Saturday, February 04, 2006
JSH: Decker example, end of an era
Now I'm going to go over the Decker example and how it refutes the standard teaching of Galois Theory, and it turns out that it shows the theory of ideals to be flawed as well, but it's more convoluted to explain why that's true and I'll leave that explanation as it is somewhat technical to others.
This example because of the use of simple quadratics takes away the objections of posters who have inexplicably refused to accept the distributive property in proofs I've shown with my own more complicated expressions.
And also I'm starting with equations put up by a Rick Decker, so it's not like I picked them personally.
The attempt will be to have everthing below in the ring of algebraic integers, so consider that the ring, but notice at a key point, we will be forced out of that ring.
Consider
7 Q(x) = 7((x^2 + x)(5^2) + (-1 + x)(5) + 7) = 7(25 x^2 + 30 x + 2)
and
7 Q(x) = (5a_1(x) + 7)(5a_2(x) + 7)
where the a's are defined by
a^2 - (x - 1)a + 7(x^2 + x) = 0
where Decker chose this example because at x=1, the middle term goes to 0 which he I guess thought was a counterexample to my ideas because at x=0, the a's are 0 and 1, indicating that 7 divides through only one of the expressions, but clearly it does not at x=1, but that special case doesn't change the refutation so I'll leave out the explanation of it
The following disproof focuses directly on Galois Theory while the relation to the theory of ideals is more complex.
The first step in the disproof is to focus on that last quadratic
a^2 - (x - 1)a + 7(x^2 + x) = 0
and write it with the focus on x, which gives a non-monic polynomial that importantly becomes monic if 'a' has 7 as a factor:
7x^2 + (7-a)x + a^2 + a = 0
and using a = 7b, you get
7x^2 + (7 - 7b)x + 49b^2 + 7b = 0
and dividing off 7, gives you
x^2 + (1 - b)x + 7b^2 + b = 0
so for ANY algebraic integer 'a' that has 7 as a factor, you get an algebraic integer x.
Also, notice that 7b^2 in the expression so that if you focus on 'b' you have a non-monic polynomial and as before, 'b' cannot be an algebraic integer if that polynomial has integer coefficients and is irreducible over Q.
BUT
a^2 - (x - 1)a + 7(x^2 + x) = 0
means that for ANY algebraic integer x, 'a' is an algebraic integer, but if 'a' does not have 7 as a factor
7x^2 + (7-a)x + a^2 + a = 0
will not allow an algebraic integer x, if 'a' is an integer and the quadratic is irreducible over Q.
So you are forced out of the ring of algebraic integers.
Algebraically you still have
x^2 + (1 - b)x + 7b^2 + b = 0
but the 'b' is outside of the ring of algebraic integers for certain algebraic integer values of x.
To save Galois Theory and the theory of ideals, the attempted defense of those ideas at this point would require that for some non-rational 'a' with partial factors of 7, as it is a root of
a^2 - (x - 1)a + 7(x^2 + x) = 0
with an integer x other than x=1, you would get a polynomial with an integer solution for x from
7x^2 + (7-a)x + a^2 + a = 0
which would remain non-monic because the factors of 7 not in common with 'a' would remain on the leading coefficient.
However, you can simply pick some x, like x=3, and solve for the a's, and then make the substitution in which will give a seemingly non-monic expression of some complexity as it will have square root radicals, but you can use basic algebraic manipulations to remove the square roots, giving you a quartic, where you will find that the leading coefficient just divides off, as one of the roots will be 3, so it will also be reducible over Q.
So Galois Theory as a way to determine where factors go is refuted, and it is proven that the ring of algebraic integers is incomplete, so that some numbers are excluded simply because they are not roots of monic polynomials with integer coefficients, which leads to the conclusion that the theory of ideals is flawed.
This post marks the end of an era in the world of mathematics.
This example because of the use of simple quadratics takes away the objections of posters who have inexplicably refused to accept the distributive property in proofs I've shown with my own more complicated expressions.
And also I'm starting with equations put up by a Rick Decker, so it's not like I picked them personally.
The attempt will be to have everthing below in the ring of algebraic integers, so consider that the ring, but notice at a key point, we will be forced out of that ring.
Consider
7 Q(x) = 7((x^2 + x)(5^2) + (-1 + x)(5) + 7) = 7(25 x^2 + 30 x + 2)
and
7 Q(x) = (5a_1(x) + 7)(5a_2(x) + 7)
where the a's are defined by
a^2 - (x - 1)a + 7(x^2 + x) = 0
where Decker chose this example because at x=1, the middle term goes to 0 which he I guess thought was a counterexample to my ideas because at x=0, the a's are 0 and 1, indicating that 7 divides through only one of the expressions, but clearly it does not at x=1, but that special case doesn't change the refutation so I'll leave out the explanation of it
The following disproof focuses directly on Galois Theory while the relation to the theory of ideals is more complex.
The first step in the disproof is to focus on that last quadratic
a^2 - (x - 1)a + 7(x^2 + x) = 0
and write it with the focus on x, which gives a non-monic polynomial that importantly becomes monic if 'a' has 7 as a factor:
7x^2 + (7-a)x + a^2 + a = 0
and using a = 7b, you get
7x^2 + (7 - 7b)x + 49b^2 + 7b = 0
and dividing off 7, gives you
x^2 + (1 - b)x + 7b^2 + b = 0
so for ANY algebraic integer 'a' that has 7 as a factor, you get an algebraic integer x.
Also, notice that 7b^2 in the expression so that if you focus on 'b' you have a non-monic polynomial and as before, 'b' cannot be an algebraic integer if that polynomial has integer coefficients and is irreducible over Q.
BUT
a^2 - (x - 1)a + 7(x^2 + x) = 0
means that for ANY algebraic integer x, 'a' is an algebraic integer, but if 'a' does not have 7 as a factor
7x^2 + (7-a)x + a^2 + a = 0
will not allow an algebraic integer x, if 'a' is an integer and the quadratic is irreducible over Q.
So you are forced out of the ring of algebraic integers.
Algebraically you still have
x^2 + (1 - b)x + 7b^2 + b = 0
but the 'b' is outside of the ring of algebraic integers for certain algebraic integer values of x.
To save Galois Theory and the theory of ideals, the attempted defense of those ideas at this point would require that for some non-rational 'a' with partial factors of 7, as it is a root of
a^2 - (x - 1)a + 7(x^2 + x) = 0
with an integer x other than x=1, you would get a polynomial with an integer solution for x from
7x^2 + (7-a)x + a^2 + a = 0
which would remain non-monic because the factors of 7 not in common with 'a' would remain on the leading coefficient.
However, you can simply pick some x, like x=3, and solve for the a's, and then make the substitution in which will give a seemingly non-monic expression of some complexity as it will have square root radicals, but you can use basic algebraic manipulations to remove the square roots, giving you a quartic, where you will find that the leading coefficient just divides off, as one of the roots will be 3, so it will also be reducible over Q.
So Galois Theory as a way to determine where factors go is refuted, and it is proven that the ring of algebraic integers is incomplete, so that some numbers are excluded simply because they are not roots of monic polynomials with integer coefficients, which leads to the conclusion that the theory of ideals is flawed.
This post marks the end of an era in the world of mathematics.
# posted by James Harris @ 16:00 
