Lemma 33.39.2. Let $(A, \mathfrak m)$ be a reduced Nagata $1$-dimensional local ring. Let $A'$ be the integral closure of $A$ in the total ring of fractions of $A$. Then $A'$ is a normal Nagata ring, $A \to A'$ is finite, and $A'/A$ has finite length as an $A$-module.

Proof. The total ring of fractions is essentially of finite type over $A$ hence $A \to A'$ is finite because $A$ is Nagata, see Algebra, Lemma 10.162.2. The ring $A'$ is normal for example by Algebra, Lemma 10.37.16 and 10.31.6. The ring $A'$ is Nagata for example by Algebra, Lemma 10.162.5. Choose $f \in \mathfrak m$ as in Lemma 33.39.1. As $A' \subset A_ f$ it is clear that $A_ f = A'_ f$. Hence the support of the finite $A$-module $A'/A$ is contained in $\{ \mathfrak m\}$. It follows that it has finite length by Algebra, Lemma 10.62.3. $\square$

There are also:

• 3 comment(s) on Section 33.39: The delta invariant

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).