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 $B \otimes _ A \kappa _ A = \kappa _ B$. $\square$
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Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.