Let $A\subset B$ Be affine domains over a field of characteristic zero, say k. We know that the integral closure of $A$ in any finite extension of $Q(A)$ is a finite $A$ module. My question is why the integral closure of $A$ in $B$ is a finite $A$ module?


I have tried to show that the integral closure is again an affine domain over k. But I couldn’t prove this.



Integral closure of affine domains
$Q(B)$ is a finitely generated field extension of $Q(A)$ and so any intermediate field is a finitely generated field extension of $Q(A)$ (see below); in particular the algebraic closure of $Q(A)$ in $Q(B)$ is a finite extension of $Q(A)$ and so the integral closure of $A$ in the algebraic closure of $Q(A)$ in $Q(B)$ is a finite $A$ module (as noted by OP).


If $K \subset L \subset M$ are fields and $M$ is a finitely generated field extension of $K$ then $L$ is a finitely generated field extension of $K$: Let $X=(X\_1,...,X\_l)$ be a (finite) transcendence basis for $L/K$ and let $Y=(Y\_1,...,Y\_m)$ be a (finite) transcendence basis for $M/L$. Then $M:K(X,Y) \lt \infty \implies L(Y):K(X,Y) \lt \infty \implies L:K(X) \lt \infty$ (because $L$ and $K(X,Y)$ are linearly disjoint over $K(X)$.


