Re: JSH: Heart of dispute, number properties

From: Nora Baron (norabaron_at_hotmail.com)
Date: 03/23/05 

Date: 23 Mar 2005 15:09:47 -0800

Arturo Magidin wrote:
> In article <1111599548.838439.15600@o13g2000cwo.googlegroups.com>,
> Nora Baron <norabaron@hotmail.com> wrote:
>
> [.snip.]
>
>
> > So you try to invent another ring which is larger than the
algebraic
> >integers, but contains a few numbers which are not algebraic
> >integers. The only units in this larger ring are 1 and -1.
> >
> > However it is easy to show that it is not possible to construct
such
> >a ring for the polynomial
> >
> > P(x) = 2 x^2 + 5 x + 1.
> >
> >Here is why. The two roots of P(x) are
> >
> > r1 = (-5 + sqrt(17))/4 and
> >
> > r2 = (-5 - sqrt(17))/4.
> >
> > Now suppose you adjoin one of those roots, say r1, to the ring A
> >of algebraic integers: i.e., consider A[r1].
> >
> > Note that 1/r1 = r2/(r1*r2) = -r2 / (1/2) = -2 r2. But
> >
> > 2 r2 = (-5 - sqrt(17))/2,
> >
> >which is an algebraic integer.
> >
> > Therefore r1 has an inverse in A[r1]; it is a unit, but it
> >is not +1 or -1.
> >
> > So if you are going to find a ring with the properties you
> >want for this polynomial, it is not going to be a ring which
contains
> >the algebraic integers.
>
> I don't think this is an accurate restatement. What is being asked is
> not that the ring contain no units other than 1 and -1 (this is
> already false in all number fields except for most of the imaginary
> quadratic number fields), but rather that the only ->rationals<-
which
> are units are 1 and -1. This is trivally equivalent to asking that R
> intersect Q be equal to Z.
>

  Arturo,

     You're right - my mistake, I should have looked up what he
said previously.

     As often happens it is a little hard to discern what Harris
is thinking. With the example P(x) = 2 x^2 + 5 x + 1, one has
the impression that he would like for one of the roots to be
an algebraic integer, and the other to be an algebraic number
which is not an algebraic integer. This would parallel the
reducible case. Letting r1 and r2 be the roots, one clearly
has

     r1 * r2 = 1/2, or

     (2r1) r2 = 1.

Thus the first term (2r1) is an algebraic integer, and the
second term (r2) is an algebraic number which can be written in
the form

      r2 = s2/2,

where s2 is an algebraic integer. Thus r2 is a kind of "fraction".
(Of course that is true for all algebraic integers). So what
Harris wants is a ring that contains r2 and in which the only
rational units are 1 and -1. Presumably this ring also contains
(2r1). Thus the minimal such ring is

      A(2r1, r2) = A(r2) (since 2r1 is in A already).

>>From the above, both 2r1 and r2 are units in this ring. Of course
neither of these is a unit in the ring of algebraic integers.

> Now, there are many, many, many such rings. Bill Dubuque and others
> have given explicit examples. Pick any algebraic number which is not
> an algebraic number and such that no power of whom is rational, call
> it r, and take A[1/r]. That ring has that property.
>
> The real difficulty lies precisely in that there are many, many, such
> rings, and that while Zorn's Lemma guarantees the existence of
maximal
> such rings, there are many such maximal rings; talking about "the"
> ring with these properties is nonsense. Your ring A[r1] is probably
> one such ring. So is the ring A[r2]. But there is no ring which
> contains BOTH r1 and r2 and satisfies the condition of having
> intersection with Q equal to Z. Because if a ring contains both r1
> and r2, then it contains r1*r2 = (25-17)/16 = 1/2. So A[r1] is
> contained in some "object ring", and A[r2] is contained in another,
> but there is no "object ring" that contains both.
>

  Agreed.

> IF you further ask, as it was at one point implicitly required, that
> the ring be ->closed under conjugates<- (if f(x) is irreducible over
> Q and has at least one root in R, then it has all its roots in R),
> then we overcome the difficulties: there ->is<- a unique largest
> subring of the algebraic numbers with that property AND with the
> property that it intersects Q at Z. But that ring is... the ring of
> all algebraic integers.
>

  Right. I don't think Harris accepted the condition of being
closed under conjugates.

  The main point in all this is, why is Harris so insistent that
there be a similarity between a reducible polynomial, like
2 x^2 + 3 x + 1, and an irreducible one like 2 x^2 + 5 x + 1?
The former has one root which is an integer and the other which
is a fraction. The latter, Harris thinks, SHOULD have one root
which is an algebraic integer and the other of which is an
algebraic fraction (i.e., an algebraic number, which of course can
be written as a quotient of an algebraic integer and an ordinary
integer) ? Why does he think that is so important?

  He quit claiming long ago that his previously announced main
result was what he said it was at first - that is, he recognizes
that neither of the roots of 2 x^2 + 5 x + 1 are algebraic integers,
and he no longer claims that if Q(x) = 65 x^3 - 12 x + 1 is factored
in the form (a x + 1)*(b x + 1)*(c x + 1), then one of a, b, or
c must be coprime to 5. But he still thinks there is something
wrong with the algebraic integers because roots of irreducible
polynomials act differently from roots of reducible polyomials.
It's no longer a matter of his thinking algebraic number theory
or Galois theory are *wrong* - it is more a matter of his just
plain *not liking* what these theories imply about factorizations.
The fact that he now refuses to stick his neck out and actually
state some definitions and theorems does not help if we want to
understand his (so-called) thinking.

  One other ingredient in all this. I don't think Harris has
ever understood, or at least has ever accepted, the fact that
his argument based on "constant terms are constant" is wrong.
This adds to the confusion. If that argument were correct,
then since it leads to a contradiction of known theorems,
mathematics would be inconsistent. Harris knows this. So he
has this problem of things not factoring the way he wants
and facts being inconsistent with the "constant terms are
constant" argument, and he somewhat illogically resolves these
two by saying the algebraic integers do not have the properties
that they "SHOULD" have. More briefly: what Harris thinks is
wrong with algebraic number theory is that it doesn't give the
results that he wants it to.

  Nora B.

> [.snip.]
>
>
> --
>
======================================================================
> "It's not denial. I'm just very selective about
> what I accept as reality."
> --- Calvin ("Calvin and Hobbes")
>
======================================================================
>
> Arturo Magidin
> magidin@math.berkeley.edu

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