# Function fields of varieties and separability

I am recently reading Linear Algebraic Groups by Humphreys, and in section 5.5, he gives a general definition of separable field extension, as follows:

Definition. A field extension $$E/F$$ is said to be separable if either $$\operatorname{char} F=0$$, or else char $$F=p$$ and $$p^{th}$$ powers of elements $$x_1,\dots, x_r \in E$$ linearly independent over $$F$$ are also linearly independent over $$F$$.

Confusion. Then he says that function fields of irreducible varieties are separable over $$K$$. In the book we assume the base field $$K$$ of the variety is algebraically closed, but I don't see why this property holds.

I have looked up some materials and I'll list something I already knew:

• This general definition of separability really generalizes the definition in the algebraic extension case.
• For finitely generated field extensions of any field, the definition of "separable" is equivalent to "separably generated".

Definition. A field $$E$$ is separably generated over $$F$$ if it has a separating transcendence basis $$\{ x_ i; i \in I\}$$ of $$E/F$$ such that the extension $$E/F(x_ i; i \in I)$$ is separable.

My doubts.

By Noether's normalization lemma, we can decompose the function fields into $$K \rightarrow K(x_1,\dots,x_d) \rightarrow K(X)$$, where $$K(X)$$ is the function field, and $$x_1,\dots,x_d$$ are the transcendence basis on which $$K(X)$$ is algebraic. If I am doing right, the algebraic extension should be separable. But I don't think it true...

Is the statement in Humphreys book really true? If so, any proofs? If not, are there any counterexamples? I'll appreciate for your answers! Please correct me if I got anything wrong!!

Suppose $$\operatorname{char} F = p$$ and $$x_1,\cdots,x_r\in E$$ are elements such that $$\sum_{i=1}^r c_ix_i^p = 0$$ in $$E$$ where $$c_i\in F$$. If $$F$$ is algebraically closed, then there exists $$d_i\in F$$ such that $$d_i^p=c_i$$, so we have $$0=\sum_{i=1}^r c_ix_i^p = \sum_{i=1}^r d_i^px_i^p = \left(\sum_{i=1}^r d_ix_i\right)^p$$ in $$E$$, demonstrating that the set $$\{x_1,\cdots,x_r\}$$ is not linearly independent in $$E$$ viewed as an $$F$$-vector space.
Additional comments: your attempt has a bit of a hole. There's no guarantee that all choices of transcendence basis for $$K(X)$$ over $$K$$ will give a separable extension $$K(x_1,\cdots,x_d)\to K(X)$$: consider $$K\subset K(t^p)\subset K(t)$$ for an accessible example. What's true is that for a perfect field $$K$$ and a finitely generated extension $$K\subset L$$, there exists a separating transcendence basis, i.e. elements $$t_1,\cdots,t_n$$ so that the extension $$K\subset K(t_1,\cdots,t_n)$$ is purely transcendental and $$K(t_1,\cdots,t_n)\subset L$$ is finite separable. See for instance Zariski-Samuel's Commutative Algebra Ch. II, theorem 31 or Matsumura's Commutative Algebra Ch. 10, the corollary on p. 194.