Re: New paper, algebraic integers, Galois Theory

From: W. Dale Hall (mailtowd-hall_at_pacbell.net)
Date: 10/10/04 

Date: Sun, 10 Oct 2004 22:39:19 GMT

James Harris wrote:
> Commented on Galois Theory at end. ___JSH
>
>
> ----------------------------------------------------------------------
>
> I. First section
>
> The following are in a commutative ring.
>
> Start with
>
> P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f)
>
> with the factorization
>
> P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)
>
> and note that at
>
> m=0, P(0) = u^2 f^2(3x + uf),
>
> which gives you terms that do not vary as m varies.
>
> So what about (a_1 x + uf), (a_2 x + uf), and (a_3 x + uf)?
>
> (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) = u^2 f^2 (3x + uf)
>
> which shows that at least two of the a's have to equal 0 at m=0, while
> one equals 3.
>
> Since, at m=0, two of the a's must equal 0, it's convenient to just
> arbitrarily select a_1 and a_2 as those two.
>
> Then you have uf for the first, uf for the second and 3x + uf for the
> third as terms that do not vary when m varies.
>
> Now then, if m=1, what are the *constant* terms?
>
> They are uf, for the first, uf for the second, and 3x + uf for the
> third.
>
> That's logical because they do not vary with m, so if m=1003909273,
> what are the constant terms?
>
> They are uf, for the first, uf for the second, and 3x + uf for the
> third.
>
> Now divide f^2 from both sides, which gives
>
> P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f
>
> P(m)/f^2 = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)/f^2
>
> and you note that P(0)/f^2 = u^2(3x + uf), which means that now your
> constant terms are u for the first, u for the second and 3x + uf for
> the third.
>
> Now then, if m=1, what are the constant terms now?
>
> They are u for the first, u for the second, and 3x + uf for the third.
>
> If m = 2938479378, what are the constant terms now?
>
> They are u for the first, u for the second, and 3x + uf for the third.
>
> How can the constant terms of the first two go from uf to u?
>
> They must be divided by f.
>
> Now, the constant term of a_1 x + uf, is uf, but when f^2 is divided
> from P(m), it is u; therefore, a_1 x + uf is divided by f, and you
> have
>
> a_1 x/f + u
>
> and the constant term of a_2 x + uf is uf, but when f^2 is divided
> from P(m), it is u; therefore, a_2 x + uf is divided by f, and you
> have
>
> a_2 x/f + u
>
> while the constant term of a_3 x + uf is 3x + uf, and after f^2 is
> divided off, it is 3x + uf, so you have
>
> a_3 x + uf
>
> so, dividing P(m) by f^2 gives
>
> P(m)/f^2 = (a_1 x/f + u)(a_2 x/f + u)(a_3 x + uf).
>
>
>
> II. Second section
>
> Now take
>
> P(m)/f^2 = (a_1 x/f + u)(a_2 x/f + u)(a_3 x + uf)
>
> and multiply inside the parentheses by f^2/(a_1 a_2 a_3), and outside
> by f^2(a_1 a_2 a_3) and you have
>
> P(m)/f^2 = ((a_1 a_2 a_3)/f^2)(x + uf/a_1)(x + uf/a_2)(x + uf/a_3)
>
> and since a_1 a_2 a_3 = f^2(m^3 f^4 - 3m^2 f^2 + 3m), that is
>
> P(m)/f^2 =
>
> (m^3 f^4 - 3m^2 f^2 + 3m)(x + uf/a_1)(x + uf/a_2)(x + uf/a_3).
>
>
> Now consider the case that m, f, and u are algebraic integers, then I
> have the ratios of algebraic integers:
>
> uf/a_1, uf/a_2, and uf/a_3,
>
> and now let
>
> v_1/w_1 = uf/a_1, v_2/w_2 = uf/a_2, and v_3/w_3 = uf/a_2
>
> where the v's and w's are algebraic integers in each case coprime to
> each other.
>
> Making the substitutions I have
>
> P(m)/f^2 =
>
> (m^3 f^4 - 3m^2 f^2 + 3m)(x + v_1/w_1)(x + v_2/w_2)(x + v_3/w_3).
>
> And I have from before that
>
> P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f
>
> so
>
> (m^3 f^4 - 3m^2 f^2 + 3m)(v_1 v_2 v_3)/(w_1 w_2 w_3) = f
>
> as that is the last coefficient from the last term u^3 f, which proves
> that
>
> (m^3 f^4 - 3m^2 f^2 + 3m) has w_1, w_2 and w_3 as factors, so let
>
> (m^3 f^4 - 3m^2 f^2 + 3m) = w_1 w_2 w_3
>
> then I have
>
> P(m)/f^2 = (w_1 x + v_1)(w_2 x + v_2)(w_3 x + v_3)
>
> but I still have that
>
> P(m)/f^2 = (a_1 x/f + u)(a_2 x/f + u)(a_3 x + uf).
>
>
>
> III. Third section
>
> So, even if a_1/f is not an algebraic integer, you can find w_1 an
> algebraic integer.
>
> But if a_1/f is an algebraic integer and w_1 is not, they cannot be
> equal.
>
> But I have
>
> P(m)/f^2 = (w_1 x + v_1)(w_2 x + v_2)(w_3 x + v_3)
>
> and
>
> P(m)/f^2 = (a_1 x/f + u)(a_2 x/f + u)(a_3 x + uf)
>
> so how do you reconcile a case where a_1 x/f is not an algebraic
> integer?
>
> There must exist z_1, z_2, and z_3 such that
>
> w_1 = (a_1 x z_1)/f, w_2 = (a_2 x z_2)/f and w_3 = a_3 x z_3
>
> and z_1 z_2 z_3 = 1,
>
> so algebraically the z's are units, but z_1, z_2 and z_3 are not units
> in the ring of algebraic integers, if a_1/f is not.
>
> I've often faced arguments over the result from Section 1, and at
> times I've dealt with people claiming that Galois Theory proves
> something about the factors of roots of monic polynomials with integer
> coefficients.
>
> The basic claim is that *each* of the roots of a monic polynomial with
> integer coefficients that is irreducible over rationals must share
> non-unit factors with ALL of the prime factors of the last
> coefficient.
>

Note that this claim does not require Galois Theory, and is easily
proven using the very basic results of field theory, together with
the elementary properties of the integers. Briefly stated, one takes
such a polynomial P (i.e., monic integral polynomial, irreducible
over rationals) and constructs the field Q(a), where a is a root of
P. This is done without reference to which root "a" really is, and
as a result, one finds that the two field extensions Q(a) and Q(b),
where a and b are distinct roots of P, are isomorphic in such a way
that the elements of Q are held fixed.

In particular, the ring of integers in Q is held fixed, and further:

        *anything you can express* using arithmetic operations and
        integers and the root "a" can be rewritten *simply by replacing
        all occurrences of the symbol "a" with the symbol "b"* to yield
        an equivalent statement about the root "b".

This directly implies that if k is an integer, then if "a" and k are
coprime, so are "b" and k. Why? Because if "a" and k are coprime, I
can find polynomials A, B, and C, with integer coefficients, such that

        A(x)*x + B(x)*k = C(x)*P(x) + 1.

If this is true, then I can substitute "b" in for the variable x in
that equation, and find integers u,v in Q(b) for which

        u b + v k = 1.

No Galois Theory, no fancy dancing with field extensions, no Galois
group. Just a pair of field extensions that we can prove directly are
isomorphic, fixing the field of rationals (and its ring of integers).

> For instance, with P(x) = x^2 + x + 6, the claim would be that the two
> roots:
>
> (-1 + sqrt(32))/2 and (-1 - sqrt(32))/2
>

Um, the roots of P(x) are given by the quadratic formula:

        For the equation

                ax^2 + bx + c = 0

        the roots are given by the formula

                x = (-b +/- sqrt(b^2 - 4ac))/2a

Here, we have a = 1, b = 1, c = 6.

The discriminant is b^2 - 4ac = 1 - 4*1*6 = 1 - 24 = -23

So the roots are

                x = (-1 +/- sqrt(-23))/2

Not, as you have incorrectly written,

> (-1 + sqrt(32))/2 and (-1 - sqrt(32))/2

> would, supposedly, each have to share non-unit factors with 2 and 3.
>

So, let's see about this. Does each share a factor with 2 and 3?

Let r be either of these roots. Note that the numbers

        g = -(r + 2)
        h = r - 1
        k = 2r + 3

are algebraic integers, and that

        g h = -(r+2)(r-1) = 2 - r - r^2
            = 2 - (r^2 + r) = 2 - (-6) = 8,
and

         g k = -(r + 2)(2r + 3)
            = -(2r^2 + 7r + 6)
            = -2r^2 - 7r - 6
            = -2(-r - 6) - 7r - 6
            = 2r + 12 - 7r - 6

So,
        g k = -5r + 6.

But, since r^2 = -r - 6, we have

        r^3 = -r^2 - 6r
            = -(-r - 6) - 6r
            = r + 6 - 6r = -5r + 6.

Thus, gk = r^3.

In other words, the numbers g,h,k are all algebraic integers,
and satisfy

        gh = 2^3,
        gk = r^3.

If we take cube roots

        u^3 = g,
        v^3 = h,
        w^3 = r,

we'll have algebraic integers u,v,w with

        uv = 2
        uw = r.

A similar exercise yields common algebraic integer
divisors for r and 3. In this case, instead of
the above choices of g,h,k, one uses the following:

        g = -2r - 3
        h = 2r - 1
        k = r + 2.

and one obtains

        gh = 3^3
        gk = r^3.

I'll leave the details up to the reader, since I'm tired of typing all
this stuff in. YOu will need to take cube roots again, as the ideals
<r,2> and <r,3> are both of order 3 in the ideal class group of the
extension Q(r).

> My works shows that it's possible that actually neither does and you
> have to check using advanced polynomial factorization techniques.
>

Sorry, wrong answer.

> Faced with the algebra, certain people simply claimed that Galois
> Theory *forces* that result, when in fact, it does not.
>

Oh, really? Why does elementary field theory force it to be
the case, then?

Where is your "fact", anyhow? All you're doing is making baseless
assertions, unsupported by *any* evidence. Show an example, why
don't you? I mean an *actual example*, with numbers for the
coefficients, and a *specific* claim.

> That's kind of obvious as consider
>
> P(x) = x^2 + 5x + 6 = (x+2)(x+3)
>
> and if Galois Theory forced the previous on irrationals, why wouldn't
> it force it on rationals as well?
>

You still haven't gotten it through that thick skull that
reducibility makes a real difference, have you?

The fact that the polynomial splits already means that the factors
can be independent in Q[x]. If the polynomial P is irreducible in Q[x],
then <P>, the ideal it generates in Q[x], is prime; Q[x] is of dimension
one, so nonzero prime ideals are maximal, thus Q[x]/<P> forms a field.
In this field, which contains a canonical copy of Q itself, the class
represented by x + <P> is a root of the polynomial P(x).

You might note that this "formal" extension of Q doesn't really depend
on which root you are considering. It shows the arithmetic properties
of Q(a) for *any* root of the irreducible polynomial P.

The same thing just does not hold for a reducible polynomial P. For one
thing, the construction above just does not produce a field. When you
perform that construction using a reducible polynomial, you actually
introduce zero-divisors and destroy any likelihood of reaching a field
until those zero-divisors are eliminated.

> It doesn't force anything on either. They were just wrong.
>

Please show my error in the arithmetic regarding the quadratic
equation above, fool.
>
> James Harris

The more you imagine you can fool anyone with your meaningless
blathering about stuff you clearly know nothing about, the less
likely it is that you'll ever get to a point of knowledge.

What's the real problem with sitting down and *learning* something?
Are you mentally deficient? Scared? Too proud of your own ignorance
and too ashamed of your own shortcomings?

Dale