Re: JSH:Understanding constant terms

From: James Harris (jstevh_at_msn.com)
Date: 11/28/04 

Date: 28 Nov 2004 06:41:13 -0800

norabaron@hotmail.com (Nora Baron) wrote in message news:<36024859.0411272103.286d4eb9@posting.google.com>...
> jstevh@msn.com (James Harris) wrote in message news:<3c65f87.0411271105.505e0fec@posting.google.com>...
> > Consider a polynomial P(x), with n factors such that
> >
> > g_1(x) g_2(x)...g_n(x) = P(x)
> >
> > and g_1(x), g_2(x),...,g_n(x) are algebraic integer functions.
> >
> > That is, for an algebraic integer x, and natural number 'a' where
> > 1<=a<=n, g_a(x) is an algebraic integer.
> >
> > Now consider setting x=0, and notice that gives
> >
> > g_1(0) g_2(0)...g_n(0) = P(0)
> >
> > and notice there is no dependency on x, as the constant term is
> > defined by terms where x is equal to 0, and that's not a trick.
> >
> > Now let c_1 = g_1(0), c_2 = g_2(0), etc. and you have
> >
> > c_1 c_2...c_n = P(0)
> >
> > and the c's are necessarily algebraic integers and factors of the
> > constant term P(0).
> >
> > Now let r_1(x) = g_1(x) - c_1, r_2(x) = g_2(x) - c_2, and so on.
> >
> > Then I can substitute and have
> >
> > (r_1(x) + c_1) (r_2(x) + c_2)...(r_n(x) + c_n) = P(x).
> >
> > Now let
> >
> > P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078
> >
> > then
> >
> > P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7)
> >
> > when the a's are roots of
> >
> > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
> >
> > so let
> >
> > g_1(x) = 5a_1(x) + 7, g_2(x) = 5a_2(x) + 7, and g_3(x) = 5a_3(x) + 7
> >
> > and c_1 is given by g_1(0), c_2 by g_2(0), and c_3 by g_3(0).
> >
> > Setting x = 0 with the cubic defining the a's gives
> >
> > a^3 - 3a^2 = 0, so two of the a's are 0, and one is 3.
> >
> > Since indices are arbitrary let the first two equal 0, and I have
> >
> > c_1 = 7, c_2 = 7, and c_3 = 22
> >
> > which is consistent with the constant term of P(x), which is 1078.
> >
> > But dividing P(x) by 49, gives
> >
> > P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22
> >
> > which means that 7 is divided from the constant terms as well, and
> > only two of the constant terms, c_1 and c_2 have 7 as a factor, so
> > necessarily 7 is divided from the factors where the constant terms are
> > 7.
> >
> > Now if you continue the analysis the conclusion that two of the
> > factors of P(x) have 7 as a factor leads to the conclusion that two of
> > the roots of
> >
> > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
> >
> > have 7 as a factor, but you can rather easily prove that with integer
> > x, if the cubic is irreducible over rationals, then that will not be
> > true in the ring of algebraic integers.
> >
> > Algebraic integers are roots of monic polynomials with integer
> > coefficients.
> >
> > For a number to be an algebraic integer, it must be the root of some
> > monic polynomial that has integer coefficients.
> >
> > Therefore, you can conclude from the mathematics that for a given x,
> > when the cubic defining the a's is irreducible over rationals then its
> > roots do not have 7 as a factor in the ring of algebraic integers.
> >
>
> Which is, of course, an out-and-out contradiction, right?

There is the appearance of contradiction, which is readily resolved by
not giving the ring of algebraic integers a special position.

So then, just because the roots do not have 7 as a factor in the ring
of algebraic integers, it's not significant.

> So unless mathematics is inconsistent, there must be something
> wrong. What is it?
>

If you overrate the ring of algebraic integers, then you can make
arguments that are wrong.

Essentially, you have to understand that the requirement that a number
be the root of a monic polynomial with integer coefficients is a
meaningless technicality, with no real mathematical importance.

There is no weight to the requirement mathematically that a number be
the root of a monic polynomial with integer coefficients.

That's what follows.

> It could be that algebraic number theory is wrong. Or possibly
> Galois theory. Or both.
>

It turns out to be a problem in algebraic number theory, which has
lead to a mis-use of Galois Theory.

It's not even really complicated, but there are reasons for people to
get emotional over the issue, as it's a mistake at the foundations of
the discipline of mathematics.

It's an error in "core".

> Or it could be that you have a logical error somewhere in what
> you have said above.
>

You can continue with that assumption. I have no problem defending
the math I outlined in my original post.

Ultimately, it boils down to constants being constant.

Specifically 7 and 22 are constants, and behave like constants.

> You concluded that c_1 = 7, c_2 = 7, and c_3 = 22. That certainly
> seems OK. I don't disagree with it.
>

Good.

> Then you noted that 49 = 7*7 must be factored out of the expression
>
> (r_1(x) + 7)(r_2(x) + 7)(r_3(x) + 22).
>
> Let's look in detail at that troublesome third term,

It's not troublesome. It's asymmetrical with regard to the rest.

The asymmetry is what blocks you out of the ring of algebraic
integers, which requires a lot of symmetry.
 
> r_3(x) + 22.
>
> The objective here is to examine whether I can divide a nonunit
> factor of 7 out of this expression, with the result after the
> factoring being an algebraic integer.

Now you're going to cheat, and I'm going to explain how you cheat
before you do it, so readers can see how it works.

A number cannot be an algebraic integer if it is not the root of some
monic polynomial with integer coefficients.

You say "nonunit" above as if that applies globally, when you're going
to rely on non-unit status in the ring of algebraic integers.

So consider a unit function u(x), which is not a unit in the ring of
algebraic integers.

Now you can multiply with that unit, and then assert that you have a
non-unit in the ring of algebraic integers, to claim that you can do
that multiplication.

But it's only a non-unit in the ring of algebraic integers because
it's not the root of some monic polynomial with integer coefficients.

Ok readers, here we go, pay careful attention...
 
> Let w be a nonunit factor of 7.
>

Notice that the poster didn't say, in the ring of algebraic integers.

It's important that you note that the claim is in the ring of
algebraic integers.

> You note that 22 is coprime to 7. Therefore 22/w cannot be an
> algebraic integer. So that is potentially a problem.
>

The poster is apparently acknowledging that 22 properly is coprime to
7. And is in fact coprime to 7 in the the ring of algebraic integers.

> But recall that r_3(x) + 22 = 5 a_3(x) + 7.
>
> Suppose I divide the latter expression through by w: I get
>
> 5 a_3(x)/w + 7/w.
>
> The term 7/w is not a problem here because I was assuming that
> w is a nonunit factor of 7. Thus 7/w *is* an algebraic integer.
>

Remember readers, non-unit in the ring of algebraic integers!

> Now, if such a w can be chosen so that a_3(x)/w is also an
> algebraic integer, I conclude that in the expression
>
> 5 a_3(x)/w + 7/w
>
> all the coefficients are algebraic integers, and the sum is
> an algebraic integer.
>
> Can such a w be defined ?
>
> If so, it is going to have to be a function of x: because when
> x = 0, w has to be 1. Whereas, for most other integers x, the
> polynomial that you mention above is irreducible, and w is a nonunit
> factor of 7.
>
> Here is how w_1(x), w_2(x), and w_3(x) need to be defined:
>
> w_1(x) = GCD(a_1(x), 7)
>
> w_2(x) = GCD(a_2(x), 7)
>
> w_3(x) = GCD(a_3(x), 7).
>
> Here GCD denotes the 'greatest common divisor' function. It
> is defined in the ring of algebraic integers by a theorem of
> Dedekind.
>
> Obviously these definitions imply that all of
>
> (5 a_1(x) + 7)/w_1(x),
>
> (5 a_2(x) + 7)/w_2(x), and
>
> (5 a_3(x) + 7)/w_3(x)
>
> are algebraic integers.
>

Notice also that you have the implication that the factors do not have
a constant term, which has to be made explicit in a bit.

Essentially the w's the poster is trying to use are unit factors, but
are not units in the ring of algebraic integers by a technicality that
they are not roots of a monic polynomial with integer coefficients.

It's a neat trick when you think about it that requires relying on the
very problem that has been outlined to try and use the ring of
algebraic integers to disprove that there is a problem with that ring!

> Now the crucial thing to show is that
>
> w_1(x) * w_2(x) * w_3(x) = 49.
>
> Note that a_1(x)*a_2(x)*a_3(x) is divisible by 49, but in general
> not by any additional factors of 7. Why? Because
>
> 2401 x^3 - 147 x^2 + 3x
>
> is coprime to 7 [except when x itself is divisible by 7], and
>
> -49*(2401 x^3 - 147 x^2 + 3x)
>
> is the constant term of the polynomial that the a's satisfy, as
> you note above.
>
> Since w_1(x), w_2(x), and w_3(x) are all greatest common divisors
> of (respectively) a_1(x), a_2(x), and a_3(x) with 7, their product
> must equal 49, as desired.
>
> What this shows is that it IS possible to divide 49 out of
>
> (r_1(x) + 7)(r_2(x) + 7)(r_3(x) + 22)
>
> in such a way that each factor is an algebraic integer.
>

Well, yes, it's possible.

> But what about the "constant terms" ?
>
> It is *not required* that the constant terms be "respected". That

Notice the poster doesn't even try at this point, simply resorting to
hand-waving by asserting that the constant terms don't need to be
respected!

The problem is that if you have a constant term that is 7, another
that is 7, and one that's 22, then to get rid of the 7's, you have to
divide out by 7, but that's algebra that's inconvenient to the poster!

So instead the poster just TELLS you that the math doesn't care.

Let's watch to see what else this poster tries on you.

> is, even though it is true that
>
> (7/w_1(x))*(7/w_2(x))*(22/w_3(x)) = 22,
>
> (as you can immediately check), there is NO REASON to require that each
> of these three terms be algebraic integers; and of course, 22/w_3(x)
> is not.
>
> Again, why? Isn't the product of the constant terms equal to the
> constant term of the product ???
>
> Yes. But here is the last key fact. The constant term of
>
> (r_2(x) + 22)/w_3(x)
>
> is NOT what you think it is. You think it must be
>
> 22/w_3(x).
>
> It isn't. You need to remember YOUR OWN DEFINITION of constant
> term. It is instead
>
> 22/w_3(0).
>
> So then you note that w_3(0) = 1, and you have no contradiction.
> Everything hangs together.
>

Notice that the poster has asserted that the constant term is constant
at x=0, but every where else it's actually a function of x.

So the full assertion is that the constant term is NOT CONSTANT, but
is instead a function of x.

Remember, when people try to fight mathematics they have to at some
point rely on something that is just wacky. Here you can see that
ultimately the poster is trying to get you to believe that a constant
is in fact a function of x.

But given g_1(x) g_2(x)...g_n(x) = P(x), where the n factors are
algebraic integer functions, then necessarily g_1(0) g_2(0)...g_n(0) =
P(0), and notice, you don't have any functions of x, as x has been set
to 0.

So the poster is using unit factors, which by a technicality are not
units in the ring of algebraic integers, unless you do wish to believe
that mathematics is inconsistent, and so what? Your belief would be
wrong.

> Again: you defined 'constant term' in a perfectly reasonable way,
> and you should have stuck with your definition. Instead you got
> confused and assumed that whatever is in the POSITION of the constant
> term *is* the constant term. But here, 22/w_3(x) is not even constant,
> because by its definition, w_3(x) is not constant.
>
> Nora B.

Well, I say you're using unit factors, which because they are not
roots of a monic polynomial with integer coefficients are not units in
the ring of algebraic integers.

How do you answer?

James Harris

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