Lemma 33.39.4. Let $A$ be a reduced Nagata local ring of dimension $1$. The $\delta $-invariant of $A$ is $0$ if and only if $A$ is a discrete valuation ring.

**Proof.**
If $A$ is a discrete valuation ring, then $A$ is normal and the ring $A'$ is equal to $A$. Conversely, if the $\delta $-invariant of $A$ is $0$, then $A$ is integrally closed in its total ring of fractions which implies that $A$ is normal (Algebra, Lemma 10.37.16) and this forces $A$ to be a discrete valuation ring by Algebra, Lemma 10.119.7.
$\square$

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