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.
I think it's possible that non-geometric extensions are indeed not as directly visualizable as geometric ones.
Some terminology: let $k$ be a field, and either assume $k$ has characteristic $0$ or beware that some separability issues are being omitted in what follows. A (one variable) function field over $k$ is a finitely generated field extension $K/k$ of transcendendence degree
one. This already allows for the possibility of a nontrivial constant extension, which is often excluded in geometric endeavors: for instance, according to this definiton, $\mathbb{C}(t)$ is a function field over $\mathbb{R}$, but a sort of weird[1] one: e.g. it has no $\mathbb{R}$-points.
One says a function field $K/k$ is regular if $k$ is algebraically closed in $K$; i.e., any element of $K$ which is algebraic over $k$ already lies in $k$ [plus separability stuff in positive characteristic]. Any function field can be made regular just by enlarging the constant field to be the algebraic closure of $k$ in $K$; e.g., the previous example is a regular function field over $\mathbb{C}$.
Regularity is what one needs to think about function fields as geometric objects: namely, there is a bijective correspondence between regular function fields $K/k$ and complete, nonsingular algebraic curves $X_{/k}$.
Now, on to covers. Let $L/K$ be a finite degree extension of function fields over $k$. One says (often; this is slightly less standard terminology) that the exension $L/K$ is geometric over $k$ if both $L$ and $K$ are regular function fields. And again, there is a bijective correspondence between geometric extensions of function fields and finite $k$-rational morphisms of algebraic curves $Y \rightarrow X$.
Assuming that the bottom function field $K$ is regular, every extension $L/K$ may be decomposed into a tower of a constant extension $lK/K$ followed by a geometric extension $L/lK$. Constant extensions have a role to play in the theory -- see for instance the chapter on constant extensions in Rosen's Number theory in function fields, but I think it is fair to describe their role as algebraic rather than geometric: at least that's the standard view.
In fact, the issue that not all extensions of regular function fields are geometric is an important technical one in the subject, because sometimes natural algebraic constructions do not preserve the class of geometric extensions.
Here is an example very close to my own heart: let $p$ be an odd prime. The elliptic modular curves $X(1)$ and $X_0(p)$ have canonical models over $\mathbb{Q}$ and there is a natural "forgetful modular" covering $X_0(p) \rightarrow X(1)$. This corresponds to a geometric extension of function fields $\mathbb{Q}(X_0(p)) / \mathbb{Q}(X(1))$. This is not a Galois extension: what is the Galois closure and what is its Galois group? If -- as was classically the case -- our constant field were $\mathbb{C}$ -- then the Galois closure is the function field of the modular curve $X(p)$ and the Galois group of the covering $X(p)/X(1)$ is
$\operatorname{PSL}_2(\mathbb{Z}/p\mathbb{Z})$. However, over $\mathbb{Q}$ the Galois closure also contains the quadratic field $\mathbb{Q}\left(\sqrt{(-1)^{\frac{p-1}{2}} p}\right)$ so is an extension of a cyclic group of order $2$ by $\operatorname{PSL}_2(\mathbb{Z}/p\mathbb{Z})$ (in fact it is $\operatorname{PGL}_2(\mathbb{Z}/p\mathbb{Z})$). Thus the extension is not geometric. This is unfortunate, because Hilbert's Irreducibility Theorem says that if one has a geometric Galois extension $L/k(t)$ with $k$ a number field, then one can realize $\operatorname{Aut}(L/k(t))$ as a Galois group over $k$. So in this case, this obtains $\operatorname{PSL}_2(\mathbb{Z}/p\mathbb{Z})$ as a Galois group over not $\mathbb{Q}$ but over the variable quadratic field given above. K.-y. Shih found a brilliant way to "tweak" this construction to realize $\operatorname{PSL}_2(\mathbb{Z}/p\mathbb{Z})$ over $\mathbb{Q}$ in certain (infinitely many) cases, and other mathematicians -- e.g. Serre, myself, my graduate student Jim Stankewicz -- have put a lot of thought into extending Shih's work, but with only very limited success.
Added: Brian's example in the comments is very nice. Maybe another remark to make is that in the arithmetic theory of coverings of curves (an active branch of arithmetic geometry) the distinction between a Galois extension and a geometrically Galois extension of fields (i.e., one which becomes Galois after base change to $\overline{k}$) is a key one: it's certainly something that many arithmetic geometer think a lot about. It just doesn't come with an obvious "visualization", at least not to me. Not everything in algebraic or arithmetic geometry can be visualized, or at least not visualized in a way common to different workers in the field. For instance, an inseparable field extension $l/k$ is by definition ramified, but I have never seen anyone describe this visually. (There are things you can say to justify that this is not a "covering map", e.g. by pointing to the nonreducedness of $l \otimes_k l$, but I don't think this is direct visualization either. Maybe some would disagree?) What you do is think of the case of a ramified cover of Riemann surfaces, and take away the (key) piece of intuition that an inseparable field extension -- which is, visually speaking, just one closed point mapping to another -- behaves like a ramified cover of Riemann surfaces in many ways. So, as Brian says, in this subject a lot of geometric reasoning proceeds by analogy. Unlike in, say, certain branches of low-dimensional topology, one does not prove a theorem by referring to (allegedly) visually apparent features of one's constructions.
[1]: Those who know me well know that I certainly don't think that a curve is weird just because it has no degree one closed points. More accurate is to say that this curve doesn't have any degree one closed points for a "weird reason".
Best Answer
Szamuely's Galois Groups and Fundamental Groups might be what you're looking for. In particular, the beginning of Chapter 2 (where the discussion switches from field theory to fundamental groups) alludes to a statement like the one you give:
Edit: I notice that this book was discussed in another MO question here: Galois Groups vs. Fundamental Groups .