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.
The original Liouville's number is probably the easiest, but most of the proofs tend to invoke calculus (because why not?), so let me try to show it in a more 7th-grade friendly way. I'll call this the swaths-of-zero approach.
So we know that Liouville's number $L$ looks like this:
.1100010000000000000000010...
with a 1 in the $n!$ places.
When we square it, we get this:
.012100220001000000000000220002...
What happens is that in the $2n!$ places we get a 1,
and in the $p!+q!$ places we get a 2.
(The great thing about this is that it can be explained using the elementary-school algorithm, the one they are all familiar with, for multiplication.)
If we multiply $L$ by an integer and write down the answer, the value of that integer will be "laid bare" as we go deeply enough into $L$'s decimal expansion, as eventually the 1s are far enough away to become that integer without stepping on each other.
Similarly, if we multiply $L^2$ by an integer, we will see that integer in some places, and 2 times that integer in others. For large enough $n,$ if we look between the $n!$ place and the $(n+1)!$ place, the last thing we'll see is that integer written at the $2n!$ place.
Thus the swaths of zero in the multiple of $L$ are, $n!-(n-1)!=(n-1)(n-1)!$ long (minus a constant), whereas the widest swaths of zero in the multiple of $L^2$ are $n!-2(n-1)!=(n-2)(n-1)!$ (minus a constant) long, which is shorter, so there is no way to add positive multiples of $L$ and $L^2$ together to clear everything after the decimal point, or find positive multiples of each so that everything after the decimal point is equal.
More generally:
Suppose $a_jL^j+...$ and $a_kL^k+...$ are integer polynomials in $L,$ where $j>k.$ We show that their values cannot match up fully past the decimal point. The swaths of zero in the first polynomial, moving back from the $n!$ spot, are a constant away from $(n-j)(n-1)!$ long (the constant being the length of the sum of the coefficients), whereas in the second they are a constant away from $(n-k)(n-1)!$ long, in the same place (moving back from the $n!$ spot).
I don't know if this explanation holds up to the standards of rigor you like to maintain when teaching them, but I think they will find it fascinating.
Best Answer
No, you cannot choose $T$ so nicely. For cardinality reasons, we can choose some $s\in S$ with $s\notin \overline{\mathbb{Q}}$. Now, consider the element $\sqrt{s}$. I'll leave it to you to consider how large this makes $T$.