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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!!

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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.

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  • $\begingroup$ Thanks! I just realized that different choices of transcendence basis do have effects on the latter separability, and the definition only needs the existence. $\endgroup$ Commented Oct 27, 2023 at 18:28

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