## 86.4 Rig-smooth algebras

As motivation for the following definition, please take a look at More on Algebra, Remark 15.83.2.

Definition 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). We say $B$ is rig-smooth over $(A, I)$ if there exists an integer $c \geq 0$ such that $I^ c$ annihilates $\mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}^\wedge , N)$ for every $B$-module $N$.

Let us work out what this means.

Lemma 86.4.2. 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-smooth over $(A, I)$,

2. the object $\mathop{N\! L}\nolimits _{B/A}^\wedge$ of $D(B)$ satisfies the equivalent conditions (1) – (4) of More on Algebra, Lemma 15.83.10 with respect to the ideal $IB$,

3. there exists a $c \geq 0$ such that for all $a \in I^ c$ there is a map $h : \bigoplus B\text{d}x_ i \to J/J^2$ such that $a : J/J^2 \to J/J^2$ is equal to $h \circ \text{d}$,

4. there exist $b_1, \ldots , b_ s \in B$ such that $V(b_1, \ldots , b_ s) \subset V(IB)$ and such that for every $l = 1, \ldots , s$ there exist $m \geq 0$, $f_1, \ldots , f_ m \in J$, and subset $T \subset \{ 1, \ldots , n\}$ with $|T| = m$ such that

1. $\det _{i \in T, j \leq m}(\partial f_ j/ \partial x_ i)$ divides $b_ l$ in $B$, and

2. $b_ l J \subset (f_1, \ldots , f_ m) + J^2$.

Proof. The equivalence of (1), (2), and (3) is immediate from More on Algebra, Lemma 15.83.10.

Assume $b_1, \ldots , b_ s$ are as in (4). Since $B$ is Noetherian the inclusion $V(b_1, \ldots , b_ s) \subset V(IB)$ implies $I^ cB \subset (b_1, \ldots , b_ s)$ for some $c \geq 0$ (for example by Algebra, Lemma 10.62.4). Pick $1 \leq l \leq s$ and $m \geq 0$ and $f_1, \ldots , f_ m \in J$ and $T \subset \{ 1, \ldots , n\}$ with $|T| = m$ satisfying (4)(a) and (b). Then if we invert $b_ l$ we see that

$\mathop{N\! L}\nolimits _{B/A}^\wedge \otimes _ B B_{b_ l} = \left( \bigoplus \nolimits _{j \leq m} B_{b_ l} f_ j \longrightarrow \bigoplus \nolimits _{i = 1, \ldots , n} B_{b_ l} \text{d}x_ i \right)$

and moreover the arrow is isomorphic to the inclusion of the direct summand $\bigoplus _{i \in T} B_{b_ l} \text{d}x_ i$. We conclude that $H^{-1}(\mathop{N\! L}\nolimits _{B/A}^\wedge )$ is $b_ l$-power torsion and that $H^0(\mathop{N\! L}\nolimits _{B/A}^\wedge )$ becomes finite free after inverting $b_ l$. Combined with the inclusion $I^ cB \subset (b_1, \ldots , b_ s)$ we see that $H^{-1}(\mathop{N\! L}\nolimits _{B/A}^\wedge )$ is $IB$-power torsion. Hence we see that condition (4) of More on Algebra, Lemma 15.83.10 holds. In this way we see that (4) implies (2).

Assume the equivalent conditions (1), (2), and (3) hold. We will prove that (4) holds, but we strongly urge the reader to convince themselves of this. The complex $\mathop{N\! L}\nolimits _{B/A}^\wedge$ determines an object of $D^ b_{\textit{Coh}}(\mathop{\mathrm{Spec}}(B))$ whose restriction to the Zariski open $U = \mathop{\mathrm{Spec}}(B) \setminus V(IB)$ is a finite locally free module $\mathcal{E}$ placed in degree $0$ (this follows for example from the the fourth equivalent condition in More on Algebra, Lemma 15.83.10). Choose generators $f_1, \ldots , f_ M$ for $J$. This determines an exact sequence

$\bigoplus \nolimits _{j = 1, \ldots , M} \mathcal{O}_ U \cdot f_ j \to \bigoplus \nolimits _{i = 1, \ldots , n} \mathcal{O}_ U \cdot \text{d}x_ i \to \mathcal{E} \to 0$

Let $U = \bigcup _{l = 1, \ldots , s} U_ l$ be a finite affine open covering such that $\mathcal{E}|_{U_ l}$ is free of rank $r_ l = n - m_ l$ for some integer $n \geq m_ l \geq 0$. After replacing each $U_ l$ by an affine open covering we may assume there exists a subset $T_ l \subset \{ 1, \ldots , n\}$ such that the elements $\text{d}x_ i$, $i \in \{ 1, \ldots , n\} \setminus T_ l$ map to a basis for $\mathcal{E}|_{U_ l}$. Repeating the argument, we may assume there exists a subset $T'_ l \subset \{ 1, \ldots , M\}$ of cardinality $m_ l$ such that $f_ j$, $j \in T'_ l$ map to a basis of the kernel of $\mathcal{O}_{U_ l} \cdot \text{d}x_ i \to \mathcal{E}|_{U_ l}$. Finally, since the open covering $U = \bigcup U_ l$ may be refined by a open covering by standard opens (Algebra, Lemma 10.17.2) we may assume $U_ l = D(g_ l)$ for some $g_ l \in B$. In particular we have $V(g_1, \ldots , g_ s) = V(IB)$. A linear algebra argument using our choices above shows that $\det _{i \in T_ l, j \in T'_ l}(\partial f_ j/ \partial x_ i)$ maps to an invertible element of $B_{b_ l}$. Similarly, the vanishing of cohomology of $\mathop{N\! L}\nolimits _{B/A}^\wedge$ in degree $-1$ over $U_ l$ shows that $J/J^2 + (f_ j; j \in T')$ is annihilated by a power of $b_ l$. After replacing each $g_ l$ by a suitable power we obtain conditions (4)(a) and (4)(b) of the lemma. Some details omitted. $\square$

Lemma 86.4.3. 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 smooth over $\mathop{\mathrm{Spec}}(A) \setminus V(I)$, then $B^\wedge$ is rig-smooth over $(A, I)$.

2. 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 86.4.2 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 15.83.10 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 86.3.2 we conclude (2) holds by More on Algebra, Lemma 15.83.7.

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

$\xi \otimes 1 \in \mathop{\mathrm{Ext}}\nolimits ^1_{B^\wedge }(\mathop{N\! L}\nolimits _{B/A} \otimes _ B B^\wedge , N \otimes _ B B^\wedge ) = \mathop{\mathrm{Ext}}\nolimits ^1_{B^\wedge }(\mathop{N\! L}\nolimits _{B/A}, N) \otimes _ B B^\wedge$

is annihilated by $I^ c$ for some integer $c \geq 0$. The equality holds for example by More on Algebra, Lemma 15.98.2 (but can also easily be deduced from the much simpler More on Algebra, Lemma 15.64.4). 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

$\xymatrix{ J/J^2 \ar[r] \ar[d]^ b & \bigoplus B\text{d}x_ i \ar@{..>}[d]^ h \\ J/J^2 \ar[r] & (J/J^2)_ g }$

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. $\square$

Lemma 86.4.4. Let $(A_1, I_1) \to (A_2, I_2)$ be as in Remark 86.2.3 with $A_1$ and $A_2$ Noetherian. Let $B_1$ be in (86.2.0.2) for $(A_1, I_1)$. Let $B_2$ be the base change of $B_1$. Let $f_1 \in B_1$ with image $f_2 \in B_2$. If $\mathop{\mathrm{Ext}}\nolimits ^1_{B_1}(\mathop{N\! L}\nolimits _{B_1/A_1}^\wedge , N_1)$ is annihilated by $f_1$ for every $B_1$-module $N_1$, then $\mathop{\mathrm{Ext}}\nolimits ^1_{B_2}(\mathop{N\! L}\nolimits _{B_2/A_2}^\wedge , N_2)$ is annihilated by $f_2$ for every $B_2$-module $N_2$.

Proof. By Lemma 86.3.4 there is a map

$\mathop{N\! L}\nolimits _{B_1/A_1} \otimes _{B_2} B_1 \to \mathop{N\! L}\nolimits _{B_2/A_2}$

which induces and isomorphism on $H^0$ and a surjection on $H^{-1}$. Thus the result by More on Algebra, Lemmas 15.83.6, 15.83.7, and 15.83.9 the last two applied with the principal ideals $(f_1) \subset B_1$ and $(f_2) \subset B_2$. $\square$

Lemma 86.4.5. Let $A_1 \to A_2$ be a map of Noetherian rings. Let $I_ i \subset A_ i$ be an ideal such that $V(I_1A_2) = V(I_2)$. Let $B_1$ be in (86.2.0.2) for $(A_1, I_1)$. Let $B_2$ be the base change of $B_1$ as in Remark 86.2.3. If $B_1$ is rig-smooth over $(A_1, I_1)$, then $B_2$ is rig-smooth over $(A_2, I_2)$.

Proof. Follows from Lemma 86.4.4 and Definition 86.4.1 and the fact that $I_2^ c$ is contained in $I_1A_2$ for some $c \geq 0$ as $A_2$ is Noetherian. $\square$

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