# Is the subextension of a purely transcendental extension purely transcendental over the base field?

Let $$K/E/F$$ be extension of fields, where $$K/F$$ is purely transcendental.

It is generally not true that $$K/E$$ is purely transcendental. For example, take $$F(x)/F(x^2)/F$$. I wonder what is the situation for $$E/F$$. Specifically, is $$E/F$$ purely transcendental?

For now, there is little technique I have learned to prove purely transcendence. Nor do I come up with a counterexample.

EDIT: I think the two answers are both fantastic. For more context, if they are necessary (Related: discussion on meta), here is how did I come up with this:

This is a question I encountered in learning infinite Galois theory, when I was trying to prove this as a lemma for an exercise, but failed (of course). I was surprised that while there are claims similar to this about inseparable extensions, nothing is spoken to transcendental extensions, at least not in my textbook.

I followed from these Lecture Notes: $$\spadesuit$$, $$\diamondsuit$$.

This is a difficult question known as the "Lüroth problem", and the answer depends on the transcendence degree of $$K/F$$. In the transcendence degree $$1$$ case, any subextension must be purely transcendental, and this is known as Lüroth's theorem. For transcendence degree $$2$$ or higher, it is possible for a subextension to not be purely transcendental. In the special case that the base field $$F$$ is algebraically closed and of characteristic $$0$$, any subextension must still be purely transcendental in the transcendence degree $$2$$ case, but there are counterexamples for transcendence degree $$3$$. In all cases except the transcendence degree $$1$$ case, the proofs of these statements are quite hard and use some heavy machinery from algebraic geometry (by thinking of the field $$E$$ as the function field of a variety over the base field $$F$$).
Let $$G\simeq\mathbb Z/p\mathbb Z$$ acts transitively on $$x_1, \cdots, x_p$$. Then $$\mathbb Q(x_1, \cdots, x_p)^G$$ is not purely transcendental over $$\mathbb Q$$ for $$p=47$$.