Lemma 33.39.5. Let $(A, \mathfrak m)$ be a strictly henselian $1$-dimensional reduced Nagata local ring. Then

If equality holds, then $A$ is a wedge of $n \geq 1$ strictly henselian discrete valuation rings.

Lemma 33.39.5. Let $(A, \mathfrak m)$ be a strictly henselian $1$-dimensional reduced Nagata local ring. Then

\[ \delta \text{-invariant of }A \geq \text{number of geometric branches of }A - 1 \]

If equality holds, then $A$ is a wedge of $n \geq 1$ strictly henselian discrete valuation rings.

**Proof.**
The number of geometric branches is equal to the number of branches of $A$ (immediate from More on Algebra, Definition 15.105.6). Let $A \to A'$ be as in Lemma 33.38.2. Observe that the number of branches of $A$ is the number of maximal ideals of $A'$, see More on Algebra, Lemma 15.105.7. There is a surjection

\[ A'/A \longrightarrow \left(\prod \nolimits _{\mathfrak m'} \kappa (\mathfrak m')\right)/ \kappa (\mathfrak m) \]

Since $\dim _{\kappa (\mathfrak m)} \prod \kappa (\mathfrak m')$ is $\geq $ the number of branches, the inequality is obvious.

If equality holds, then $\kappa (\mathfrak m') = \kappa (\mathfrak m)$ for all $\mathfrak m' \subset A'$ and the displayed arrow above is an isomorphism. Since $A$ is henselian and $A \to A'$ is finite, we see that $A'$ is a product of local henselian rings, see Algebra, Lemma 10.153.4. The factors are the local rings $A'_{\mathfrak m'}$ and as $A'$ is normal, these factors are discrete valuation rings (Algebra, Lemma 10.119.7). Since the displayed arrow is an isomorphism we see that $A$ is indeed the wedge of these local rings. $\square$

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