# Integral closure of affine domains

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.

New contributor
N12 is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.

## 1 Answer

$$Q(B)$$ 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 $$K(X)$$.