The Stacks project

Proof. Let $T$ be the $m \times m$ matrix with entries $\partial f_ j/\partial x_ i$ for $1 \leq i, j \leq n$. Let $K \in D(B)$ be represented by the complex $T : B^{\oplus m} \to B^{\oplus m}$ with terms sitting in degrees $-1$ and $0$. By More on Algebra, Lemmas 15.84.12 we have $g : K \to K$ is zero in $D(B)$. Set $J = (f_1, \ldots , f_ m)$. Recall that $\mathop{N\! L}\nolimits _{B/A}$ is homotopy equivalent to $J/J^2 \to \bigoplus _{i = 1, \ldots , n} B\text{d}x_ i$, see Algebra, Section 10.134. Denote $L$ the complex $J/J^2 \to \bigoplus _{i = 1, \ldots , m} B\text{d}x_ i$ to that we have the quotient map $\mathop{N\! L}\nolimits _{B/A} \to L$. We also have a surjective map of complexes $K \to L$ by sending the $j$th basis element in the term $B^{\oplus m}$ in degree $-1$ to the class of $f_ j$ in $J/J^2$. Picture

From More on Algebra, Lemma 15.84.8 we conclude that multiplication by $g$ on $L$ is $0$ in $D(B)$. On the other hand, the distinguished triangle $B^{\oplus n - m}[0] \to \mathop{N\! L}\nolimits _{B/A} \to L$ shows that $\mathop{\mathrm{Ext}}\nolimits ^1_ B(L, N) \to \mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}, N)$ is surjective for every $B$-module $N$ and hence annihilated by $g$. This proves part (1). If $n = m$ then $\mathop{N\! L}\nolimits _{B/A} = L$ and we see that (2) holds. $\square$

Proof. The conditions imply that $B$ is rig-smooth over $(A, I)$, see Lemma 88.4.2. Write $b' \Delta = b$ in $B$ for some $b' \in B$. Say $I = (a_1, \ldots , a_ t)$. Since $V(b) \subset V(IB)$ there exists an integer $c \geq 0$ such that $I^ cB \subset bB$. Write $bb_ i = a_ i^ c$ in $B$ for some $b_ i \in B$.

Choose an integer $n \gg 0$ (we will see later how large). Choose polynomials $f'_1, \ldots , f'_ m \in A[x_1, \ldots , x_ r]$ such that $f_ i - f'_ i \in I^ nA[x_1, \ldots , x_ r]^\wedge $. We set $\Delta ' = \det _{1 \leq i, j \leq m}(\partial f'_ j/\partial x_ i)$ and we consider the finite type $A$-algebra

We will apply Lemma 88.7.1 to $C$. We compute

Hence we see that $\mathop{\mathrm{Ext}}\nolimits ^1_ C(\mathop{N\! L}\nolimits _{C/A}, N)$ is annihilated by $(\Delta ')^{t + 1}$ for all $C$-modules $N$. Since $a_ i^ c$ is divisible by $\Delta '$ in $C$ we see that $a_ i^{(t + 1)c}$ annihilates these $\mathop{\mathrm{Ext}}\nolimits ^1$'s also. Thus $I^{c_1}$ annihilates $\mathop{\mathrm{Ext}}\nolimits ^1_ C(\mathop{N\! L}\nolimits _{C/A}, N)$ for all $C$-modules $N$ where $c_1 = 1 + t((t + 1)c - 1)$. The exact value of $c_1$ doesn't matter for the rest of the argument; what matters is that it is independent of $n$.

Since $\mathop{N\! L}\nolimits _{C^\wedge /A}^\wedge = \mathop{N\! L}\nolimits _{C/A} \otimes _ C C^\wedge $ by Lemma 88.3.2 we conclude that multiplication by $I^{c_1}$ is zero on $\mathop{\mathrm{Ext}}\nolimits ^1_{C^\wedge }(\mathop{N\! L}\nolimits _{C^\wedge /A}^\wedge , N)$ for any $C^\wedge $-module $N$ as well, see More on Algebra, Lemmas 15.84.7 and 15.84.6. In particular $C^\wedge $ is rig-smooth over $(A, I)$.

Observe that we have a surjective $A$-algebra homomorphism

sending the class of $x_ i$ to the class of $x_ i$ and sending the class of $z_ i$ to the class of $b_ ib'$. This works because of our choices of $b'$ and $b_ i$ in the first paragraph of the proof.

Let $d = d(\text{Gr}_ I(B))$ and $q_0 = q(\text{Gr}_ I(B))$ be the integers found in Local Cohomology, Section 51.22. By Lemma 88.5.3 if we take $n \geq \max (q_0 + (d + 1)c_1, 2(d + 1)c_1 + 1)$ we can find a homomorphism $\varphi : C^\wedge \to B$ of $A$-algebras which is congruent to $\psi _ n$ modulo $I^{n - (d + 1)c_1}B$.

Since $\varphi : C^\wedge \to B$ is surjective modulo $I$ we see that it is surjective (for example use Algebra, Lemma 10.96.1). To finish the proof it suffices to show that $\mathop{\mathrm{Ker}}(\varphi )/\mathop{\mathrm{Ker}}(\varphi )^2$ is annihilated by a power of $I$, see More on Algebra, Lemma 15.108.4.

Since $\varphi $ is surjective we see that $\mathop{N\! L}\nolimits _{B/C^{\wedge }}^\wedge $ has cohomology modules $H^0(\mathop{N\! L}\nolimits _{B/C^{\wedge }}^\wedge ) = 0$ and $H^{-1}(\mathop{N\! L}\nolimits _{B/C^{\wedge }}^\wedge ) = \mathop{\mathrm{Ker}}(\varphi )/\mathop{\mathrm{Ker}}(\varphi )^2$. We have an exact sequence

by Lemma 88.3.5. The first two modules are annihilated by a power of $I$ as $B$ and $C^\wedge $ are rig-smooth over $(A, I)$. Hence it suffices to show that the kernel of the surjective map $H^0(\mathop{N\! L}\nolimits _{C^\wedge /A}^\wedge \otimes _{C^\wedge } B) \to H^0(\mathop{N\! L}\nolimits _{B/A}^\wedge )$ is annihilated by a power of $I$. For this it suffices to show that it is annihilated by a power of $b$. In other words, it suffices to show that

is an isomorphism. However, both are free $B[1/b]$ modules of rank $r - m$ with basis $\text{d}x_{m + 1}, \ldots , \text{d}x_ r$ and we conclude the proof. $\square$