The Stacks project

Proof. Choose a finite type ring map $A \to C$ and a finite $C$-module $N$ and a prime $\mathfrak q$ of $C$ such that $B = C_{\mathfrak q}$ and $M = N_{\mathfrak q}$. In the following, when we say “the theorem holds for $(N/C/A, \mathfrak q)$ we mean that it holds for $(A \to B, M)$ where $B = C_{\mathfrak q}$ and $M = N_{\mathfrak q}$. By Lemma 38.20.6 the functor $F_{lf}$ is unchanged if we replace $B$ by a local ring flat over $B$. Hence, since $A$ is henselian, we may apply Lemma 38.6.6 and assume that there exists a complete dévissage of $N/C/A$ at $\mathfrak q$.

Let $(A_ i, B_ i, M_ i, \alpha _ i, \mathfrak q_ i)_{i = 1, \ldots , n}$ be such a complete dévissage of $N/C/A$ at $\mathfrak q$. Let $\mathfrak q'_ i \subset A_ i$ be the unique prime lying over $\mathfrak q_ i \subset B_ i$ as in Definition 38.6.4. Since $C \to A_1$ is surjective and $N \cong M_1$ as $C$-modules, we see by Lemma 38.20.5 it suffices to prove the theorem holds for $(M_1/A_1/A, \mathfrak q'_1)$. Since $B_1 \to A_1$ is finite and $\mathfrak q_1$ is the only prime of $B_1$ over $\mathfrak q'_1$ we see that $(A_1)_{\mathfrak q'_1} \to (B_1)_{\mathfrak q_1}$ is finite (see Algebra, Lemma 10.41.11 or More on Morphisms, Lemma 37.47.4). Hence by Lemma 38.20.5 it suffices to prove the theorem holds for $(M_1/B_1/A, \mathfrak q_1)$.

At this point we may assume, by induction on the length $n$ of the dévissage, that the theorem holds for $(M_2/B_2/A, \mathfrak q_2)$. (If $n = 1$, then $M_2 = 0$ which is flat over $A$.) Reversing the last couple of steps of the previous paragraph, using that $M_2 \cong \mathop{\mathrm{Coker}}(\alpha _2)$ as $B_1$-modules, we see that the theorem holds for $(\mathop{\mathrm{Coker}}(\alpha _1)/B_1/A, \mathfrak q_1)$.

Let $A'$ be an object of $\mathcal{C}$. At this point we use Lemma 38.10.1 to see that if $(M_1 \otimes _ A A')_{\mathfrak q'}$ is flat over $A'$ for a prime $\mathfrak q'$ of $B_1 \otimes _ A A'$ lying over $\mathfrak m_{A'}$, then $(\mathop{\mathrm{Coker}}(\alpha _1) \otimes _ A A')_{\mathfrak q'}$ is flat over $A'$. Hence we conclude that $F_{lf}$ is a subfunctor of the functor $F'_{lf}$ associated to the module $\mathop{\mathrm{Coker}}(\alpha _1)_{\mathfrak q_1}$ over $(B_1)_{\mathfrak q_1}$. By the previous paragraph we know $F'_{lf}$ is corepresented by $A/J$ for some ideal $J \subset A$. Hence we may replace $A$ by $A/J$ and assume that $\mathop{\mathrm{Coker}}(\alpha _1)_{\mathfrak q_1}$ is flat over $A$.

Since $\mathop{\mathrm{Coker}}(\alpha _1)$ is a $B_1$-module for which there exist a complete dévissage of $N_1/B_1/A$ at $\mathfrak q_1$ and since $\mathop{\mathrm{Coker}}(\alpha _1)_{\mathfrak q_1}$ is flat over $A$ by Lemma 38.10.2 we see that $\mathop{\mathrm{Coker}}(\alpha _1)$ is free as an $A$-module, in particular flat as an $A$-module. Hence Lemma 38.10.1 implies $F_{lf}(A')$ is nonempty if and only if $\alpha \otimes 1_{A'}$ is injective. Let $N_1 = \mathop{\mathrm{Im}}(\alpha _1) \subset M_1$ so that we have exact sequences

The flatness of $\mathop{\mathrm{Coker}}(\alpha _1)$ implies the first sequence is universally exact (see Algebra, Lemma 10.82.5). Hence $\alpha \otimes 1_{A'}$ is injective if and only if $B_1^{\oplus r_1} \otimes _ A A' \to N_1 \otimes _ A A'$ is an isomorphism. Finally, Theorem 38.23.3 applies to show this functor is corepresentable by $A/I$ for some ideal $I$ and we conclude $F_{lf}$ is corepresentable by $A/I$ also.

To prove the final statement, suppose that $A \to B$ is essentially of finite presentation and $M$ of finite presentation over $B$. Let $I \subset A$ be the ideal such that $F_{lf}$ is corepresented by $A/I$. Write $I = \bigcup I_\lambda$ where $I_\lambda$ ranges over the finitely generated ideals contained in $I$. Then, since $F_{lf}(A/I) = \{ *\}$ we see that $F_{lf}(A/I_\lambda ) = \{ *\}$ for some $\lambda$, see Lemma 38.20.4 part (2). Clearly this implies that $I = I_\lambda$. $\square$