TITLE: Integral Closures and Affine Curves
QUESTION [7 upvotes]: Let $C$ be an irreducible affine curve with singular points, and let $A$ be its ring of regular functions.  Since $C$ has singular points, $A$ is not integrally closed in its field of fractions, $K$.  I'd like to compute $\overline{A}^K$, the integral closure of $A$ in $K$ by resolving the singularities of $C$.
Suppose by blowing up the singularities of $C$ I obtain a non-singular curve $C'$ which happens to be affine itself, and a map $C'\twoheadrightarrow C$ of varieties.  This induces an injection $A\hookrightarrow A'$, where $A'$ is the ring of regular functions of $C'$.  Since $C'$ is non-singular, $A'$ is integrally closed, and I suspect is isomorphic to $\overline{A}^K$.  To verify my suspicion, I would need to show that all elements of $A'$ are integral over $A$.
Can someone help me see why this is true or not true?  I understand that one implication of my suspicion is that if a birational equivalence class of curves has a non-singular affine curve, then it is unique up to isomorphism, but this doesn't seem so hard to believe.

REPLY [3 votes]: The blowing-up morphism $C'\to C$ is projective by definition. It is quasi-finite because $\dim C'=1$ which excludes the possibility for the excetional locus to have positive dimension. So it is finite. This means $A\to A'$ is module-finite, hence integral.