Lemma 33.41.2. Let $X$ be a reduced Nagata scheme of dimension $1$. Let $\nu : X^\nu \to X$ be the normalization. Let $x \in X$ denote a closed point. Then

1. $\nu : X^\nu \to X$ is finite, surjective, and birational,

2. $\mathcal{O}_ X \subset \nu _*\mathcal{O}_{X^\nu }$ and $\nu _*\mathcal{O}_{X^\nu }/\mathcal{O}_ X$ is a direct sum of skyscraper sheaves $\mathcal{Q}_ x$ in the singular points $x$ of $X$,

3. $A' = (\nu _*\mathcal{O}_{X^\nu })_ x$ is the integral closure of $A = \mathcal{O}_{X, x}$ in its total ring of fractions,

4. $\mathcal{Q}_ x = A'/A$ has finite length equal to the $\delta$-invariant of $X$ at $x$,

5. $A'$ is a semi-local ring which is a finite product of Dedekind domains,

6. $A^\wedge$ is a reduced Noetherian complete local ring of dimension $1$,

7. $(A')^\wedge$ is the integral closure of $A^\wedge$ in its total ring of fractions,

8. $(A')^\wedge$ is a finite product of complete discrete valuation rings, and

9. $A'/A \cong (A')^\wedge /A^\wedge$.

Proof. We may and will use all the results of Lemma 33.41.1. Finiteness of $\nu$ follows from Morphisms, Lemma 29.54.10. Since $X$ is reduced, Nagata, of dimension $1$, we see that the regular locus is a dense open $U \subset X$ by More on Algebra, Proposition 15.48.7. Since a regular scheme is normal, this shows that $\nu$ is an isomorphism over $U$. Since $\dim (X) \leq 1$ this implies that $\nu$ is not an isomorphism over a discrete set of closed points $x \in X$. In particular we see that we have a short exact sequence

$0 \to \mathcal{O}_ X \to \nu _*\mathcal{O}_{X^\nu } \to \bigoplus \nolimits _{x \in X \setminus U} \mathcal{Q}_ x \to 0$

As we have the description of the stalks of $\nu _*\mathcal{O}_{X^\nu }$ by Lemma 33.41.1, we conclude that $Q_ x = A'/A$ indeed has length equal to the $\delta$-invariant of $X$ at $x$. Note that $Q_ x \not= 0$ exactly when $x$ is a singular point for example by Lemma 33.39.4. The description of $A'$ as a product of semi-local Dedekind domains follows from Lemma 33.41.1 as well. The relationship between $A$, $A'$, and $(A')^\wedge$ we have see in Lemma 33.39.5 (and its proof). $\square$

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).