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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.

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    $\begingroup$ 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 $\endgroup$
    – Asvin
    Commented Mar 20, 2020 at 6:10
  • $\begingroup$ For me an $R$-module $M$ is faithfully flat if it is flat over $R$ and for any $R$-module $N$ such that $M\otimes N=\{0\}$, $N=\{0\}$. $\endgroup$ Commented Mar 20, 2020 at 16:21
  • $\begingroup$ Look at kernels and cokernels. Both properties are preserved by tensoring (due to flatness and right exactness) and are zero after tensoring (by assumption). $\endgroup$
    – Asvin
    Commented Mar 20, 2020 at 16:28

1 Answer 1

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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.

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