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 $B \otimes _ A \kappa _ A = \kappa _ B$. $\square$

Comment #8392 by Xiaolong Liu on

It's better to use $B'/B = (A'/A) \otimes _ A B$ instead of $B'/B = A'/A \otimes _ A B$ as the latter may be confused with $B'/B = A'/(A \otimes _ A B)$.

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