# Any transcendence basis is separable trancendence basis in separable transcendental extension

Suppose $$k$$ is a perfect field and $$K/k$$ is finitely generated separable transcendental extension i.e. there is a purely transcendental extension $$k(x_1, x_2,.., x_r)/k$$ such that $$K/k(x_1, x_2,.., x_r)$$ is separable algebraic extension. In such case $$\{x_1, x_2, .., x_r\}$$ is separable transcendence basis of $$K$$ over $$k$$. My question is whether any transcendence basis is separable i.e if there is tower of extensions $$k \subset k(x_1^{\prime}, x_2^{\prime},.., x_r^{\prime}) \subset L$$ such that $$k(x_1^{\prime}, x_2^{\prime},.., x_r^{\prime})/k$$ is purely transcendental and $$L/k(x_1^{\prime}, x_2^{\prime},.., x_r^{\prime})$$ is algebraic extension, then is $$L/k(x_1^{\prime}, x_2^{\prime},.., x_r^{\prime})$$ is separable algebraic extension?

I think this is true. Indeed over a perfect field any subextension is separably generated. If $$\alpha \in K$$, $$k(x_1^{\prime}, x_2^{\prime},.., x_r^{\prime}, \alpha)/k$$ is separably generated. So $$k(x_1^{\prime}, x_2^{\prime},.., x_r^{\prime}, \alpha)/k$$ has separable transcendence basis. I am trying to show $$\{x_1^{\prime}, x_2^{\prime},.., x_r^{\prime,})\}$$ is a separable transcendence basis of $$k(x_1^{\prime}, x_2^{\prime},.., x_r^{\prime})/k$$, However I can't get it.

I have arrived at this problem in the context of the following algebraic geometry problem: suppose there is a dominant morphism $$\phi: \mathbb{A}_k \times C \to X$$ between smooth affine surfaces over an algebraically closed field $$k$$. Suppose moreover $$X$$ is a rational surface. Then I want to show $$\phi$$ is a separable morphism i.e.the induced morphism between the residue fields $$\phi^*: k(X) \to k(\mathbb{A}^1_k \times C)$$ is separable extension. Since $$C$$ is smooth curve over $$k$$, so $$k(\mathbb{A}^1_k \times C)/k$$ is separable transcendental extension. Since both are surfaces, $$\phi^*$$ is an algebraic extension. From here I thought that $$\phi^*$$ is a separable algebraic extension since $$X$$ is a rational. However I can't prove it. Please clarify in case I am wrong. Thanks in advance.

• Welcome to MSE - this is in my mind a natural question to ask (+1) given some of the material you may have read around the concept (particular theorems guaranteeing the existence of separable transcendence bases, for instance). This sort of motivation could be made more explicit in your post, though, and it would improve your post a bit - I'd suggest adding it with an edit. Commented Nov 8, 2022 at 15:11

Unfortunately this is not true. Let $$k=\overline{\Bbb F_p}$$ and $$K=k(t)$$, which is separably generated by $$t$$. Then consider the tower $$k\subset k(t^p)\subset k(t)=K$$. The extension $$k\subset k(t^p)$$ is purely transcendental, but the algebraic extension $$k(t^p)\subset k(t)$$ is purely inseparable.