The Stacks project

Lemma 91.8.7. Let $A \to B$ be a local ring homomorphism of local rings. Let $A^ h \to B^ h$, resp. $A^{sh} \to B^{sh}$ be the induced maps of henselizations, resp. strict henselizations. Then

\[ L_{B^ h/A^ h} = L_{B^ h/A} = L_{B/A} \otimes _ B^\mathbf {L} B^ h \quad \text{resp.}\quad L_{B^{sh}/A^{sh}} = L_{B^{sh}/A} = L_{B/A} \otimes _ B^\mathbf {L} B^{sh} \]

in $D(B^ h)$, resp. $D(B^{sh})$.

Proof. The complexes $L_{A^ h/A}$, $L_{A^{sh}/A}$, $L_{B^ h/B}$, and $L_{B^{sh}/B}$ are all zero by Lemma 91.8.4. Using the fundamental distinguished triangle ( for $A \to B \to B^ h$ we obtain $L_{B^ h/A} = L_{B/A} \otimes _ B^\mathbf {L} B^ h$. Using the fundamental triangle for $A \to A^ h \to B^ h$ we obtain $L_{B^ h/A^ h} = L_{B^ h/A}$. Similarly for strict henselizations. $\square$

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Your email address will not be published.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 08UN. Beware of the difference between the letter 'O' and the digit '0'.