# Is the function field of a variety a function field?

Let $X$ be an integral Noetherian scheme of dimension $n$ over a field $k$ (arbitrary field).

• The function field of $X$ is defined as $K(X):=\mathcal O_{X,\eta}$ where $\eta$ is the generic point of $X$.

• A field $K\supset k$ is called an algebraic function field in $n$ variables over $k$ if the extension $K|k$ is finitely generated, regular (it means that $k$ is algebraically closed in $K$) and of transcendence degree $n$.

How can I prove that $K(X)$ is a an algebraic function field in $n$ variables over $k$? Also just a reference for the proof will be appreciated.

Important edit: Ok, this fact is well shown in literature but there is still a problem.

Usually, for an algebraic function field $K|k$ one requires that $k$ is algebraically closed in $K$ (namely $k$ is the field of constants). When is this true for the function field $K(X)|k$? Which are the conditions on $X$ and $k$?

• Can you do this for varieties? – Mariano Suárez-Álvarez Feb 20 '16 at 1:53
• I mean, this should be the case precisely when $X$ is a geometrically integral $k$-variety. – Alex Youcis Feb 20 '16 at 2:08
• I guess that I should comment that my above comment only holds when $k$ is perfect. – Alex Youcis Feb 20 '16 at 9:03
• So, should I assume that $k$ is perfect? – Dubious Feb 20 '16 at 16:15
• This is shown, for example, in Hartshorne, Prop. Thm. I.4.4, or Görtz/Wedhorn, Prop. 9.35. – Takumi Murayama Feb 22 '16 at 4:16