The Stacks project

Proof. Choose a surjection $D''' = D''[x_1, \ldots , x_ n] \to D'$ with finitely generated kernel $J$. By Algebra, Lemma 10.36.23 it suffices to show that $N'$ is finitely presented as a $D'''$-module. Moreover, $D''' \otimes _{B'} B \to D' \otimes _{B'} B = D$ and $D''' \otimes _{B'} A' \to D' \otimes _{B'} A' = C'$ are surjections whose kernels are generated by the image of $J$, hence $N$ is a finitely presented $D''' \otimes _{B'} B$-module and $M'$ is a finitely presented $D''' \otimes _{B'} A'$-module by Algebra, Lemma 10.36.23 again. Thus we may replace $D'$ by $D'''$ and $D$ by $D''' \otimes _{B'} B$, etc. Since $D'''$ is flat over $B'$, it follows that we may assume that $B' \to D'$ is flat.

Assume $B' \to D'$ is flat. By Lemma 15.7.4 the module $N'$ is finite over $D'$. Choose a surjection $(D')^{\oplus n} \to N'$ with kernel $K'$. By base change we obtain maps $D^{\oplus n} \to N$, $(C')^{\oplus n} \to M'$, and $C^{\oplus n} \to M$ with kernels $K_ D$, $K_{C'}$, and $K_ C$. There is a canonical map

On the other hand, since $N' = N \times _ M M'$ and $D' = D \times _ C C'$ (by Lemma 15.6.8; applied to the flat $B'$-module $D'$) there is also a canonical map $K_ D \times _{K_ C} K_{C'} \to K'$ inverse to the displayed arrow. Hence the displayed map is an isomorphism. By Algebra, Lemma 10.5.3 the modules $K_ D$ and $K_{C'}$ are finite. We conclude from Lemma 15.7.4 that $K'$ is a finite $D'$-module provided that $K_ D \to K_ C$ and $K_{C'} \to K_ C$ induce isomorphisms $K_ D \otimes _ B A = K_ C = K_{C'} \otimes _{A'} A$. This is true because the flatness assumptions implies the sequences

stay exact upon tensoring, see Algebra, Lemma 10.39.12. $\square$