### Thursday, September 02, 2010

## JSH: Broken symmetry and algebraic integers

The ring of algebraic integers has special properties which force it to require symmetry, so breaking symmetry with it, can produce mathematical results of interest. Here is simple broken symmetry fun with which to experiment:

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 polynomial with integer coefficients, and g_1(0) = g_2(0) = 0, and at least one of the f's equals 0 as well when x = 0.

Serious math students realize immediately that if you split the 7 on the left into complex conjugates you have an issue getting 7 again in the middle, as if its splits into r_1 and r_2, where y^2 + by + 7 = 0, where b is a non-zero integer, non-rational solutions will require BOTH r_1 and r_2 for you to get 7.

Let's see it in action!

7(g_1(x) + 1)(g_2(x) + 2) = 7(g_1(x)*r_1 + r_1)(g_2(x)*r_2 + 2r_2) = (f_1(x) + 7)(f_2(x) + 7) = 7*P(x)

There is clearly a problem! How do we get to 7? Obviously we need r_2 in the first and r_1 in the second, like so:

7(g_1(x) + 1)(g_2(x) + 2) = 7(g_1(x) - r_2 + 1 + r_2)(g_2(x) - r_1 + 2 + r_1) = (g_1(x)*r_1 - 7+ r_1 + 7)(g_2(x)*r_2 - 7 + 2*r_2 + 7)

So:

(g_1(x)*r_1 - 7+ r_1 + 7)(g_2(x)*r_2 - 7 + 2*r_2 + 7) = (f_1(x) + 7)(f_2(x) + 7) = 7*P(x)

which REQUIRES that

f_1(x) = g_1(x)*r_1 - 7+ r_1

and

f_2(x) = g_2(x)*r_2 - 7 + 2*r_2

But now there is a problem with the restriction that one of the f's equals 0—if neither of the r's is an integer!!!

f_1(0) = g_1(0)*r_1 - 7+ r_1 = -7 + r_1

and

f_2(0) = g_2(0)*r_2 - 7 + 2*r_2 = -7 + 2*r_2

So that restriction requires, if integers only, that r_1 = 7.

So you now know that:

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

is excluded from the ring of algebraic integers, if 7 itself is not a factor of one of the f's.

Cool, huh? I think you should puzzle over all of the above for a bit. Could I have done it any other way? Is it possible to split that 7 up with the rules in some other way than I've said?

What are the consequences then of that result!

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 polynomial with integer coefficients, and g_1(0) = g_2(0) = 0, and at least one of the f's equals 0 as well when x = 0.

Serious math students realize immediately that if you split the 7 on the left into complex conjugates you have an issue getting 7 again in the middle, as if its splits into r_1 and r_2, where y^2 + by + 7 = 0, where b is a non-zero integer, non-rational solutions will require BOTH r_1 and r_2 for you to get 7.

Let's see it in action!

7(g_1(x) + 1)(g_2(x) + 2) = 7(g_1(x)*r_1 + r_1)(g_2(x)*r_2 + 2r_2) = (f_1(x) + 7)(f_2(x) + 7) = 7*P(x)

There is clearly a problem! How do we get to 7? Obviously we need r_2 in the first and r_1 in the second, like so:

7(g_1(x) + 1)(g_2(x) + 2) = 7(g_1(x) - r_2 + 1 + r_2)(g_2(x) - r_1 + 2 + r_1) = (g_1(x)*r_1 - 7+ r_1 + 7)(g_2(x)*r_2 - 7 + 2*r_2 + 7)

So:

(g_1(x)*r_1 - 7+ r_1 + 7)(g_2(x)*r_2 - 7 + 2*r_2 + 7) = (f_1(x) + 7)(f_2(x) + 7) = 7*P(x)

which REQUIRES that

f_1(x) = g_1(x)*r_1 - 7+ r_1

and

f_2(x) = g_2(x)*r_2 - 7 + 2*r_2

But now there is a problem with the restriction that one of the f's equals 0—if neither of the r's is an integer!!!

f_1(0) = g_1(0)*r_1 - 7+ r_1 = -7 + r_1

and

f_2(0) = g_2(0)*r_2 - 7 + 2*r_2 = -7 + 2*r_2

So that restriction requires, if integers only, that r_1 = 7.

So you now know that:

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

is excluded from the ring of algebraic integers, if 7 itself is not a factor of one of the f's.

Cool, huh? I think you should puzzle over all of the above for a bit. Could I have done it any other way? Is it possible to split that 7 up with the rules in some other way than I've said?

What are the consequences then of that result!