transcendental Galois theory

Suppose we define an arbitrary field extension $K/F$ to be Galois if, for all subextensions $L$ of $K/F$, we have $K^{\operatorname{Aut}(K/L)} = L$. In words: for any element $x$ of $K \setminus L$, there exists an automorphism $s$ of $K$ such that $s(l) = l$ for all $l$ in $L$, but $s(x) \neq x$. (Note that in case $K/F$ is algebraic, this is indeed a characteristic property of Galois extensions.) What are the transcendental Galois extensions?

In my rough notes Transcendental Galois Theory, I show that if $F$ has characteristic $0$ and $K$ is algebraically closed, then $K/F$ is Galois in the above sense. [Actually, these notes are somewhat incomplete. Having been unable to complete the proof of the conjecture below, I left out some of the more straightforward details. If anyone wants to see more detail on anything in these notes, please let me know.]

I also conjectured: if $K/F$ is Galois, then either $K/F$ is algebraic, normal and separable, or $F$ has characteristic $0$ and $K$ is algebraically closed. Is this true?

Comment: It is easy to see that if $K/F$ is not algebraic, then $K$ must have characteristic $0$. It is possible to modify the question a bit so that the positive characteristic case is not ruled out, but I would like to understand what's going on in characteristic $0$ first!

In my notes, I show that an affirmative answer follows from a certain (arguably) less weird conjecture about Galois closures of subfields of rational function fields. If there is any interest, I will reproduce this conjecture here explicitly.

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I suppose for transcendental extensions Artin's theorem on linear independence of characters will break down, and without this powerful theorem it will be hard to draw any consequences. Are there deeper reasons for you getting interested in transcendental Galois extensions, in addition to it being a curiosity? –  Anweshi Jan 16 '10 at 14:10
I would argue that curiosity is one of the deepest reasons to be interested in something! But if you are asking whether there is some other conjecture up my sleeve that would follow from this: no, it is not related to any of my "real" work. –  Pete L. Clark Feb 20 '10 at 7:06
I would argue that once one moves beyond etale extensions that one should not look at discrete Galois groups. E.g. Jacobson's theory for exponent purely inseparable extensions uses Lie algebras. For a Galois theory for extensions $E/F$ I would be more inclined to look at subalgebras of the derivations (or of the differential operators) of $E$ over $F$. See e.g. "On Galois correspondence between intermediate fields and closed derivation subalgebras" Teppei Kikuchi J. Math. Kyoto Univ. 23, (1983), 281-287. –  user10849 Jan 6 '11 at 21:36
That's a very intriguing question... An exercise in Bourbaki Algebra (chapter 5 I believe) calls "Dedekind extensions" your Galois extensions. Unfortunately, they don't propose much in the interesting direction, except the following: "If E is a Dedekind extension of K and L a pure transcendental subextension of E, distinct from K and such that E is an algebraic extension of L, show that E has infinite degree over L." I tried to use this after reading your very interesting notes, but without success... –  Sylvain Bonnot Jun 10 '11 at 20:55
@Sylvain: so I checked, and indeed there it is on p. 173 of the English copy of Bourbaki's "Algebra II". By the way, I think that what I do in my note does lead to a solution of this exercise: the definition of Dedekind allows us to replace $K$ with any larger subextension, so we may assume that $L = K(t)$ and then if $E/L$ were finite then $E/K$ would be the function field of an algebraic curve: I do that case! –  Pete L. Clark Aug 2 '11 at 5:29

Even in our day of sophisticated search engines, it still seems that the success of a search often turns on knowing exactly the right keyword.

I just followed up on Sylvain Bonnot's comment above. The property of a field extension $K/F$ that for all subextensions $L$ we have $K^{\operatorname{Aut}(K/L)} = L$ is apparently most commonly called Dedekind. This terminology appears in Exercise V.9 of Bourbaki's Algebra II, where the reader is asked to show that if $L/K$ is a nonalgebraic Dedekind extension and $T$ is a transcendence basis, then $L/K(T)$ must have infinite degree. Ironically, this is exactly what I could show in my note. One can (in the general case, even...) immediately reduce to the case $T = \{t\}$ and then the exercise is saying that the function field $K(C)$ of an algebraic curve (again, it is no loss of generality to assume the function field is regular by enlarging $K$) is not Dedekind over $K$. This is kind of a strange coincidence! [However, the proof I give is openly geometric so is probably not the one that N.B. had in mind...]

It also appears in

MR0067098 (16,669f) Barbilian, D. Solution exhaustive du problème de Steinitz. (Romanian. Russian, French summary) Acad. Repub. Pop. Române. Stud. Cerc. Mat. 2, (1951). 195–259 (misprinted 189–253).

In this paper, the author shows that $L/K$ is a Dedekind extension iff for all subextensions $M$, the algebraic closure $M^*$ of $M$ in $L$ is such that $M^*/M$ is Galois in the usual sense: i.e., normal and separable. (This is a nice fact, I suppose, and I didn't know it before, but it seems that the author regarded this as a solution of the problem of which extensions are Dedekind. I don't agree with that, since it doesn't answer my question!)

Apparently one is not supposed to read the above paper but rather this one:

MR0056588 (15,97b) Krull, Wolfgang Über eine Verallgemeinerung des Normalkörperbegriffs. (German) J. Reine Angew. Math. 191, (1953). 54–63.

Here is the MathSciNet review by E.R. Kolchin (who knew something about transcendental Galois extensions!):

The author reviews a definition and some results of D. Barbilian [Solutia exhaustiva a problemai lui Steinitz, Acad. Repub. Pop. Române. Stud. Cerc. Mat. 2, 189--253 (1950), unavailable in this country], providing proofs which are said to be simpler, and further results. Let L be an extension of a field K. Then L is called normal over K if for every intermediate field M the relative algebraic closure M∗ of M in L is normal (in the usual sense) over M. If L has the property that every M is uniquely determined by the automorphism group U(M) of L over M, then L is normal over K and, if the characteristic p=0, conversely; if p>0 the converse fails but a certain weaker conclusion is obtained. Various further results are found, and constructive aspects of normal extensions are explored. Some open questions are discussed, the most important one being: Do there exist transcendental normal extensions which are not algebraically closed?

So it seems that my question is a nearly 60 year-old problem which was considered but left unsolved by Krull. I am tempted to officially give up at this point, and perhaps write up an expository note informing (and warning?) contemporary readers about this circle of ideas. Comments, suggestions and/or advice would be most welcome...

P.S.: Thanks very much to M. Bonnot.

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There is a discussion of the concept of "transcendental Galois extensions" in the paper "On the transcendental Galois extensions" by Feng-Wen An: http://arxiv.org/abs/1004.5036. In the terminology of this paper, an arbitrary field extension $L/K$ is said to be Galois over $K$ iff $L^{\operatorname{Aut}(L/K)}=K$. A field extension $L/K$ is called absolute Galois over $K$ iff for every subextension $L\supseteq F\supseteq K$ the extension $L/F$ is Galois over $F$. The latter corresponds to your definition of Galois extension for arbitrary fields.

According to this paper, the answer to your second question appears to be negative: the "absolute Galois" field extension $\mathbb{C}(t)/\mathbb{C}$ is a counter-example as $\mathbb{C}(t)$ is neither algebraically closed, nor algebraic over $\mathbb{C}$.

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@ex falso: $\mathbb{C}(t)/\mathbb{C}$ is definitely not a Galois extension according to my definition. My definition of Galois implies that if $L/K$ is Galois and $M$ is any subextension, then $L/M$ is also Galois. Taking $M$ to have finite index in $\mathbb{C}(t)$ and applying Luroth's theorem, if this were a Galois extension then every finite covering of $\mathbb{P}^1$ by $\mathbb{P}^1$ would be Galois. But this is false: for instance there are well-known to be $A_5$ Galois covers of $\mathbb{P}^1$ over $\mathbb{P}^1$, and $A_5$ has non-normal subgroups... –  Pete L. Clark Jan 31 '11 at 2:27
Let me also say -- and here let me be careful with what I am saying -- that I have glanced at this paper before as well as others by the same author. The claimed results are interesting, but I've never been able to follow the proofs. –  Pete L. Clark Jan 31 '11 at 2:29
Well, in particular, it is claimed in this paper (Remark 1.6) that if $L$ is a proper subfield of $\mathbb{C}(t)$ containing $\mathbb{C}$, then $L$ is not isomorphic to $\mathbb{C}(t)$. Not a good sign. –  Laurent Moret-Bailly Jan 31 '11 at 11:58
On a second glance over the paper, my guess is that certain statements are meant as yet to be proved or disproved, rather than as true or false statements (especially considering the fact that the paper rather appears to be a draft). I guess I should have added a disclaimer in my answer, lol. Anyways, when I googled "transcendental galois theory", the only two meaningful results appeared to be that thread (without a real answer to it) and the aforementioned draft paper. –  efq Feb 13 '11 at 15:16