Tuesday, December 09, 2008
JSH: Then consider this example
It seems that I haven't convinced with my talk of the distributive property so here is an example which should test your understanding of Galois Theory to the limit.
In the ring of algebraic integers, let x=1+sqrt(-6), with
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.
I picked x so that it is a factor of 7, as (1+sqrt(-6))(1-sqrt(-6)) = 7. Also notice that despite x having a factor in common with 7:
7*(175x^2 - 15x + 2)
still only has 7 as a factor.
Further note that a_1(x)*a_2(x) has 7(1+sqrt(-6)) as a factor, which is the point, as if the a's do not share factors in common with 7 carefully, then you contradict with only 7 being a factor of
(5a_1(x) + 7)(5a_2(x)+ 7).
Give the a's and solve for the factors in common with the a's for this value of x.
Have fun! Preserve your deluded view of Galois Theory—if you can.
The challenge is in front of you, are any of you good enough? Smart enough?
In the ring of algebraic integers, let x=1+sqrt(-6), with
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.
I picked x so that it is a factor of 7, as (1+sqrt(-6))(1-sqrt(-6)) = 7. Also notice that despite x having a factor in common with 7:
7*(175x^2 - 15x + 2)
still only has 7 as a factor.
Further note that a_1(x)*a_2(x) has 7(1+sqrt(-6)) as a factor, which is the point, as if the a's do not share factors in common with 7 carefully, then you contradict with only 7 being a factor of
(5a_1(x) + 7)(5a_2(x)+ 7).
Give the a's and solve for the factors in common with the a's for this value of x.
Have fun! Preserve your deluded view of Galois Theory—if you can.
The challenge is in front of you, are any of you good enough? Smart enough?
# posted by James Harris @ 11:37 
