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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: $\spadesuit$, $\diamondsuit$.

fantasie
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2 Answers2

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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 $F$).

Eric Wofsey
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The answer is no in general. This type of questions are related to birational geometry.

See https://en.wikipedia.org/wiki/Rational_variety

Where it links to a paper by Swan, which shows that

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 $p=47$.

I haven't read this paper, and I don't know whether there is an easier construction.

Just a user
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