Re: Question on algebraic numbers

In article
<8d104490-5657-41db-9dd9-63022ec1b060@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
dmr5713@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?

Here's some history. Gauss showed that the equation x^17 - 1 can be
solved by radicals (in fact, just using square roots, thus showing that
a regular 17-sided polygon can be constructed by ruler and compass).
Lagrange gave an analysis of why equations of degree less than 5 could
be solved (investigating how the roots could be permuted to yield
solutions by radicals). Abel proved that the general quintic cannot be
solved by radicals (with techniques, I believe, that he applied to his
study of elliptic integrals and which Riemann used in his study of
Riemann surfaces). Galois used a different technique (group theory) to
give necessary and sufficient conditions that a polynomial equation can
be solved by radicals.

Galois thus investigates a particular, concrete, specific equation and
determines whether it can be solved by radicals. It can be solved by
radicals if and only if the Galois group of that equation is a solvable
group.

How does this relate to the solvability of the general equation? One
text book that I like approached this question as follows. Let's
consider the general quadratic equation ax^2 + bx + c. We need some
framework to analyze this equation. The text book proceeds as follows.
Let F = Q(a,b,c) be the field of the three indeterminates a, b, c over
the field Q of rational numbers. Some elements of F are a + 5b*c^2 + 13
and (b^2 - 4ac) / (2a). The field F is a specific, concrete field
(despite the appearance of the indeterminates a, b and c). The
polynomial ax^2 + bx + c is a specific element of F[x]. It can be shown
that the Galois group of ax^2 + bx + c is the symmetric group S_2, and
thus is a solvable group, whence the polynmial ax^2 + bx + c in F[x] is
solvable by radicals.

Likewise, we can consider the specific field F = Q(a,b,c,d,e,f) and the
specific polynomial f(x) = ax^5 + bx^4 + cx^3 + dx^2 + ex + f in F[x].
It can be shown that the Galois group of f(x) is the symmetric group
S_5, which is not a solvable group. Thus, f(x) is not solvable by
radicals (ie, there is not a field E containing F such that f(x) splits
in E and E is a radical extension of F).

One now needs to convince oneself that this f(x) captures the idea of
"general" quintic polynomial (or change your intuition to match this
framework presented above).


2) What is prosolvability?
.

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