Lemma 88.4.3. Let $A$ be a Noetherian ring and let $I$ be an ideal. Let $B$ be a finite type $A$-algebra.
If $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is smooth over $\mathop{\mathrm{Spec}}(A) \setminus V(I)$, then $B^\wedge $ is rig-smooth over $(A, I)$.
If $B^\wedge $ is rig-smooth over $(A, I)$, then there exists $g \in 1 + IB$ such that $\mathop{\mathrm{Spec}}(B_ g)$ is smooth over $\mathop{\mathrm{Spec}}(A) \setminus V(I)$.
Proof. We will use Lemma 88.4.2 without further mention.
Assume (1). Recall that formation of $\mathop{N\! L}\nolimits _{B/A}$ commutes with localization, see Algebra, Lemma 10.134.13. Hence by the very definition of smooth ring maps (in terms of the naive cotangent complex being quasi-isomorphic to a finite projective module placed in degree $0$), we see that $\mathop{N\! L}\nolimits _{B/A}$ satisfies the fourth equivalent condition of More on Algebra, Lemma 15.84.10 with respect to the ideal $IB$ (small detail omitted). Since $\mathop{N\! L}\nolimits _{B^\wedge /A}^\wedge = \mathop{N\! L}\nolimits _{B/A} \otimes _ B B^\wedge $ by Lemma 88.3.2 we conclude (2) holds by More on Algebra, Lemma 15.84.7.
Assume (2). Choose a presentation $B = A[x_1, \ldots , x_ n]/J$, set $N = J/J^2$, and consider the element $\xi \in \mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}, J/J^2)$ determined by the identity map on $J/J^2$. Using again that $\mathop{N\! L}\nolimits _{B^\wedge /A}^\wedge = \mathop{N\! L}\nolimits _{B/A} \otimes _ B B^\wedge $ we find that our assumption implies the image
is annihilated by $I^ c$ for some integer $c \geq 0$. The equality holds for example by More on Algebra, Lemma 15.99.2 (but can also easily be deduced from the much simpler More on Algebra, Lemma 15.65.4). Thus $M = I^ cB\xi \subset \mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}, N)$ is a finite submodule which maps to zero in $\mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}, N) \otimes _ B B^\wedge $. Since $B \to B^\wedge $ is flat this means that $M \otimes _ B B^\wedge $ is zero. By Nakayama's lemma (Algebra, Lemma 10.20.1) this means that $M = I^ cB\xi $ is annihilated by an element of the form $g = 1 + x$ with $x \in IB$. This implies that for every $b \in I^ cB$ there is a $B$-linear dotted arrow making the diagram commute
Thus $(\mathop{N\! L}\nolimits _{B/A})_{gb}$ is quasi-isomorphic to a finite projective module; small detail omitted. Since $(\mathop{N\! L}\nolimits _{B/A})_{gb} = \mathop{N\! L}\nolimits _{B_{gb}/A}$ in $D(B_{gb})$ this shows that $B_{gb}$ is smooth over $\mathop{\mathrm{Spec}}(A)$. As this holds for all $b \in I^ cB$ we conclude that $\mathop{\mathrm{Spec}}(B_ g) \to \mathop{\mathrm{Spec}}(A)$ is smooth over $\mathop{\mathrm{Spec}}(A) \setminus V(I)$ as desired. $\square$
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