Lemma 53.16.1. Let $k$ be a separably closed field. Let $A$ be a $1$-dimensional reduced Nagata local $k$-algebra with residue field $k$. Then

$\delta \text{-invariant }A \geq \text{number of branches of }A - 1$

If equality holds, then $A^\wedge$ is as in (53.16.0.1).

Proof. Since the residue field of $A$ is separably closed, the number of branches of $A$ is equal to the number of geometric branches of $A$, see More on Algebra, Definition 15.106.6. The inequality holds by Varieties, Lemma 33.40.6. Assume equality holds. We may replace $A$ by the completion of $A$; this does not change the number of branches or the $\delta$-invariant, see More on Algebra, Lemma 15.108.7 and Varieties, Lemma 33.39.6. Then $A$ is strictly henselian, see Algebra, Lemma 10.153.9. By Varieties, Lemma 33.40.5 we see that $A$ is a wedge of complete discrete valuation rings. Each of these is isomorphic to $k[[t]]$ by Algebra, Lemma 10.160.10. Hence $A$ is as in (53.16.0.1). $\square$

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