The Stacks project

Lemma 15.33.8. Let $A \to B$ be a local homomorphism of local rings. Let $A^ h \to B^ h$, resp. $A^{sh} \to B^{sh}$ be the induced map on henselizations, resp. strict henselizations (Algebra, Lemma 10.155.6, resp. Lemma 10.155.10). Then $\mathop{N\! L}\nolimits _{B/A} \otimes _ B B^ h \to \mathop{N\! L}\nolimits _{B^ h/A^ h}$ and $\mathop{N\! L}\nolimits _{B/A} \otimes _ B B^{sh} \to \mathop{N\! L}\nolimits _{B^{sh}/A^{sh}}$ induce isomorphisms on cohomology groups.

Proof. Since $A^ h$ is a filtered colimit of étale algebras over $A$ we see that $\mathop{N\! L}\nolimits _{A^ h/A}$ is an acyclic complex by Algebra, Lemma 10.134.9 and Algebra, Definition 10.143.1. The same is true for $B^ h/B$. Using the Jacobi-Zariski sequence (Algebra, Lemma 10.134.4) for $A \to A^ h \to B^ h$ we find that $\mathop{N\! L}\nolimits _{B^ h/A} \to \mathop{N\! L}\nolimits _{B^ h/A^ h}$ induces isomorphisms on cohomology groups. Moreover, an étale ring map is a local complete intersection as it is even a global complete intersection, see Algebra, Lemma 10.143.2. By Lemma 15.33.7 we get a six term exact Jacobi-Zariski sequence associated to $A \to B \to B^ h$ which proves that $\mathop{N\! L}\nolimits _{B/A} \otimes _ B B^ h \to \mathop{N\! L}\nolimits _{B^ h/A}$ induces isomorphisms on cohomology groups. This finishes the proof in the case of the map on henselizations. The case of strict henselization is proved in exactly the same manner. $\square$


Comments (0)


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Your email address will not be published.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.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0D08. Beware of the difference between the letter 'O' and the digit '0'.