The Stacks project

Proof. This proof is exactly the same as the proof of Lemma 88.17.3. 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-étale $\Rightarrow \varphi '$ is rig-étale,

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

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

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-étale” 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-étale. Then there exists a $c \geq 0$ such that multiplication by $a \in I^ c$ is zero on $\mathop{N\! L}\nolimits _{B/A}^\wedge $ in $D(B)$. This property is preserved under base change by $B \to (B')^\wedge $, see More on Algebra, Lemmas 15.84.6. By the isomorphism above we find that $\varphi '$ is rig-étale. This proves (1).

To prove (2) assume $B \to B'$ is faithfully flat and that $\varphi '$ is rig-étale. Then there exists a $c \geq 0$ such that multiplication by $a \in I^ c$ is zero on $\mathop{N\! L}\nolimits _{(B')^\wedge /(A')^\wedge }^\wedge $ in $D((B')^\wedge )$. By the isomorphism above we see that $a^ c$ annihilates the cohomology modules of $\mathop{N\! L}\nolimits _{B/A}^\wedge \otimes _ B (B')^\wedge $. The composition $B \to (B')^\wedge $ is faithfully flat by our assumption that $B \to B'$ is faithfully flat, see Formal Spaces, Lemma 87.19.14. Hence the cohomology modules of $\mathop{N\! L}\nolimits _{B/A}^\wedge $ are annihilated by $I^ c$. It follows from Lemma 88.8.2 that $\varphi $ is rig-étale. 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. $\square$