Lemma 86.4.2. Let $A$ be a Noetherian ring and let $I$ be an ideal. Let $B$ be a finite type $A$-algebra.

1. If $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is étale over $\mathop{\mathrm{Spec}}(A) \setminus V(I)$, then $B^\wedge$ satisfies the equivalent conditions of Lemma 86.4.1.

2. If $B^\wedge$ satisfies the equivalent conditions of Lemma 86.4.1, then there exists $g \in 1 + IB$ such that $\mathop{\mathrm{Spec}}(B_ g)$ is étale over $\mathop{\mathrm{Spec}}(A) \setminus V(I)$.

Proof. Assume $B^\wedge$ satisfies the equivalent conditions of Lemma 86.4.1. The naive cotangent complex $\mathop{N\! L}\nolimits _{B/A}$ is a complex of finite type $B$-modules and hence $H^{-1}$ and $H^0$ are finite $B$-modules. Completion is an exact functor on finite $B$-modules (Algebra, Lemma 10.96.2) and $\mathop{N\! L}\nolimits ^\wedge _{B^\wedge /A}$ is the completion of the complex $\mathop{N\! L}\nolimits _{B/A}$ (this is easy to see by choosing presentations). Hence the assumption implies there exists a $c \geq 0$ such that $H^{-1}/I^ nH^{-1}$ and $H^0/I^ nH^0$ are annihilated by $I^ c$ for all $n$. By Nakayama's lemma (Algebra, Lemma 10.19.1) this means that $I^ cH^{-1}$ and $I^ cH^0$ are annihilated by an element of the form $g = 1 + x$ with $x \in IB$. After inverting $g$ (which does not change the quotients $B/I^ nB$) we see that $\mathop{N\! L}\nolimits _{B/A}$ has cohomology annihilated by $I^ c$. Thus $A \to B$ is étale at any prime of $B$ not lying over $V(I)$ by the definition of étale ring maps, see Algebra, Definition 10.142.1.

Conversely, assume that $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is étale over $\mathop{\mathrm{Spec}}(A) \setminus V(I)$. Then for every $a \in I$ there exists a $c \geq 0$ such that multiplication by $a^ c$ is zero $\mathop{N\! L}\nolimits _{B/A}$. Since $\mathop{N\! L}\nolimits ^\wedge _{B^\wedge /A}$ is the derived completion of $\mathop{N\! L}\nolimits _{B/A}$ (see Lemma 86.3.1) it follows that $B^\wedge$ satisfies the equivalent conditions of Lemma 86.4.1. $\square$

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