Lemma 86.16.3. The property $P(\varphi )=$“$\varphi $ is rig-étale” on arrows of $\textit{WAdm}^{Noeth}$ is a local property as defined in Formal Spaces, Remark 85.17.5.

**Proof.**
This proof is exactly the same as the proof of Lemma 86.14.3. 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 85.17.1. We have to show

$\varphi $ is rig-étale $\Rightarrow \varphi '$ is rig-étale,

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

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 86.11.1 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 86.11.1 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 (86.2.0.2) 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 86.3.2. Applying Lemma 86.3.5 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 86.3.5 and 86.3.6 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 15.83.6. 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 85.15.14. Hence the cohomology modules of $\mathop{N\! L}\nolimits _{B/A}^\wedge $ are annihilated by $I^ c$. It follows from Lemma 86.8.2 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. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)