Normalization of affine variety

Question

Let $x$ be a point in an irreducible algebraic variety $X$. Is the following equality hold? $$ \mathcal{O}_{\overline{X},x}=\overline{\mathcal{O}_{X,x}} $$ Where $\bar X$ is the normalization of the variety $X$ and $\overline{\mathcal{O}_{X,x}}$ is the integral closure of ${\mathcal{O}_{X,x}}$ in its field of fractions.

Answer 1

Let's consider the affine case: $A$ is an integral domain which is finitely generated as a $k$-algebra, with quotient field $K$. Let $B$ be the integral closure of $A$ in $K$. If we set $X = \operatorname{Spec}A, \overline{X} = \operatorname{Spec}B$, then the inclusion $A \subseteq B$ gives a dominant morphism of varieties $f: \overline{X} \rightarrow X$. This morphism is also closed (hence surjective) with finite fibers.

If $\mathfrak P = \bar x \in \overline{X}$, and $\mathfrak p = x = f(\bar x) = \mathfrak P \cap A$, then your question is whether $B_{\mathfrak P}$ is the integral closure of $A_{\mathfrak p}$ in $K$.

The answer turns out to be yes if and only if $\mathfrak P$ is the only prime of $B$ lying over $\mathfrak p$, i.e. if the preimage $f^{-1}(x)$ contains a single element.

To see this, let $S = A - \mathfrak p$. Since $B$ is the integral closure of $A$ in $K$, $S^{-1}B$ is the integral closure of $S^{-1}A = A_{\mathfrak p}$ in $K$, and we can identify $\operatorname{Spec} S^{-1}B$ with those primes $\mathfrak Q$ of $B$ for which $\mathfrak Q \cap A \subseteq \mathfrak p$. In particular, by incomparability/going up/going down, the maximal ideals of $S^{-1}B$ are those primes of $B$ which lie over $\mathfrak p$.

Now, by the transitivity of localization, $B_{\mathfrak P}$ is the localization of $S^{-1}B$ at the prime $\mathfrak P$.

Thus, $B_{\mathfrak P}$ is the integral closure of of $A_{\mathfrak p}$ if and only if $B_{\mathfrak P} = S^{-1}B$, if and only if $\mathfrak P$ is the only prime of $B$ lying over $\mathfrak p$.

If you are just sticking to closed points (maximal ideals), you can probably simplify this argument.

And by covering a general irreducible variety $X$ with affine open sets $U$, using the fact that $f^{-1}(U)$ corresponds to the integral closure of $U$, you can reduce to the affine case to say the same thing about nonaffine varieties.