### Monday, November 30, 2009

## Trashing Galois Theory

Turns out you can destroy the underpinnings of modern number theory with a simple mathematical example:

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.

The 7 on the left of the equals with the first expression cannot in general be removed on the right hand side IN THE RING OF ALGEBRAIC INTEGERS.

The a's are the roots of the second expression, so you can use the quadratic formula to solve for them:

a_1(x) = ((7x-1)+/- sqrt((7x-1)^2 - 4(49x^2 - 14x)))/2, a_2(x) = ((7x-1) -/+ sqrt((7x-1)^2 - 4(49x^2 - 14x)))/2

I've brought this issue up time and time again, and one objection I've heard repeatedly is that you CAN divide the 7 off for INDIVIDUAL CASES, like for x=1, you can do some convoluted crap using Galois Theory to find the "factors" of 7 in the ring of algebraic integers that would divide off FOR THAT SPECIAL CASE.

But hey, isn't this algebra? Why don't you have special case for:

7(x^2 + 3x + 2) = (7x + 7)(x + 2)?

Why can I just divide THAT 7 off to get: x^2 + 3x + 2 = (x + 1)(x + 2)?

The answer is: the ring of algebraic integers is CRAP. It was crap years ago when I found this problem and it's still crap now though YOUR STUPID PROFESSORS STILL TEACH YOU CRAP.

Yes, they are stupid for teaching you crap and you are stupid if you keep learning it after this example.

Or are you such a moron that you don't know that if you multiply something times say, 7, you should be able to divide the freaking 7 back off?

I don't care if your professor is at Harvard. If he's teaching you stupid crap it's still stupid crap.

It doesn't matter if she is at Oxford. If she is teaching you stupid crap it's still stupid crap.

And you're still stupid for learning it.

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.

The 7 on the left of the equals with the first expression cannot in general be removed on the right hand side IN THE RING OF ALGEBRAIC INTEGERS.

The a's are the roots of the second expression, so you can use the quadratic formula to solve for them:

a_1(x) = ((7x-1)+/- sqrt((7x-1)^2 - 4(49x^2 - 14x)))/2, a_2(x) = ((7x-1) -/+ sqrt((7x-1)^2 - 4(49x^2 - 14x)))/2

I've brought this issue up time and time again, and one objection I've heard repeatedly is that you CAN divide the 7 off for INDIVIDUAL CASES, like for x=1, you can do some convoluted crap using Galois Theory to find the "factors" of 7 in the ring of algebraic integers that would divide off FOR THAT SPECIAL CASE.

But hey, isn't this algebra? Why don't you have special case for:

7(x^2 + 3x + 2) = (7x + 7)(x + 2)?

Why can I just divide THAT 7 off to get: x^2 + 3x + 2 = (x + 1)(x + 2)?

The answer is: the ring of algebraic integers is CRAP. It was crap years ago when I found this problem and it's still crap now though YOUR STUPID PROFESSORS STILL TEACH YOU CRAP.

Yes, they are stupid for teaching you crap and you are stupid if you keep learning it after this example.

Or are you such a moron that you don't know that if you multiply something times say, 7, you should be able to divide the freaking 7 back off?

I don't care if your professor is at Harvard. If he's teaching you stupid crap it's still stupid crap.

It doesn't matter if she is at Oxford. If she is teaching you stupid crap it's still stupid crap.

And you're still stupid for learning it.