### Saturday, September 04, 2010

## JSH: Algebraic integers and Galois Theory

So I've used a simple expression to try and help math students comprehend a VERY important error in "core".

7(g_1(x) + 1)(g_2(x) + 2) = (f_1(x) + 7)(f_2(x) + 7) = 7*P(x)

where P(x) is a quadratic with integer coefficients, g_1(0) = g_2(0) = 0, and one of the f's equals 0 at x=0.

A simple exercise reveals that the 7 cannot split up in any way to give:

(f_1(x) + 7)(f_2(x) + 7)

in the middle, which is a BIG DEAL because the f's can be algebraic integers that are roots of a polynomial irreducible over Q, and the result should force only one of them to have 7 as a factor, but PROVABLY neither can in that situation in the ring of algebraic integers!!!

For years posters insulted me in this situation claiming they could arithmetically show that I was wrong, as they'd trot out results where the 7 had been divided off, they thought—having split up in a way that is commonly believed.

But the correct answer requires a complete ring, which is my ring of objects. I discovered the ring of objects BECAUSE of this problem. Now consider unit factors u_1 and u_2 in the ring of objects which are NOT units in the ring of algebraic integers. If you say, but that's not right, see why it is the only possibility allowed, as now you can have:

7(g_1(x) + 1)(g_2(x) + 2) = (f_1(x)*u_1 + 7*u_1)(f_2(x)*u_2 + 7*u_2) = 7*P(x)

as being unit factors they can spontaneously arise, and NOW divide off the 7 to get:

(g_1(x) + 1)(g_2(x) + 2) = (f_1(x)*u_1/7 + u_1)(f_2(x)*u_2 + 7*u_2) = P(x)

and u_1 is an algebraic integer that would be found as a solution, which is NOT a unit in the ring of algebraic integers, while 7u_2 IS an algebraic integer, but u_2 is NOT an algebraic integer, so you have the 7 wrapped up neatly.

You know that is the only possibility because if the 7 cannot split up, and 7 cannot be a factor in the ring of algebraic integers for a particular case then that's the only way mathematically to avoid a contradiction. The 7 gets wrapped up.

And you need unit factors to do it!

So why is u_1 not a unit in the ring of algebraic integers? Because u_1 and 7u_2 are roots of a monic quadratic with integer coefficients! That's why!

Turns out that the ring of algebraic integers will not allow a number to be a unit unless it is a root of a monic polynomial with integer coefficients that has 1 or -1 as the last coefficient, so it throws out u_2 from the ring, and doesn't consider u_1 to be a unit within the ring.

Clever.

All of the above has implications for Galois Theory as then, Galois Theory doesn't tell you when 7 really is or is not a factor of one of the roots! And all the machinery of Galois Theory isn't giving you useful information when it looks like the 7 is "split up" as you consider different algebraic integer factors appearing as you try to divide that 7 off.

What you are seeing then are UNIT FACTORS that are appearing as needed to wrap up the actual factors. So the real factors are being wrapped up and you learn NOTHING from Galois Theory.

When I figured this out several years ago I noted that you can do the same things with integer solutions to understand what Galois Theory is actually telling you—by not taking the square root.

For instance with x^2 +3x + 2 = 0, you'd have

x = (-3 +/- sqrt(1))/2

and if you refuse to take sqrt(1)—I know sounds silly but use your imagination—then you can use all the mechanisms of Galois theory and get a class whatever, and do all those things, but notice if you do the same with:

x^2 + 7x + 10 = 0, you'd have

x = (-7 +/- sqrt(9))/2

and a different class number, but so what? Remember NOT taking the square root.

So that is MEANINGLESS. It's just noticing that you're NOT taking the square root, and ignores the reality that one of the x's is the same in both cases.

With non-rationals where you can't take the square root the mathematics is simply acknowledging that, and giving you no additional information.

So Galois Theory tells you nothing about the underlying factors. Nothing at all.

And supposed solutions where you try to divide off integers like 7, are simply unit factors wrapping around the underlying factors, which remain hidden.

Years of arguing with posters who tried to fight the distributive property. And claimed they didn't. But to try and attack the above— serious math people are welcome and encouraged to try—you need what I call coupling.

So with:

7(g_1(x) + 1)(g_2(x) + 2) = (f_1(x)*u_1 + 7*u_1)(f_2(x)*u_2 + 7*u_2) = 7*P(x)

you need (g_1(x) + 1)(g_2(x) + 2) to somehow push BACK out against the 7 and tell it how to multiply in order to get:

(f_1(x)*u_1 + 7*u_1)(f_2(x)*u_2 + 7*u_2)

doing so as a function of x, as at x=0, you're forced to have 7(g_1(x) + 1) = 7*g_1(x) + 7, by the distributive property, as at that value g_1(0) = 0, and one of the f's equals 0, which requires that solution!!!

So it should be set, right? Well they'd argue that it was a special case!!! And that the 7 multiplied *differently* for differing values of x.

Which I call coupling. The 7 then becomes a function of x, though some weird mechanism they argued about endlessly which would frustrate me so much that I'd come up with names for it, like voodoo mathematics.

7(g_1(x) + 1)(g_2(x) + 2) = (f_1(x) + 7)(f_2(x) + 7) = 7*P(x)

where P(x) is a quadratic with integer coefficients, g_1(0) = g_2(0) = 0, and one of the f's equals 0 at x=0.

A simple exercise reveals that the 7 cannot split up in any way to give:

(f_1(x) + 7)(f_2(x) + 7)

in the middle, which is a BIG DEAL because the f's can be algebraic integers that are roots of a polynomial irreducible over Q, and the result should force only one of them to have 7 as a factor, but PROVABLY neither can in that situation in the ring of algebraic integers!!!

For years posters insulted me in this situation claiming they could arithmetically show that I was wrong, as they'd trot out results where the 7 had been divided off, they thought—having split up in a way that is commonly believed.

But the correct answer requires a complete ring, which is my ring of objects. I discovered the ring of objects BECAUSE of this problem. Now consider unit factors u_1 and u_2 in the ring of objects which are NOT units in the ring of algebraic integers. If you say, but that's not right, see why it is the only possibility allowed, as now you can have:

7(g_1(x) + 1)(g_2(x) + 2) = (f_1(x)*u_1 + 7*u_1)(f_2(x)*u_2 + 7*u_2) = 7*P(x)

as being unit factors they can spontaneously arise, and NOW divide off the 7 to get:

(g_1(x) + 1)(g_2(x) + 2) = (f_1(x)*u_1/7 + u_1)(f_2(x)*u_2 + 7*u_2) = P(x)

and u_1 is an algebraic integer that would be found as a solution, which is NOT a unit in the ring of algebraic integers, while 7u_2 IS an algebraic integer, but u_2 is NOT an algebraic integer, so you have the 7 wrapped up neatly.

You know that is the only possibility because if the 7 cannot split up, and 7 cannot be a factor in the ring of algebraic integers for a particular case then that's the only way mathematically to avoid a contradiction. The 7 gets wrapped up.

And you need unit factors to do it!

So why is u_1 not a unit in the ring of algebraic integers? Because u_1 and 7u_2 are roots of a monic quadratic with integer coefficients! That's why!

Turns out that the ring of algebraic integers will not allow a number to be a unit unless it is a root of a monic polynomial with integer coefficients that has 1 or -1 as the last coefficient, so it throws out u_2 from the ring, and doesn't consider u_1 to be a unit within the ring.

Clever.

All of the above has implications for Galois Theory as then, Galois Theory doesn't tell you when 7 really is or is not a factor of one of the roots! And all the machinery of Galois Theory isn't giving you useful information when it looks like the 7 is "split up" as you consider different algebraic integer factors appearing as you try to divide that 7 off.

What you are seeing then are UNIT FACTORS that are appearing as needed to wrap up the actual factors. So the real factors are being wrapped up and you learn NOTHING from Galois Theory.

When I figured this out several years ago I noted that you can do the same things with integer solutions to understand what Galois Theory is actually telling you—by not taking the square root.

For instance with x^2 +3x + 2 = 0, you'd have

x = (-3 +/- sqrt(1))/2

and if you refuse to take sqrt(1)—I know sounds silly but use your imagination—then you can use all the mechanisms of Galois theory and get a class whatever, and do all those things, but notice if you do the same with:

x^2 + 7x + 10 = 0, you'd have

x = (-7 +/- sqrt(9))/2

and a different class number, but so what? Remember NOT taking the square root.

So that is MEANINGLESS. It's just noticing that you're NOT taking the square root, and ignores the reality that one of the x's is the same in both cases.

With non-rationals where you can't take the square root the mathematics is simply acknowledging that, and giving you no additional information.

So Galois Theory tells you nothing about the underlying factors. Nothing at all.

And supposed solutions where you try to divide off integers like 7, are simply unit factors wrapping around the underlying factors, which remain hidden.

Years of arguing with posters who tried to fight the distributive property. And claimed they didn't. But to try and attack the above— serious math people are welcome and encouraged to try—you need what I call coupling.

So with:

7(g_1(x) + 1)(g_2(x) + 2) = (f_1(x)*u_1 + 7*u_1)(f_2(x)*u_2 + 7*u_2) = 7*P(x)

you need (g_1(x) + 1)(g_2(x) + 2) to somehow push BACK out against the 7 and tell it how to multiply in order to get:

(f_1(x)*u_1 + 7*u_1)(f_2(x)*u_2 + 7*u_2)

doing so as a function of x, as at x=0, you're forced to have 7(g_1(x) + 1) = 7*g_1(x) + 7, by the distributive property, as at that value g_1(0) = 0, and one of the f's equals 0, which requires that solution!!!

So it should be set, right? Well they'd argue that it was a special case!!! And that the 7 multiplied *differently* for differing values of x.

Which I call coupling. The 7 then becomes a function of x, though some weird mechanism they argued about endlessly which would frustrate me so much that I'd come up with names for it, like voodoo mathematics.