Lemma 88.17.3. The property $P(\varphi )=$“$\varphi $ is rig-smooth” on arrows of $\textit{WAdm}^{Noeth}$ is a local property as defined in Formal Spaces, Remark 87.21.5.
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
\[ \xymatrix{ B \ar[r] & (B')^\wedge \\ A \ar[r] \ar[u]^\varphi & (A')^\wedge \ar[u]_{\varphi '} } \]
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
$\varphi $ is rig-smooth $\Rightarrow \varphi '$ is rig-smooth,
if $B \to B'$ faithfully flat, then $\varphi '$ is rig-smooth $\Rightarrow \varphi $ is rig-smooth, and
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
\[ H^ i(\mathop{N\! L}\nolimits _{(B')^\wedge /(A')^\wedge }^\wedge ) \to H^ i(\mathop{N\! L}\nolimits _{(B')^\wedge /A}^\wedge ) \]
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
\[ H^ i(\mathop{N\! L}\nolimits _{B/A}^\wedge \otimes _ B (B')^\wedge ) \to H^ i(\mathop{N\! L}\nolimits _{(B')^\wedge /A}^\wedge ) \]
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
\[ \mathop{N\! L}\nolimits _{(B')^\wedge /(A')^\wedge }^\wedge \cong \mathop{N\! L}\nolimits _{B/A}^\wedge \otimes _ B (B')^\wedge \]
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
\[ \mathop{\mathrm{Ext}}\nolimits ^1_ B(\mathop{N\! L}\nolimits _{B/A}^\wedge , N) \otimes _ B (B')^\wedge = \mathop{\mathrm{Ext}}\nolimits ^1_{(B')^\wedge }(\mathop{N\! L}\nolimits _{B/A}^\wedge \otimes _ B (B')^\wedge , N \otimes _ B (B')^\wedge ) \]
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. $\square$
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