transcendental Galois theory - MathOverflow

transcendental Galois theory

2010-01-16T04:40:18Z

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.

Answer by none for transcendental Galois theory

2010-02-19T18:38:57Z

For some reason I was not able to add this comment to the thread about snappy proofs in model theory, but actually my response sort of fits here as well. The notes of Zilber are a chapter in a new edition of a book Manin, Course in Mathematical Logic for Mathematicians

You might want to look at a ( http://people.maths.ox.ac.uk/~bays/dist/thesis )DPhil thesis of a student of Zilber for a nice model theory explanation of some arithmetics of elliptic curves, and references therein, ( arxiv:0704.3561)this one and ( DOI 10.1007/s10977-007-9015-0 ) this one mentioning Shimura curves.

In particular, these papers deal with automorphisms of infinite extensions of algebraically closed fields, proving things by involved model-theoretic induction etc. Perhaps you might find the model theory methods helpful to whatever your interests are in the transcendental Galois extensions...

Answer by ex falso quodlibet for transcendental Galois theory

2011-01-30T21:57:42Z

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 $\mathbb{C}$.

Answer by Pete L. Clark for transcendental Galois theory

2011-08-02T05:57:27Z

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.