The Stacks project

Proof. Let $I$ and $I'$ be ideals of definitions of $A$. Then there exists an integer $c \geq 0$ such that $I^ c \subset I'$ and $(I')^ c \subset I$. Hence $B$ is rig-smooth over $(A, I)$ if and only if $B$ is rig-smooth over $(A, I')$. This follows from Definition 88.4.1, the inclusions $I^ c \subset I'$ and $(I')^ c \subset I$, and the fact that the naive cotangent complex $\mathop{N\! L}\nolimits _{B/A}^\wedge $ is independent of the choice of ideal of definition of $A$ by Remark 88.11.2. $\square$

Proof. Let us recall what the statement signifies. First, $\textit{WAdm}^{Noeth}$ is the category whose objects are adic Noetherian topological rings and whose morphisms are continuous ring homomorphisms. Consider a commutative diagram

satisfying the following conditions: $A$ and $B$ are adic Noetherian topological rings, $A \to A'$ and $B \to B'$ are étale ring maps, $(A')^\wedge = \mathop{\mathrm{lim}}\nolimits A'/I^ nA'$ for some ideal of definition $I \subset A$, $(B')^\wedge = \mathop{\mathrm{lim}}\nolimits B'/J^ nB'$ for some ideal of definition $J \subset B$, and $\varphi : A \to B$ and $\varphi ' : (A')^\wedge \to (B')^\wedge $ are continuous. Note that $(A')^\wedge $ and $(B')^\wedge $ are adic Noetherian topological rings by Formal Spaces, Lemma 87.21.1. We have to show

1. $\varphi $ is rig-smooth $\Rightarrow \varphi '$ is rig-smooth,

2. if $B \to B'$ faithfully flat, then $\varphi '$ is rig-smooth $\Rightarrow \varphi $ is rig-smooth, and

3. if $A \to B_ i$ is rig-smooth for $i = 1, \ldots , n$, then $A \to \prod _{i = 1, \ldots , n} B_ i$ is rig-smooth.

The equivalent conditions of Lemma 88.11.1 satisfy conditions (1), (2), and (3). Thus in verifying (1), (2), and (3) for the property “rig-smooth” we may already assume our ring maps satisfy the equivalent conditions of Lemma 88.11.1 in each case.

Pick an ideal of definition $I \subset A$. By the remarks above the topology on each ring in the diagram is the $I$-adic topology and $B$, $(A')^\wedge $, and $(B')^\wedge $ are in the category (88.2.0.2) for $(A, I)$. Since $A \to A'$ and $B \to B'$ are étale the complexes $\mathop{N\! L}\nolimits _{A'/A}$ and $\mathop{N\! L}\nolimits _{B'/B}$ are zero and hence $\mathop{N\! L}\nolimits _{(A')^\wedge /A}^\wedge $ and $\mathop{N\! L}\nolimits _{(B')^\wedge /B}^\wedge $ are zero by Lemma 88.3.2. Applying Lemma 88.3.5 to $A \to (A')^\wedge \to (B')^\wedge $ we get isomorphisms

Thus $\mathop{N\! L}\nolimits _{(B')^\wedge /A}^\wedge \to \mathop{N\! L}\nolimits _{(B')^\wedge /(A')^\wedge }$ is a quasi-isomorphism. The ring maps $B/I^ nB \to B'/I^ nB'$ are étale and hence are local complete intersections (Algebra, Lemma 10.143.2). Hence we may apply Lemmas 88.3.5 and 88.3.6 to $A \to B \to (B')^\wedge $ and we get isomorphisms

We conclude that $\mathop{N\! L}\nolimits _{B/A}^\wedge \otimes _ B (B')^\wedge \to \mathop{N\! L}\nolimits _{(B')^\wedge /A}^\wedge $ is a quasi-isomorphism. Combining these two observations we obtain that

in $D((B')^\wedge )$. With these preparations out of the way we can start the actual proof.

Proof of (1). Assume $\varphi $ is rig-smooth. Then there exists a $c \geq 0$ such that $\mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}^\wedge , N)$ is annihilated by $I^ c$ for every $B$-module $N$. By More on Algebra, Lemmas 15.84.6 and 15.84.7 this property is preserved under base change by $B \to (B')^\wedge $. Hence $\mathop{\mathrm{Ext}}\nolimits ^1_{(B')^\wedge }(\mathop{N\! L}\nolimits _{(B')^\wedge /(A')^\wedge }^\wedge , N)$ is annihilated by $I^ c(A')^\wedge $ for all $(B')^\wedge $-modules $N$ which tells us that $\varphi '$ is rig-smooth. This proves (1).

To prove (2) assume $B \to B'$ is faithfully flat and that $\varphi '$ is rig-smooth. Then there exists a $c \geq 0$ such that $\mathop{\mathrm{Ext}}\nolimits ^1_{(B')^\wedge }(\mathop{N\! L}\nolimits _{(B')^\wedge /(A')^\wedge }^\wedge , N')$ is annihilated by $I^ c(B')^\wedge $ for every $(B')^\wedge $-module $N'$. The composition $B \to B' \to (B')^\wedge $ is flat (Algebra, Lemma 10.97.2) hence for any $B$-module $N$ we have

by More on Algebra, Lemma 15.99.2 part (3) (minor details omitted). Thus we see that this module is annihilated by $I^ c$. However, $B \to (B')^\wedge $ is actually faithfully flat by our assumption that $B \to B'$ is faithfully flat (Formal Spaces, Lemma 87.19.14). Thus we conclude that $\mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}^\wedge , N)$ is annihilated by $I^ c$. Hence $\varphi $ is rig-smooth. This proves (2).

To prove (3), setting $B = \prod _{i = 1, \ldots , n} B_ i$ we just observe that $\mathop{N\! L}\nolimits _{B/A}^\wedge $ is the direct sum of the complexes $\mathop{N\! L}\nolimits _{B_ i/A}^\wedge $ viewed as complexes of $B$-modules. $\square$

Proof. The statement makes sense by Lemma 88.17.1. To see that it is true assume we have morphisms $B \to A$ and $B \to C$ in $\textit{WAdm}^{Noeth}$ and that $B \to A$ is rig-smooth and $B \to C$ is adic (Formal Spaces, Definition 87.6.1). Then we can choose an ideal of definition $I \subset B$ such that the topology on $A$ and $C$ is the $I$-adic topology. In this situation it follows immediately that $A \widehat{\otimes }_ B C$ is rig-smooth over $(C, IC)$ by Lemma 88.4.5. $\square$

Proof. We strongly urge the reader to find their own proof and not read the proof that follows. The statement makes sense by Lemma 88.17.1. To see that it is true assume we have rig-smooth morphisms $A \to B$ and $B \to C$ in $\textit{WAdm}^{Noeth}$. Then we can choose an ideal of definition $I \subset A$ such that the topology on $C$ and $B$ is the $I$-adic topology. By Lemma 88.3.5 we obtain an exact sequence

Observe that $H^{-1}(\mathop{N\! L}\nolimits _{B/A}^\wedge \otimes _ B C)$ and $H^{-1}(\mathop{N\! L}\nolimits _{C/B}^\wedge )$ are annihilated by a power of $I$; this follows from Lemma 88.4.2 part (2) combined with More on Algebra, Lemmas 15.84.6 and 15.84.7 (to deal with the base change by $B \to C$). Hence $H^{-1}(\mathop{N\! L}\nolimits _{C/A}^\wedge )$ is annihilated by a power of $I$. Next, by the characterization of rig-smooth algebras in Lemma 88.4.2 part (2) which in turn refers to More on Algebra, Lemma 15.84.10 part (5) we can choose $f_1, \ldots , f_ s \in IB$ and $g_1, \ldots , g_ t \in IC$ such that $V(f_1, \ldots , f_ s) = V(IB)$ and $V(g_1, \ldots , g_ t) = V(IC)$ and such that $H^0(\mathop{N\! L}\nolimits _{B/A}^\wedge )_{f_ i}$ is a finite projective $B_{f_ i}$-module and $H^0(\mathop{N\! L}\nolimits _{C/B}^\wedge )_{g_ j}$ is a finite projective $C_{g_ j}$-module. Since the cohomologies in degree $-1$ vanish upon localization at $f_ ig_ j$ we get a short exact sequence

and we conclude that $H^0(\mathop{N\! L}\nolimits _{C/A}^\wedge )_{f_ ig_ j}$ is a finite projective $C_{f_ ig_ j}$-module as an extension of same. Thus by the criterion in Lemma 88.4.2 part (2) and via that the criterion in More on Algebra, Lemma 15.84.10 part (4) we conclude that $C$ is rig-smooth over $(A, I)$. $\square$

Proof. Let $f \in A$. To prove that $\varphi $ is rig-flat we have to show that $\varphi _{\{ f\} } : A_{\{ f\} } \to B_{\{ f\} }$ is naively rig-flat. Now either by viewing $\varphi _{\{ f\} }$ as a base change of $\varphi $ and using Lemma 88.17.4 or by using the fact that being rig-smooth is a local property (Lemma 88.17.3) we see that $\varphi _{\{ f\} }$ is rig-smooth. Hence it suffices to show that $\varphi $ is naively rig-flat.

Choose a presentation $B = A\{ x_1, \ldots , x_ n\} /J$. In order to check the second part of the lemma it suffices to check $J_\mathfrak q \subset A\{ x_1, \ldots , x_ n\} _\mathfrak q$ is generated by a regular sequence for $J \subset \mathfrak q$ for $\mathfrak q$ maximal with respect to not containing an ideal of definition, see Algebra, Lemma 10.68.6 (which shows that the set of primes in $V(J)$ where there is a regular sequence generating $J$ is open). In other words, we may assume $\mathfrak q$ is rig-closed in $A\{ x_1, \ldots , x_ n\} $. And to check that $B$ is naively rig-flat, it also suffices to check that the corresponding localizations $B_\mathfrak q$ are flat over $A$.

Let $\mathfrak q \subset A\{ x_1, \ldots , x_ n\} $ be rig-closed with $J \subset \mathfrak q$. By Lemma 88.14.9 we may choose an $f \in A$ mapping to a unit in $A\{ x_1, \ldots , x_ n\} /\mathfrak q$ and such that the prime ideal $\mathfrak p'$ in $A_{\{ f\} }$ induced is rig-closed. Below we will use that $A_{\{ f\} }\{ x_1, \ldots , x_ n\} = A\{ x_1, \ldots , x_ n\} _{\{ f\} }$; details omitted. Consider the diagram

The middle horizontal arrow is faithfully flat as it is a local homomorphism of local rings and flat as $A_{\{ f\} }\{ x_1, \ldots , x_ n\} $ is the completion of a localization of the Noetherian ring $A\{ x_1, \ldots , x_ n\} $. Similarly the bottom horizontal arrow is flat. Hence to show that $J_\mathfrak q$ is generated by a regular sequence and that $A \to A\{ x_1, \ldots , x_ n\} _{\mathfrak q} / J_\mathfrak q$ is flat, it suffices to prove the same things for $J A_{\{ f\} }\{ x_1, \ldots , x_ n\} _{\mathfrak q'}$ and $A_{\{ f\} } \to A_{\{ f\} }\{ x_1, \ldots , x_ n\} _{\mathfrak q'}/ J A_{\{ f\} }\{ x_1, \ldots , x_ n\} _{\mathfrak q'}$. See Algebra, Lemma 10.68.5 or More on Algebra, Lemma 15.32.4 for the statement on regular sequences. Finally, we have already seen that $A_{\{ f\} } \to B_{\{ f\} }$ is rig-smooth. This reduces us to the case discussed in the next paragraph.

Let $\mathfrak q \subset A\{ x_1, \ldots , x_ n\} $ be rig-closed with $J \subset \mathfrak q$ such that moreover $\mathfrak p = A \cap \mathfrak q$ is rig-closed as well. By the characterization of rig-smooth algebras given in Lemma 88.4.2 after reordering the variables $x_1, \ldots , x_ n$ we can find $m \geq 0$ and $f_1, \ldots , f_ m \in J$ such that

1. $J_\mathfrak q$ is generated by $f_1, \ldots , f_ m$, and

2. $\det _{1 \leq i, j \leq m}(\partial f_ j/ \partial x_ i)$ maps to a unit in $A\{ x_1, \ldots , x_ n\} _\mathfrak q$.

By Lemma 88.14.12 the fibre ring

is regular. Observe that the $A$-derivations $\partial / \partial x_ i$ extend (uniquely) to derivations $D_ i : F \to F$. By More on Algebra, Lemma 15.48.3 we see that $f_1, \ldots , f_ m$ map to a regular sequence in $F_\mathfrak q$. By flatness of $A \to A\{ x_1, \ldots , x_ n\} $ and Algebra, Lemma 10.99.3 this shows that $f_1, \ldots , f_ m$ map to a regular sequence in $A\{ x_1, \ldots , x_ m\} _\mathfrak q$ and the quotient by these elements is flat over $A$. This finishes the proof. $\square$

Proof. Let $\overline{\mathfrak q} \subset C$ be a prime ideal which does not contain an ideal of definition. Since the modules in question are finite it suffices to show that

is injective. As in the proof of Lemma 88.3.5 choose presentations $B = A\{ x_1, \ldots , x_ r\} /J$, $C = B\{ y_1, \ldots , y_ s\} /J'$, and $C = A\{ x_1, \ldots , x_ r, y_1, \ldots , y_ s\} /K$. Looking at the diagram in the proof of Lemma 88.3.5 we see that it suffices to show that $J/J^2 \otimes _ B C \to K/K^2$ is injective after localization at the prime ideal $\mathfrak q \subset A\{ x_1, \ldots , x_ r, y_1, \ldots , y_ s\} $ corresponding to $\overline{\mathfrak q}$. Please compare with More on Algebra, Lemma 15.33.6 and its proof. This is the same as asking $J/KJ \to K/K^2$ to be injective after localization at $\mathfrak q$. Equivalently, we have to show that $J_\mathfrak q \cap K^2_\mathfrak q = (KJ)_\mathfrak q$. By Lemma 88.17.6 we know that $(K/J)_\mathfrak q = J'_\mathfrak q$ is generated by a regular sequence. Hence the desired intersection property follows from More on Algebra, Lemma 15.32.5 (and the fact that an ideal generated by a regular sequence is $H_1$-regular, see More on Algebra, Section 15.32). $\square$