# Is every algebraic extension of a field of absolute transcendence degree one a separable extension of a purely inseparable extension?

Any decent course on field theory will state that in characteristic $p$ an extension of fields $k\subset K$ canonically decomposes as the tower $k\subset K_{sep}\subset K$ with $K$ purely inseparable over its subfield $K_{sep}$ of elements separable over $k$.
The alert student will then ask if the order can be reversed, that is whether $K$ is separable over its subfield $K_{perfect}$ of elements purely inseparable over $k$, and be told that the the answer is no.

However in all the counterexamples I am aware of, the base field is of absolute transcendence degree two: $trdeg_{\mathbb F_p}(k)=2$ ( usually $k=\mathbb F_p(x,y)$, the purely transcendental field in two indeterminates over $\mathbb F_p$).
Of course no counterexample can be found with $k$ of absolute transcendence degree zero, because $k$ would then be algebraic over $\mathbb F_p$ and thus perfect. I have the suspicion that no counterexample can be found with $k$ of absolute transcendence degree one either and that is what I'd like to ask here:

Question Let $k$ be a field with $trdeg_{\mathbb F_p}(k)=1$ and $k\subset K$ an algebraic extension. Is the field $K$ separable over its subfield $K_{perfect}$ in the tower $k\subset K_{perfect} \subset K$ ?

Edit ulrich has solved the question ( with even more general hypotheses!) in the affirmative: congratulations and thank you, ulrich!
For reference purposes and because much information is in the comments to his answer, I'd like to sum up and display the main points of his subtle proof.

General result Let $F\subset k$ be an extension of fields with $F$ perfect and $trdeg_{F}(k)=1$.Then for any algebraic extension $k\subset K$ the field $K$ is separable over its subfield $K_{perfect}$ of purely inseparable elements over $k$.

[My question was the case $F=\mathbb F_p$]

Core result Let $F\subset E$ be an extension of fields with $F$ perfect and $trdeg_{F}(E)=1$. Then an algebraic extension $E\subset K$ is separable as soon as the only elements of $K$ purely inseparable over $E$ are already in $E$.

[This clearly follows from the general result and conversely ulrich reduces the general case to this core result by taking $E=K_{perfect}$ . This core result is false for transcendence degree $trdeg_{F}E \geq2$: cf. my answer, lifted from Bourbaki, here ]

Dimensional lemma If a field $L$ has transcendence degree $1$ over a perfect field $F$, then the purely inseparable extension $L\subset L^{1/p}$ has dimension $1$ or $p$ according as $L$ is perfect or not.

[Ulrich uses it to prove the core result. He considers the tower $E\subset L=E^{sep} \subset K$ and proves that if $L\neq K$, then we would have $L\subsetneq L^{1/p} \subset K$ thanks to the lemma . From there the contradiction to the hypothesis of the core lemma $E \subsetneq E^{1/p} \subset L^{1/p} \subset K$ obtains. (Notice that non-perfection of $L$ implies that of $E$, hence the asserted strict inclusion $E \subsetneq E^{1/p}$.) This dimensional lemma was proved in a particular case in this link furnished by Martin, and generalized by ulrich. This key technical lemma is exactly the reason why the transcendence degree one hypothesis is needed: the lemma is false for higher transcendence degree.]

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Dear Georges, I would just like to say that the result is probably known to many people though I have not seen it anywhere in print. I had worked out the finitely generated case a few years ago when I was asked about the corresponding statement for curves by a colleague. –  ulrich Jul 12 '11 at 6:43
If you are looking for a reference, check out Deuring, Lectures on the theory of algebraic functions of one variable, Springer LNM 314. If I recall correctly, this is proved there (but I don't have it handy, so I am not sure). The result is probably much older than that. –  Felipe Voloch Jul 12 '11 at 11:27
Thanks for the reference,Felipe. –  Georges Elencwajg Jul 12 '11 at 16:44

Yes. One may also replace $\mathbb{F}_p$ by any perfect field $F$.
The reason is that any non-perfect algebraic extension $L$ of $k$ has a unique inseparable extension of degree $p$, i.e. $L^{1/p}$, the field obtained from $L$ by adjoining all $p$'th roots. This follows from the fact that $L^{1/p}$ is of degree at most $p$ over $L$.
Assuming this one can prove the claim as follows: Suppose it is not true. Then we can find subfields $L,M$ of $K$ so that we have inclusions $K_{perfect} \subset L \subset M \subset K$ so that $M$ is inseparable of degree $p$ over $L$. By the above remark we must have $M = L^{1/p}$ so we must have $K_{perfect}^{1/p} \subset L^{1/p}\subset K$. If $K_{perfect}^{1/p} = K_{perfect}$, then it is perfect so $L,M$ as above cannot exist. If not, then it contradicts the definition of $K_{perfect}$.
Dear ulrich, could you please explain why $L^{1/p}$ has degree at most $p$ over $L$? –  Georges Elencwajg Jul 11 '11 at 11:30
Also, I don't understand at all the nature of $L$ and $M$: are they somehow canonically associated to the inclusion $K_{perf}\subset K$? –  Georges Elencwajg Jul 11 '11 at 11:50
For the first question math.stackexchange.com/questions/33286/degree-of-frobenius helps if $L/k$ is finite. –  Martin Brandenburg Jul 11 '11 at 11:55
@Georges: For your first question, the link given by Martin gives the argument if $L/k$ is finite. For a general algebraic extension $L/k$, if the claim is not true then there exist $x,y \in L$ such that $L \subsetneq L(x^{1/p}) \subsetneq L(x^{1/p},y^{1/p})$. Replacing $L$ by the finite extension $k(x,y)$ of k$we see that this cannot happen. – ulrich Jul 11 '11 at 12:29 @Georges: For your second question,$L$and$M$are not canonical in any way. All I am using is that if$K_{perfect} \subset K$is not separable, then there must be some purely inseparable extension, so one of degree$p\$, sandwiched in between. –  ulrich Jul 11 '11 at 12:32