TITLE: $0 \rightarrow M' \rightarrow M \rightarrow M'' \rightarrow 0$ exact, $M''$ flat. Why is $M$ flat $\Leftrightarrow M'$ flat?
QUESTION [7 upvotes]: Let $A$ be a commutative ring with identity, and let
  \begin{align}
0 \rightarrow M' \rightarrow M \rightarrow M'' \rightarrow 0
\end{align}
  be an exact sequence of $A$-modules with $M''$ flat.
I want to show: $M$ flat $\Leftrightarrow M'$ flat.

However, I don't manage to show this. I try to use the fact that $M / M'$ is isomorphic to the flat module $M''$, but I don't know how this helps. Any help is appreciated.

REPLY [5 votes]: If $N' \to N$ is a monomorphism, we have a commutative diagram:
$\begin{array}{c} & & & & & &  0 & \\ &&&&&& \downarrow \\ 0 & \rightarrow & M' \otimes N' & \rightarrow & M \otimes N' & \rightarrow & M'' \otimes N' & \rightarrow & 0 \\ && \downarrow && \downarrow &&\downarrow && \\0 & \rightarrow & M' \otimes N & \rightarrow & M \otimes N & \rightarrow & M'' \otimes N & \rightarrow & 0\end{array}$
The rows and columns are exact because $M''$ is flat. The Snake Lemma yields an exact sequence
$$0 \to \ker(M' \otimes N' \to M' \otimes N) \to \ker(M \otimes N' \to M \otimes N) \to 0,$$
and the claim follows. This proof works in a quite general setting, namely in so-called tame abelian $\otimes$-categories. See Jacob Lurie's Tannaka duality for geometric stacks, Lemma 5.4.