Saturday, December 13, 2008


JSH: Breaking Galois Theory

In the ring of algebraic integers consider the special construction:

7*(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7)

where the a's are roots of

a^2 - (7x-1)a + (49x^2 - 14x) = 0

with x=1+sqrt(-6) which is chosen because it is a factor of 7, as (1+sqrt(-6))(1-sqrt(-6)) = 7.

Notice that

7*(175x^2 - 15x + 2) has 7(1-sqrt(-6)) as a factor


7*(175x^2 - 15x + 2) = (5a_1(x) + 7)(5a_2(x)+ 7)

where a_1(x)*a_2(x) has 7(1+sqrt(-6)) as a factor, but a_1(x) is coprime to a_2(x), because

a_1(x) + a_2(x) = 7x-1

as I remind that the a's are roots of a^2 - (7x-1)a + (49x^2 - 14x) = 0.

There are multiple apparent contradictions with this example.

If you divide 7 from both sides, then 1-sqrt(-6) remains a factor of both sides, but then factors of 7 that remain for BOTH a's are coprime to it, while also being coprime to each other, but in each case with

(5a_1(x) + 7) and (5a_2(x)+ 7)

the 7 on the right in each the result after dividing 7 from both sides must be coprime to the 'a' opposite it, but then in at least one case what remains must have factors in common with 1-sqrt(-6).

Those attacking the validity of this example as a counter-example to standard Galois Theory should give the factors of the a's in common with 7.

Oh, the resolution? It's easy. There is no actual contradiction.

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