The Stacks project

Lemma 33.39.6. Let $A$ be a reduced Nagata local ring of dimension $1$. The $\delta $-invariant of $A$ is the same as the $\delta $-invariant of the henselization, strict henselization, or the completion of $A$.

Proof. Let us do this in case of the completion $B = A^\wedge $; the other cases are proved in exactly the same manner. Let $A'$, resp. $B'$ be the integral closure of $A$, resp. $B$ in its total ring of fractions. Then $B' = A' \otimes _ A B$ by Lemma 33.39.5. Hence $B'/B = (A'/A) \otimes _ A B$. The equality now follows from Algebra, Lemma 10.52.13 and the fact that $B \otimes _ A \kappa _ A = \kappa _ B$. $\square$


Comments (2)

Comment #8392 by Xiaolong Liu on

It's better to use instead of as the latter may be confused with .

There are also:

  • 3 comment(s) on Section 33.39: The delta invariant

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Your email address will not be published.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 0C3W. Beware of the difference between the letter 'O' and the digit '0'.