# Algebraic closure of a subfield of the field of fraction of a variety

I think that there is the following claim in Basic Algebraic geometry of Shafarevich.

Let $X\to Y$ be a dominant morphism of varieties over an alebraically closed field $k$ of $char=0$. Let $\varphi: k(Y)\to k(X)$ be the corresponding embedding. Denote by $F$ the subfield of $k(X)$ of all elements algebraic over $\varphi(k(Y))$. Then $F$ is a of field "finite type", i.e., the field of rational functions of a (finite-dimensional affine) variety over $k$.

How does one prove such a statement?

## 1 Answer

$K(X)$ is a finitely generated field extension of $k$, so any field between $k$ and $K(X)$ is also such an extension. This is believable but harder to prove than one might expect. See, for example, this MO answer.

• Thank you for sharing the link. I wonder if there is some classical book where this proof is written down. Apr 14 '14 at 16:50
• @aglearner Georges gives a pointer to Isaacs's Algebra here.
– Hoot
Apr 14 '14 at 17:21