N''\longrightarrow 0$$ is exact. The only thing I have left to show is that the $$\psi$$ is surjective but I don't know how.

• This is sometimes taken as the definition of faithfully flat. What definition are you working with? Anyway here is a nice exposition: ayoucis.wordpress.com/2014/03/12/flat-morphisms-and-flatness Commented Mar 20, 2020 at 6:10
• For me an $R$-module $M$ is faithfully flat if it is flat over $R$ and for any $R$-module $N$ such that $M\otimes N=\{0\}$, $N=\{0\}$. Commented Mar 20, 2020 at 16:21
• Look at kernels and cokernels. Both properties are preserved by tensoring (due to flatness and right exactness) and are zero after tensoring (by assumption). Commented Mar 20, 2020 at 16:28

As mentioned by Asvin, this is often taken as definition of faithful flatness. To develop on this, $$M$$ is a faithfully flat module if it is flat and the functor $$- \otimes M$$ is faithful, namely, $$f \otimes id_M = 0$$ implies $$f = 0$$ for all $$R$$-module homomorphisms $$f$$. In particular we can show that $$\psi \otimes id_M$$ surjective implies $$\psi$$ surjective using only faithfulness (no flatness required).
It is convenient to use the category-theoretic generalisation of surjectiveness called epimorphism. Assume $$f, g : N'' \rightarrow X$$ are morphisms such that $$f \circ \psi = g \circ \psi$$. Then $$(f \otimes id_M)\circ(\psi\otimes id_M) = (g \otimes id_M)\circ(\psi\otimes id_M)$$ and since $$\psi$$ is surjective by assumption, we have $$f \otimes id_M = g \otimes id_M$$. Using the group structure of module homomorphisms, we obtain $$(f-g)\otimes id_M$$ and hence $$f-g = 0$$ by faithfulness of the $$- \otimes M$$ functor. Hence $$f=g$$ as required, so $$\psi$$ is surjective.