Lemma 86.8.2. Let $A$ be a Noetherian ring and let $I \subset A$ be an ideal. Let $B$ be an object of (86.2.0.2). Write $B = A[x_1, \ldots , x_ r]^\wedge /J$ (Lemma 86.2.2) and let $\mathop{N\! L}\nolimits _{B/A}^\wedge = (J/J^2 \to \bigoplus B\text{d}x_ i)$ be its naive cotangent complex (86.3.0.1). The following are equivalent

1. $B$ is rig-étale over $(A, I)$,

2. there exists a $c \geq 0$ such that for all $a \in I^ c$ multiplication by $a$ on $\mathop{N\! L}\nolimits _{B/A}^\wedge$ is zero in $D(B)$,

3. there exits a $c \geq 0$ such that $H^ i(\mathop{N\! L}\nolimits _{B/A}^\wedge )$, $i = -1, 0$ is annihilated by $I^ c$,

4. there exists a $c \geq 0$ such that $H^ i(\mathop{N\! L}\nolimits _{B_ n/A_ n})$, $i = -1, 0$ is annihilated by $I^ c$ for all $n \geq 1$ where $A_ n = A/I^ n$ and $B_ n = B/I^ nB$,

5. for every $a \in I$ there exists a $c \geq 0$ such that

1. $a^ c$ annihilates $H^0(\mathop{N\! L}\nolimits _{B/A}^\wedge )$, and

2. there exist $f_1, \ldots , f_ r \in J$ such that $a^ c J \subset (f_1, \ldots , f_ r) + J^2$.

6. for every $a \in I$ there exist $f_1, \ldots , f_ r \in J$ and $c \geq 0$ such that

1. $\det _{1 \leq i, j \leq r}(\partial f_ j/\partial x_ i)$ divides $a^ c$ in $B$, and

2. $a^ c J \subset (f_1, \ldots , f_ r) + J^2$.

7. choosing generaters $f_1, \ldots , f_ t$ for $J$ we have

1. the Jacobian ideal of $B$ over $A$, namely the ideal in $B$ generated by the $r \times r$ minors of the matrx $(\partial f_ j/\partial x_ i)_{1 \leq i \leq r, 1 \leq j \leq t}$, contains the ideal $I^ cB$ for some $c$, and

2. the Cramer ideal of $B$ over $A$, namely the ideal in $B$ generated by the image in $B$ of the $r$th Fitting ideal of $J$ as an $A[x_1, \ldots , x_ r]^\wedge$-module, contains $I^ cB$ for some $c$.

Proof. The equivalence of (1) and (2) is a restatement of Definition 86.8.1.

The equivalence of (2) and (3) follows from More on Algebra, Lemma 15.83.11.

The equivalence of (3) and (4) follows from the fact that the systems $\{ \mathop{N\! L}\nolimits _{B_ n/A_ n}\}$ and $\mathop{N\! L}\nolimits _{B/A}^\wedge \otimes _ B B_ n$ are strictly isomorphic, see Lemma 86.3.3. Some details omitted.

Assume (2). Let $a \in I$. Let $c$ be such that multiplication by $a^ c$ is zero on $\mathop{N\! L}\nolimits _{B/A}^\wedge$. By More on Algebra, Lemma 15.83.4 part (1) there exists a map $\alpha : \bigoplus B\text{d}x_ i \to J/J^2$ such that $\text{d} \circ \alpha$ and $\alpha \circ \text{d}$ are both multiplication by $a^ c$. Let $f_ i \in J$ be an element whose class modulo $J^2$ is equal to $\alpha (\text{d}x_ i)$. A simple calculation gives that (6)(a), (b) hold.

We omit the verification that (6) implies (5); it is just a statement on two term complexes over $B$ of the form $M \to B^{\oplus r}$.

Assume (5) holds. Say $I = (a_1, \ldots , a_ t)$. Let $c_ i \geq 0$ be the integer such that (5)(a), (b) hold for $a_ i^{c_ i}$. Then we see that $I^{\sum c_ i}$ annihilates $H^0(\mathop{N\! L}\nolimits _{B/A}^\wedge )$. Let $f_{i, 1}, \ldots , f_{i, r} \in J$ be as in (5)(b) for $a_ i$. Consider the composition

$B^{\oplus r} \to J/J^2 \to \bigoplus B\text{d}x_ i$

where the $j$th basis vector is mapped to the class of $f_{i, j}$ in $J/J^2$. By (5)(a) and (b) the cokernel of the composition is annihilated by $a_ i^{2c_ i}$. Thus this map is surjective after inverting $a_ i^{c_ i}$, and hence an isomorphism (Algebra, Lemma 10.16.4). Thus the kernel of $B^{\oplus r} \to \bigoplus B\text{d}x_ i$ is $a_ i$-power torsion, and hence $H^{-1}(\mathop{N\! L}\nolimits _{B/A}^\wedge ) = \mathop{\mathrm{Ker}}(J/J^2 \to \bigoplus B\text{d}x_ i)$ is $a_ i$-power torsion. Since $B$ is Noetherian (Lemma 86.2.2), all modules including $H^{-1}(\mathop{N\! L}\nolimits _{B/A}^\wedge )$ are finite. Thus $a_ i^{d_ i}$ annihilates $H^{-1}(\mathop{N\! L}\nolimits _{B/A}^\wedge )$ for some $d_ i \geq 0$. It follows that $I^{\sum d_ i}$ annihilates $H^{-1}(\mathop{N\! L}\nolimits _{B/A}^\wedge )$ and we see that (3) holds.

Thus conditions (2), (3), (4), (5), and (6) are equivalent. Thus it remains to show that these conditions are equivalent with (7). Observe that the Cramer ideal $\text{Fit}_ r(J) B$ is equal to $\text{Fit}_ r(J/J^2)$ as $J/J^2 = J \otimes _{A[x_1, \ldots , x_ r]^\wedge } B$, see More on Algebra, Lemma 15.8.4 part (3). Also, observe that the Jacobian ideal is just $\text{Fit}_0(H^0(\mathop{N\! L}\nolimits _{B/A}^\wedge ))$. Thus we see that the equivalence of (3) and (7) is a purely algebraic question which we discuss in the next paragraph.

Let $R$ be a Noetherian ring and let $I \subset R$ be an ideal. Let $M \xrightarrow {d} R^{\oplus r}$ be a two term complex. We have to show that the following are equivalent

1. the cohomology of $M \to R^{\oplus r}$ is annihilated by a power of $I$, and

2. the ideals $\text{Fit}_ r(M)$ and $\text{Fit}_0(\text{Coker}(d))$ contain a power of $I$.

Since $R$ is Noetherian, we can reformulate part (2) as an inclusion of the corresponding closed subschemes, see Algebra, Lemmas 10.17.2 and 10.32.5. On the other hand, over the complement of $V(\text{Fit}_0(\mathop{\mathrm{Coker}}(d)))$ the cokernel of $d$ vanishes and over the complement of $V(\text{Fit}_ r(M))$ the module $M$ is locally generated by $r$ elements, see More on Algebra, Lemma 15.8.6. Thus (B) is equivalent to

1. away from $V(I)$ the cokernel of $d$ vanishes and the module $M$ is locally generated by $\leq r$ elements.

Of course this is equivalent to the condition that $M \to R^{\oplus r}$ has vanishing cohomology over $\mathop{\mathrm{Spec}}(R) \setminus V(I)$ which in turn is equivalent to (A). This finishes the proof. $\square$

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