My question is based on exercise II.6.9 in Hartshorne's Algebraic Geometry.
Suppose that $X$ is a projective curve over $k$ (an algebraically closed field). Let $\tilde{X}$ be its normalization and $\pi: \tilde{X}\rightarrow X$ be the projection map. For each point $P\in X$, let $\mathscr{O}_P$ be its local ring and $\tilde{\mathscr{O}_P}$ be the integral closure of $\mathscr{O}_P$. Then my question is:
Is there a relationship between the stalk of $\pi_*\mathscr{O}_\tilde{X}$ at $P$ and the local ring $\tilde{\mathscr{O}_P}$? Are they equal? If it's true, how to prove it?
My attempt.
In the affine case, suppose $X=\operatorname{Spec}A$ for some domain $A$. I know that $\pi$ is just the ring extension $A\hookrightarrow\tilde{A}$, where $\tilde{A}$ is the integral closure of $A$ in its field of fractions. And $P$ corresponds to a prime ideal $\mathfrak{p}$ of $A$. Then $\tilde{\mathscr{O}_P}$ is the integral closure of $A_\mathfrak{p}$. The stalk of $\pi_*\mathscr{O}_\tilde{X}$ at $P$ can be finally expressed as a direct limit of local rings $\{\tilde{A}_f\}$, where $f$ is taken over all elements in $A\setminus\mathfrak{p}$. So the question is translated into a problem of commutative algebra. That is, is it true for $\tilde{A_\mathfrak{p}}=\varinjlim\tilde{A}_f$?
Maybe we need something given by the assumption on $X$, i.e. $A$ is noetherian and of dimension $1$. But I still have no strategy to deal with it. Could anyone help me? Thanks in advance!
Best Answer
Yes, they are. The formation of integral closure commutes with localization. See Proposition 5.12 in Atiyah-MacDonald's Commutative Algebra book.