Re: Question on algebraic numbers

On Feb 2, 1:45 pm, dmr5...@xxxxxxxxx wrote:
On Feb 1, 8:53 pm, Rupert <rupertmccal...@xxxxxxxxx> wrote:





On Feb 1, 3:54 pm, dmr5...@xxxxxxxxx wrote:

The algebraic closure of Q (the rationals) is, of course, formed by
adjoining to Q the roots of all polynomials over Q.

Consider now the field consisting of all numbers that can be written
as finite expressions involving addition, subtraction, multiplication,
and division of integers, raising to rational powers, and compositions
of these operations.  This is obviously a subfield of the algebraic
closure of Q.

Question: is it a proper subfield?  

Yes, it is. What you have described is the maximal prosolvable
extension of Q. That is, it is the direct limit of all the finite
extensions of Q which have a solvable Galois group.

None of the roots of the polynomial x^5-4x+2 would be included in the
field you described. The Galois group of that polynomial over Q is
known to be S_5, which is not solvable.

OK, so let me make sure I understand this.  My original question in
effect was comparing two ways of extending the rationals.  One is the
standard algebraic closure.  The other is what a smart high school
student who had learned what a field is and had seen irrational roots
and complex numbers might think of as the largest extension of Q, i.e.
the collection of all expressions that could be formed finitely using
the operations of arithmetic along with rational powers.

My understanding (admittedly that of somehow who only majored in math
as an undergraduate) of the insolubility of the quintic and higher
polynomials has been only that there is no general formula that allows
one to express the roots of arbitrary quintics in terms of the
coefficients.  That seems to me (unless I am missing something) to be
different from (and weaker than) saying that there exist algebraic
numbers (say, any of the roots of x^5-4x+2) that cannot be written in
any way as finite combinations of integers using the arithmetic
operations and rational powers.  The absence of a systematic procedure
for algebraically deriving the roots from the coefficients of an
arbitrary quintic or higher degree polynomial does not appear to me to
imply that those roots are not somehow expressible as such finite
combinations of integers and algebraic operations.

So what Rupert is telling me (and any of the rest of us who are
reading) is that the apparently stronger result (that there exist
algebraic numbers that cannot be written in such finite terms at all)
is true.  Thus, more questions:

1) Is this apparently stronger result an easy consequence of the
insolubility of the quintic and higher polynomials that I am
(obtusely) not seeing?  If not an easy consequence, what is its
history?


It does not obviously follow from the insolubility of the general
quintic that there exist specific quintics over Q which are not
solvable by radicals.

Abel was the first to prove the insolubility of the general quintic. I
believe that Galois was the first to produce specific quintics with
coefficients with Q which are not solvable in radicals (although Ted
Hwa is telling us differently). In any event Galois was the first to
provide a general algorithm for determining whether a particular
quintic in Q is solvable by radicals or not.

2) What is prosolvability?

When one considers a Galois extension of infinite degree, the Galois
group of this extension is what is known as a profinite group. A good
discussion can be found in the first section of Jurgen Neukirch's
"Class Field Theory", and I expect you could read about it on
Wikipedia if you Google for "profinite group". The profinite group has
a topology on it known as the Krull topology, and the closed normal
subgroups are in one-to-one correspondence with the Galois
subextensions of the original field extension, in a generalisation of
the fundamental theorem of Galois theory. Furthermore the profinite
group is the direct limit of its finite quotients; an element of the
profinite group is determined by its images under the projections to
all the finite quotients. The profinite group is said to be
prosolvable if all its finite quotients are solvable. So by "the
maximal prosolvable extension of Q" I simply mean the union of all the
finite solvable extensions of Q. This is the maximal extension of Q by
radicals.

One interesting property the p-adic number fields have is that every
finite Galois extension is solvable. I once attended a talk by a
number theorist who said that you could measure the complexity of the
arithmetic in a given field by how complex its absolute Galois group
was. (The absolute Galois group is the Galois group of the algebraic
closure.) The absolute Galois group of a finite field is procyclic;
that is, every finite Galois extension is cyclic. The absolute Galois
group of a local field of characteristic zero (which is the same as a
finite algebraic extension of a p-adic number field) is prosolvable;
that is, every finite Galois extension is solvable. The absolute
Galois group of Q is very complex indeed and many questions about it
are unsolved. It is known that it is not topologically finitely
generated but many questions are unsolved. In particular, it is not
known whether any finite group is isomorphic to a finite quotient of
the absolute Galois group; that is, it is not known whether any finite
group can be the Galois group of a finite extension of Q. This is the
inverse Galois problem. The number theorist who was giving the talk
was saying that this is a good measure of how complicated the number
theory is in the field in question.
.

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