Lemma 86.4.1. Let $A$ be a Noetherian ring and let $I \subset A$ be an ideal. Let $B$ be an object of (86.2.0.2). The following are equivalent

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

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

3. 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 $n \geq 1$,

4. $B = A[x_1, \ldots , x_ r]^\wedge /J$ and for every $a \in I$ there exists a $c \geq 0$ such that

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

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

Proof. The equivalence of (1) and (2) follows from Lemma 86.3.5. The equivalence of (1) $+$ (2) and (3) follows from Lemma 86.3.1. Some details omitted.

Assume the equivalent conditions (1), (2), (3) holds and let $B = A[x_1, \ldots , x_ r]^\wedge /J$ be a presentation (see Lemma 86.2.1). Let $a \in I$. Let $c$ be such that multiplication by $a^ c$ is zero on $\mathop{N\! L}\nolimits ^\wedge _{B/A}$ which exists by (1). By Lemma 86.3.4 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 $\alpha (\text{d}x_ i)$. Then we see that (4)(a), (b) hold.

Assume (4) holds. Say $I = (a_1, \ldots , a_ t)$. Let $c_ i \geq 0$ be the integer such that (4)(a), (b) hold for $a_ i^{c_ i}$. Then we see that $I^{\sum c_ i}$ annihilates $H^0(\mathop{N\! L}\nolimits ^\wedge _{B/A})$. Let $f_{i, 1}, \ldots , f_{i, r} \in J$ be as in (4)(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 (4)(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 10.15.4). Thus the kernel of $B^{\oplus r} \to \bigoplus B\text{d}x_ i$ is $a_ i$-power torsion, and hence $H^{-1}(\mathop{N\! L}\nolimits ^\wedge _{B/A}) = \mathop{\mathrm{Ker}}(J/J^2 \to \bigoplus B\text{d}x_ i)$ is $a_ i$-power torsion. Since $B$ is Noetherian (Lemma 86.2.2), all modules including $H^{-1}(\mathop{N\! L}\nolimits ^\wedge _{B/A})$ are finite. Thus $a_ i^{d_ i}$ annihilates $H^{-1}(\mathop{N\! L}\nolimits ^\wedge _{B/A})$ for some $d_ i \geq 0$. It follows that $I^{\sum d_ i}$ annihilates $H^{-1}(\mathop{N\! L}\nolimits ^\wedge _{B/A})$ and we see that (2) holds. $\square$

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