## Thursday, October 30, 2008

### Easy proof of problem with ring of algebraic integers

Some simple algebra with a basic polynomial reveals a serious problem with the naive use of the ring of algebraic integers, bringing into question over a hundred years of algebraic number theory.

In an integral domain, consider a simple polynomial

P(x) = 175x2 - 15x + 2.

Multiply it times 7, to get

7*P(x) = 1225x2 - 105x + 14. Cleverly re-group terms:

1225x2 - 105x + 14 = (49x2 - 14x)52 + (7x-1)(7)(5) + 72

and now factor into non-polynomials:

7*(175x2 - 15x + 2) = (5a1(x) + 7)(5a2(x)+ 7)

where you'll note that the a's are functions of x that are roots of

a2 - (7x-1)a + (49x2 - 14x) = 0.

But now consider our polynomial again with a factorization before any
multiplying by 7:

175x2 - 15x + 2 = (5b1(x) + 2)(5b2(x)+ 1)

Now multiply by 7, to get

7*(175x2 - 15x + 2) = (5b1(x) + 2)(5(7*b2(x))+ 7)

and use the substitutions b1(x) = c1(x) + 1, and 7*b2(x) = c2(x), and you have

7*(175x2 - 15x + 2) = (5c1(x) + 7)(5c2(x)+ 7)

and of course if c1(x) = a1(x) and c2(x) = a2(x), I have my original factorization, but in so doing I'm PICKING that 7 multiplies times just one of the factors of 175x2 - 15x + 2, but what if I picked wrong?

For instance, consider again

175x2 - 15x + 2 = (5b1(x) + 2)(5b2(x)+ 1)

and again multiply times 7, but split it up so that each factor is multiplied times sqrt(7):

7*(175x2 - 15x + 2) = (5*sqrt(7)*b1(x) + 2*sqrt(7))(5*sqrt(7)b2(x)+ sqrt(7))

but there's an immediate problem!

If you let x=0, then you have the factorization:

7*(2) = (5*sqrt(7)*b1(0) + 2*sqrt(7))(5*sqrt(7)b2(0)+ sqrt(7))

which contradicts at x= 0 with

7*(175x2 - 15x + 2) = (5a1(x) + 7)(5a2(x)+ 7)

where the a's are functions of x that are roots of

a2 - (7x-1)a + (49x2 - 14x) = 0

unless b_2(0) divides off sqrt(7), or b1(0) divides off sqrt(7), as

7*(2) = (5a1(0) + 7)(5a2(0)+ 7) = (0 + 7)(-5+ 7)

because then the a's are roots of

a2 + a = 0.

Therefore, there is no other way to multiply

175x2 - 15x + 2 = (5b1(x) + 2)(5b2(x)+ 1)

by 7, and get the factorization

7*(175x2 - 15x + 2) = (5a1(x) + 7)(5a2(x)+ 7)

where the a's are functions of x that are roots of

a2 - (7x-1)a + (49x2 - 14x) = 0

as ANY other way other than multiplying (5b2(x)+ 1) by 7, will contradict with

7*(175x2 - 15x + 2) = (5a1(x) + 7)(5a2(x)+ 7)

at x=0, as demonstrated above with sqrt(7).

Therefore, one of the a's must have 7 as a factor for all x, but it is trivial to show that NEITHER of them can have 7 as a factor for any integer x, for which the a's are not rational, in the ring of algebraic integers, so there is proven a problem with that ring.