The Stacks project

Proof. Let $(A_ i, B_ i, M_ i, \alpha _ i, \mathfrak q_ i)_{i = 1, \ldots , n}$ be the given complete dévissage. We prove the lemma by induction on $n$. Note that $N$ is finitely presented as an $S$-module if and only if $M_1$ is finitely presented as an $B_1$-module, see Remark 38.6.3. Note that $N_{\mathfrak q} \cong (M_1)_{\mathfrak q_1}$ as $R$-modules because (a) $N_{\mathfrak q} \cong (M_1)_{\mathfrak q'_1}$ where $\mathfrak q'_1$ is the unique prime in $A_1$ lying over $\mathfrak q_1$ and (b) $(A_1)_{\mathfrak q'_1} = (A_1)_{\mathfrak q_1}$ by Algebra, Lemma 10.41.11, so (c) $(M_1)_{\mathfrak q'_1} \cong (M_1)_{\mathfrak q_1}$. Hence $(M_1)_{\mathfrak q_1}$ is a flat $R$-module. Thus we may replace $(S, N)$ by $(B_1, M_1)$ in order to prove the lemma. By Lemma 38.10.1 the map $\alpha _1 : B_1^{\oplus r_1} \to M_1$ is $R$-universally injective and $\mathop{\mathrm{Coker}}(\alpha _1)_{\mathfrak q}$ is $R$-flat. Note that $(A_ i, B_ i, M_ i, \alpha _ i, \mathfrak q_ i)_{i = 2, \ldots , n}$ is a complete dévissage of $\mathop{\mathrm{Coker}}(\alpha _1)/B_1/R$ at $\mathfrak q_1$. Hence the induction hypothesis implies that $\mathop{\mathrm{Coker}}(\alpha _1)$ is finitely presented as a $B_1$-module, free as an $R$-module, and has a decomposition as in the lemma. This implies that $M_1$ is finitely presented as a $B_1$-module, see Algebra, Lemma 10.5.3. It further implies that $M_1 \cong B_1^{\oplus r_1} \oplus \mathop{\mathrm{Coker}}(\alpha _1)$ as $R$-modules, hence a decomposition as in the lemma. Finally, $B_1$ is projective as an $R$-module by Lemma 38.9.3 hence free as an $R$-module by Algebra, Theorem 10.85.4. This finishes the proof. $\square$