Question on algebraic numbers
dmr...@gmail.com
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? The fact that one cannot write any
algebraic formula for the roots of polynomials over Q of degree
greater than or equal to five does not mean that such roots cannot be
written as some finite expressions of integers. (To put that
differently, the insolvability of a polynomial over Q does not mean
that its roots may not happen to be expressible as finite expressions
involving integers that bear no generalizable algebraic relation to
the coefficients of the polynomial.)
In other words, are there algebraic numbers that not only cannot be
written as general finite expressions in the coefficients of
polynomials that may define them, but that cannot be written at all as
finite expressions involving integers, the arithmetic operations, and
raising to rational powers?
William Hale
<c5d16525-3304-4d5a...@q25g2000vbn.googlegroups.com>,
dmr...@gmail.com 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? The fact that one cannot write any
> algebraic formula for the roots of polynomials over Q of degree
> greater than or equal to five does not mean that such roots cannot be
> written as some finite expressions of integers. (To put that
> differently, the insolvability of a polynomial over Q does not mean
> that its roots may not happen to be expressible as finite expressions
> involving integers that bear no generalizable algebraic relation to
> the coefficients of the polynomial.)
I think the standard proofs can be seen to say that the roots of a
polynomial cannot be expressible in the way you describe. The theorem is
not usually stated as such, but the proof actually proves that any
particular polynomial cannot be expressed as finite expressions
involving integers (the algebraic relations to the coefficients of the
polynomial is a red herring).
Mariano Suárez-Alvarez
> The algebraic closure of Q (the rationals) is, of course, formed by
> adjoining to Q the roots of all polynomials over Q.
In principle, that's really just the first step in the construction,
which gives you a field F1. Then you construct a new field F2
adjoining to F1 all the roots of polynomials with coefficients in F1,
and so on.
If you start with a general field, you usually need to iterate
that construction transfinitely to get an algebracally closed
field.
I do not know if starting with Q a single step is suffient.
Usually one sees in an introductory course on Galois
theory that the general equation is not solvable by radicals,
but in fact there are *specific* equations which are not solvalble
by radicals, in the sense that they admit no splitting field
contained in an extension of Q which can be constructed iteratively
by adding radicals.
So the answer to your question is yes.
-- m
Mariano Suárez-Alvarez
> In article
> <c5d16525-3304-4d5a-8777-9b5b04e2c...@q25g2000vbn.googlegroups.com>,
>
>
>
> dmr5...@gmail.com 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? The fact that one cannot write any
> > algebraic formula for the roots of polynomials over Q of degree
> > greater than or equal to five does not mean that such roots cannot be
> > written as some finite expressions of integers. (To put that
> > differently, the insolvability of a polynomial over Q does not mean
> > that its roots may not happen to be expressible as finite expressions
> > involving integers that bear no generalizable algebraic relation to
> > the coefficients of the polynomial.)
>
> I think the standard proofs can be seen to say that the roots of a
> polynomial cannot be expressible in the way you describe. The theorem is
> not usually stated as such, but the proof actually proves that any
> particular polynomial cannot be expressed as finite expressions
> involving integers (the algebraic relations to the coefficients of the
> polynomial is a red herring).
Theproov cannot prove that because that is not true. There *are*
polynomials of all degrees whose roots are expresible as finite
expressions involving integers. The standard proof only proves that
a poynomial whose Galois group is not solvable does not have
that property.
-- m
Robert Israel
> On Feb 1, 9:54=A0pm, dmr5...@gmail.com wrote:
> > The algebraic closure of Q (the rationals) is, of course, formed by
> > adjoining to Q the roots of all polynomials over Q.
>
> In principle, that's really just the first step in the construction,
> which gives you a field F1. Then you construct a new field F2
> adjoining to F1 all the roots of polynomials with coefficients in F1,
> and so on.
>
> If you start with a general field, you usually need to iterate
> that construction transfinitely to get an algebracally closed
> field.
No, you don't.
> I do not know if starting with Q a single step is suffient.
Yes it is: an algebraic extension of an algebraic extension is an
algebraic extension. See e.g. Herstein, Topics in Algebra, Theorem 5.E.
So the roots of polynomials with coefficients in F1 are roots of polynomials
with coefficients in Q.
--
Robert Israel isr...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
Mariano Suárez-Alvarez
<isr...@math.MyUniversitysInitials.ca> wrote:
> writes:
>
> > On Feb 1, 9:54=A0pm, dmr5...@gmail.com wrote:
> > > The algebraic closure of Q (the rationals) is, of course, formed by
> > > adjoining to Q the roots of all polynomials over Q.
>
> > In principle, that's really just the first step in the construction,
> > which gives you a field F1. Then you construct a new field F2
> > adjoining to F1 all the roots of polynomials with coefficients in F1,
> > and so on.
>
> > If you start with a general field, you usually need to iterate
> > that construction transfinitely to get an algebracally closed
> > field.
>
> No, you don't.
I should know better than trust my memory. What I had in mind
is the construction of the algebraic closure of a field F due
to Emil Artin, as given in Atiyah-Macdonald's little book on
commutative algebra:
If K is a field, let P be the set of monic irreducible polynomials
in K[x], and let K[X] be the ring of polynomials with
coefficients in K and variables indexed by the elements of P.
Let I be the ideal in K[X] generated by the elements f(X_f)
where f is in P and X_f is the variable in K[X] corresponding
to the polynomial f. Let M be a maximal ideal in K[X] containing
the ideal I, and let K_1 = K[X] / M
Iterate this construction, and consider the union
L = union_{n\geq 0} K_n
and let T be the set of elements in L which are
algebraic over the starting field K. Then T is
an algebraic closure.
Not quite the same as what the OP wrote.
I suppose the reason Artin went through all that is that it
is otherwise a bit annoying to make the "add to Q all the roots
of all the polynomials over Q" precise, without having a big field
which contains both Q and those roots (For Q one can of course
use C---for general fields it is less obvious...)
-- m
Rupert
> 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.
dmr...@gmail.com
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?
2) What is prosolvability?
Ted Hwa
: 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?
All proofs of the insolubility of the quintic are actually proofs
that some specific quintic has no solution in radical form. The
proof of the general insolubility of the quintic follows from that,
not the other way around.
Ted
Rupert
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.
William Hale
<8d104490-5657-41db...@r41g2000yqm.googlegroups.com>,
dmr...@gmail.com wrote:
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?
Bill Dubuque
Although your question has already been answered, below I add
some further related remarks that I suspect may be of interest.
First, below is appended an excerpt from Franz Lemmermeyer's
12: The failure of unique factorization.
As a matter of fact, however, E. Heine defined algebraic integers in
[Hei] as numbers that can be constructed from the rational integers
by addition, multiplication, and raising to m/n-th powers with m,n
positive integers. He then goes on to show that every such number is
integral in the modern sense, i.e. it is a root of a monic polynomial
with integral coefficients. The converse is, of course, false, as
Heine's construction gives only algebraic numbers that are solvable,
i.e., that can be expressed in terms of radicals, and Heine's claim
that any solvable root of a monic polynomial with integral coefficients
is integral in his sense is still an open problem (see Exercise 12.17)
since his proof is not valid.
12.17 Let L be a number field such that its normal closure N has a
solvable Galois group. Is it true that the ring of integers O_L
coincides with the subset of L consisting of algebraic integers in
Heine's sense?
[Hei] E. Heine, Der Eisensteinsche Satz uber die Reihenntwickelung
algebraischer Functionen, J. Reine Angew. Math. 45 (1853), 285-302
Second, the buzzword you're looking for is root-closure / root-closed,
e.g. see the reviews below.
--Bill Dubuque
------------------------------------------------------------------------------
92f:13006 13B22
Anderson, David F.(1-TN); Dobbs, David E.(1-TN); Roitman, Moshe(IL-HAIF)
Root closure in commutative rings.
Ann. Sci. Univ. Clermont-Ferrand II Math. No. 26 (1990), 1--11.
------------------------------------------------------------------------------
Let S be a subset of N. If A < B are rings, A is said to be S-root closed in B
if, whenever b in B and b^n in A for some n in S , then b in A . The smallest
subring of B which contains A and is S-root closed in B is called the total S
-root closure of A in B and is denoted by R^S_oo(A,B) . It is shown that
R^S_oo(A,B) = \/ {R^S_m(A,B) : 0 <= m <= oo}, where R^S_0(A,B) = A and,
for m>0, R^S_m(A,B) is the subring of B generated by R^S_{m-1}(A,B) and
the elements b in B such that b^n in R^S_{m-1}(A,B) for some n in S .
If A is a domain, if B is the quotient field of A and S = N, then
R^S_oo(A,B) = R_oo(A) = \/{R_m(A) : 0 <= m <= oo} is the total root-closure
of A. This paper deals with the question of how many steps are needed to obtain
the total root closure of a domain. An example is given of a (quasilocal,
one-dimensional, seminormal) non-Noetherian domain A such that for all m,
R_m(A) != R_{m+1}(A) and, for each positive integer m , an example of a
(quasilocal, one-dimensional, seminormal) domain A such that R_oo(A) =
R_m(A) != R_{m-1}(A). The authors also conjecture that the total root closure
of a Noetherian domain A need not be obtainable in finitely many steps.
Reviewed by Valentina Barucci
------------------------------------------------------------------------------
98h:13008 13B22 (13F20 13G05)
Roitman, Moshe(IL-HAIF)
On root closure in Noetherian domains. (English. English summary)
Factorization in integral domains (Iowa City, IA, 1996), 417--428,
Lecture Notes in Pure and Appl. Math., 189, Dekker, New York, 1997.
------------------------------------------------------------------------------
For a subring A of a commutative ring B and an integer n >= 1 , we say
that A is n-root closed in B if b^n in A with b in B implies b in A.
Then C(A,B) = {n | n >= 1 and A is n-root closed in B} is a multiplicative
submonoid of positive integers generated by primes. In this paper, the author
shows that any multiplicative submonoid of positive integers generated by
primes may be realized as C(A,K) for a Noetherian domain A with quotient
field K. This answers a question raised by the reviewer [Glasgow Math. J. 31
(1989), no. 1, 127--130; MR 90b:13008]. For commutative rings A < B, the root
closure of A in B may be constructed as \/ A_n, where A_0 = A and
A_{n+1} = A_n[{b in B| b^m in A_n for some m>=1}]. He shows that for each
positive integer d , there is a d-dimensional Noetherian domain A such
that each A_n < A_{n+1} (in this case B is the quotient field of A).
This answers a question raised by D. E. Dobbs, the author, and the reviewer in
[Ann. Sci. Univ. Clermont-Ferrand II Math. No. 26 (1990), 1-11; MR 92f:13006].
He also gives a simple proof of the Brewer-Costa-McCrimmon result [J. W.
Brewer, D. L. Costa, and K. McCrimmon, J. Algebra 58 (1979), no. 1, 217--226;
MR 80e:13002] that C(A[X],B[X]) = C(A,B).
Reviewed by David F. Anderson
Dave L. Renfro
> 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.
This is exactly a question I had for a long time and
I found it amazing how difficult it was (and still is)
to find a "no-nonsense jargon-free" answer when looking
through textbooks and discussions in semi-technical
"popular" math books.
First, an algebraic number is defined to be any complex
number that is a zero to some nonzero polynomial with
integer coefficients, and an algebraic real number is
any algebraic number that is also a real number. It is
not difficult to see that both notions are the same if
we replace "integer coefficients" with "(real) rational
coefficients. Moreover, nothing new (either real algebraic
or arbitrary algebraic) is gained if we now include all zeros
of nonzero polynomials with real algebraic number coefficients
or if we include all zeros of nonzero polynomials with (complex)
algebraic number coefficients. There are 4 different statements
being asserted in the previous sentence, by the way, and none
of these 4 statements is straightforward to prove (at least,
independently of the other 3 statements) like the case of
replacing "integer coefficients" with "rational coefficients".
Below is something I wrote several years ago, which I'd
probably rewrite now if I had the time (and interest),
and note also that I posted two follow-ups/corrections
in the same thread, but it goes into exactly the kinds
of issues you raised. By "explicit algebraic number",
I mean any real number (I only deal with real numbers
below) that cannot be expressed as a finite combination
of arithmetical operations and non-iterated rational powers
(i.e. while sqrt(2 + sqrt(2)) is allowed, 2 ^ (2 ^ (1/2))
is not allowed), beginning with the integers. This is a
little vague, I know. A little better might be to say
"finite combination of arithmetical operations and integer
root extractions".
------------------------------------------------------
sci.math.symbolic -- Unsolubility by radicals for n>4 and
computer algebra (14 January 2001)
http://groups.google.com/group/sci.math.symbolic/msg/a9efefac3d56e906
http://mathforum.org/kb/message.jspa?messageID=1565341
I also know very little algebra and have long been confused by
discussions of this topic in books. For example, here's what
one finds on page 125 of James Pierpont "Lectures on the Theory
of Functions" (volume I, 1905), which is typical of what I'm
talking about:
"... equations of degree n > 4 cannot, in general, be solved by
the extraction of roots, or, as we say, do not admit of an
algebraic solution."
The two versions that David Ullrich brings up seem very different
to me, but the type of comment I quoted leads me to wonder if
these authors are even aware of the possibilities. [They probably
are, and I'm sure Pierpont was, but I wonder if they've thought
through the logical implications of what they're writing from
the standpoint of a beginner.]
Here are several interpretations of what Pierpont is saying:
Version #1: For some n > 4, including n=5 and perhaps some other
values of n as well, there exists no general explicit
"algebraic formula" for the roots of an arbitrary
n'th degree polynomial in terms of its coefficients.
Version #2: For each n > 4, there exists no general explicit
"algebraic formula" for the roots of an arbitrary
n'th degree polynomial in terms of its coefficients.
Version #3: For some n > 4, including n=5 and perhaps some other
values of n as well, there exists a specific n'th degree
polynomial with integer coefficients that has at
least one real root that cannot be expressed using
a finite sequence of additions, subtractions,
multiplications, divisions, and positive integer
root extractions applied to the coefficients of
the polynomial in question.
Version #4: For some n > 4, including n=5 and perhaps some other
values of n as well, there exists a specific n'th degree
polynomial with integer coefficients that has at
least one real root that cannot be expressed using
a finite sequence of additions, subtractions,
multiplications, divisions, and positive integer
root extractions applied to the set of integers.
Version #5: For each n > 4, there exists a specific n'th degree
polynomial with integer coefficients that has at
least one real root that cannot be expressed using
a finite sequence of additions, subtractions,
multiplications, divisions, and positive integer
root extractions applied to the coefficients of
the polynomial in question.
Version #6: For each n > 4, there exists a specific n'th degree
polynomial with integer coefficients that has at
least one real root that cannot be expressed using
a finite sequence of additions, subtractions,
multiplications, divisions, and positive integer
root extractions applied to the set of integers.
Versions 1 and 2 are equivalent. Obviously, Version 2
implies Version 1. For the other direction, suppose there
was a general formula for the roots of all (say) 9'th degree
polynomials. Then we could get a formula for the roots of all
5'th degree polynomials by applying the 9'th degree formula
to the special case of (x^4)*(ax^5 + bx^4 + cx^3 + dx^2 + ex + f),
contradicting Version 1. This argument obviously generalizes from
n = 9 to any n > 5.
The same idea also shows that Version 3 is equivalent to Version 5,
and that Version 4 is equivalent to Version 6.
Version 3 is equivalent to Version 4, and Version 5 is equivalent
to Version 6. As long as we have a nonzero polynomial, we can
divide one of its nonzero coefficients by itself to get 1. From 1
we can get any integer by finitely many (depending on the integer,
of course) additions and/or subractions of 1. So if there were a
"finite representation using integers" of all solutions to a
given nonzero polynomial, we could also come up with a finite
representation of each solution using only the coefficients of
that particular polynomial.
Therefore, the six versions I gave reduce to the two versions
David Ullrich mentioned, Version 2 and Version 6.
Unless I'm greatly mistaken, I'm fairly sure that Version 6 (which
clearly implies all the other versions) is true. I think the
solvability of the Galois group for the polynomial (the meaning
of which I don't remember any details of), which can sometimes be
proved for certain polynomials (but which I also suspect there's no
constructive way of doing this in general), corresponds to all
the real roots (Or is it all the real and complex roots?) of that
polynomial being "finitely expressible using integers".
------------------------------------------------------
Dave L. Renfro