Show that given \(\pi: X \rightarrow Y\), \(Y\) is integrally closed in \(K(X)\) over \(K(Y)\) if and only if \(\OO_{Y} \rightarrow \pi_{*} \OO_{X}\) is an isomorphism (i.e., \(\pi\) is \(\OO\)-connected).
The problem states no hypotheses on \(X\), \(Y\), or \(\pi\). The solution assumes \(X\) and \(Y\) integral Noetherian, \(\pi\) proper and dominant, so that \(K(Y) \subseteq K(X)\), and \(X\) normal. The implication from \(\OO_Y \cong \pi_* \OO_X\) to integral closedness fails without normality of \(X\). Let \(Y\) be the cuspidal cubic \(V(y^2 - x^3)\) and \(\pi = \id_Y\). Then \(\OO_Y \to \pi_* \OO_X\) is an isomorphism, but \(t = y/x \in K(Y)\) satisfies \(t^2 = x\), so it is integral over \(\OO_Y\) and does not lie in \(\OO_Y\).
Solution
Assume that \(X\) and \(Y\) are integral Noetherian schemes, \(\pi:X\to Y\) is proper and dominant, and \(X\) is normal. Say that \(Y\) is integrally closed in \(K(X)\) if \(\OO_Y(V)\) is integrally closed in \(K(X)\) for every affine open \(V\subseteq Y\). Dominance gives an inclusion \[ K(Y)\subseteq K(X). \]
1. Let\[ V=\Spec A\subseteq Y \]be a nonempty affine open and put\[ U=\pi^{-1}(V), \qquad B=\Gamma(U,\OO_X). \]Then \(B\) is a finite \(A\)-algebra contained in \(K(X)\).
The restricted morphism \[ \pi_U:U\longrightarrow V \] is proper. By proper pushforward of coherent sheaves, or equivalently the finite part of Stein factorisation T-MORSTEIN, \[ (\pi_U)_*\OO_U \] is a coherent \(\OO_V\)-algebra. Since \(V\) is affine, this means \[ B=\Gamma(U,\OO_U) \] is a finite \(A\)-module.
Because \(\pi\) is dominant and \(V\) is nonempty, \(U\) is a nonempty open subset of the integral scheme \(X\). Every regular function on \(U\) determines a rational function on \(X\), giving an injection \[ B\injects K(X). \]
2. Every element of \(B\) is integral over \(A\).
By step 1, \(B\) is finite as an \(A\)-module. Every element of a finite \(A\)-algebra is integral over \(A\). Therefore \[ B \subseteq \overline A^{\,K(X)}, \] where \(\overline A^{\,K(X)}\) denotes the integral closure of \(A\) in \(K(X)\).
3. Since \(X\) is normal, every element of \(K(X)\) integral over \(A\)belongs to \(B\).
Let \[ \alpha\in K(X) \] be integral over \(A\). Thus \(\alpha\) satisfies a monic equation \[ \alpha^n+a_{n-1}\alpha^{n-1}+\cdots+a_0=0, \qquad a_i\in A. \]
Fix \(x\in U\). The map \[ A\longrightarrow\OO_{X,x} \] induced by \(\pi\) carries the same equation into \(\OO_{X,x}\). Hence \(\alpha\) is integral over \(\OO_{X,x}\). Since \(X\) is normal, \(\OO_{X,x}\) is integrally closed in \(K(X)\), so \[ \alpha\in\OO_{X,x}. \]
This holds for every \(x\in U\). Thus \(\alpha\) is regular in a neighborhood of every point of \(U\). These local representatives are all the same rational function and therefore agree on overlaps, so they glue to a section of \(\OO_X(U)\). Hence \[ \alpha\in B. \] Therefore \[ \overline A^{\,K(X)} \subseteq B. \]
4. Consequently,\[ \boxed{ \Gamma(\pi^{-1}V,\OO_X) = \overline{\Gamma(V,\OO_Y)}^{\,K(X)} } \]for every nonempty affine open \(V\subseteq Y\).
Step 2 gives \[ B\subseteq\overline A^{\,K(X)}, \] and step 3 gives the reverse inclusion.
5. If \(Y\) is integrally closed in \(K(X)\), then\[ \OO_Y\xrightarrow{\sim}\pi_*\OO_X. \]
By hypothesis, for every nonempty affine open \[ V=\Spec A\subseteq Y, \] the ring \(A\) is integrally closed in \(K(X)\): \[ A=\overline A^{\,K(X)}. \] Step 4 therefore gives \[ \Gamma(V,\OO_Y) = A = \Gamma(\pi^{-1}V,\OO_X) = \Gamma(V,\pi_*\OO_X). \] These identifications are induced by the natural sheaf map and are compatible with restriction. Hence that map is an isomorphism.
6. Conversely, if\[ \OO_Y\xrightarrow{\sim}\pi_*\OO_X, \]then \(Y\) is integrally closed in \(K(X)\).
Let \[ V=\Spec A\subseteq Y \] be nonempty affine. The sheaf isomorphism gives \[ A = \Gamma(V,\OO_Y) \cong \Gamma(\pi^{-1}V,\OO_X) =B. \] By step 4, \[ B=\overline A^{\,K(X)}. \] Therefore \[ A=\overline A^{\,K(X)}. \] Since this holds on every affine open, \(Y\) is integrally closed in \(K(X)\).
7. Under these hypotheses,\[ \boxed{ Y\text{ is integrally closed in }K(X) \iff \OO_Y\xrightarrow{\sim}\pi_*\OO_X. } \]
Step 5 proves the forward implication and step 6 proves the reverse implication.
8. Q.E.D.
Step 7 is the asserted equivalence.