### Tuesday, August 24, 2010

## JSH: Expression blocking by ring of algebraic integers

A simple way to see a problem with the ring of algebraic integers is by compare and contrast, as consider:

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

where P(x) is a quadratic irreducible over Q, P(0)=1, and none of the functions are constants. That is ALWAYS available in the ring of algebraic integers without restrictions. Now in contrast consider:

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

where also P(x) is a quadratic irreducible over Q, but P(0) = 2, but still none of the functions are constants. That is NOT available in the ring of algebraic integers.

That expression blocking is peculiar because one wonders, why did changing a 1 to a 2, blow an expression out of the ring of algebraic integers for certain cases? Answer: symmetry breaking

If one considers those expressions carefully you DO find that in BOTH cases the f's can be algebraic integer functions. And it turns out that one of the g's can always be an algebraic integer function as well, but one cannot in the second example, while BOTH can in the first example.

What is happening is that the rules of the ring of algebraic integers do not allow it to allow the second example in certain particular circumstances, so it blocks it. It doesn't allow those particular expressions to exist as given.

And all it takes is changing a 1 to a 2.

One way to look at the issue is that the ring of algebraic integers has symmetry requirements that allow it to specify to algebra itself that certain algebraic expressions are not allowed within that ring, under highly particular circumstances, unlike any other ring known.

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

where P(x) is a quadratic irreducible over Q, P(0)=1, and none of the functions are constants. That is ALWAYS available in the ring of algebraic integers without restrictions. Now in contrast consider:

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

where also P(x) is a quadratic irreducible over Q, but P(0) = 2, but still none of the functions are constants. That is NOT available in the ring of algebraic integers.

That expression blocking is peculiar because one wonders, why did changing a 1 to a 2, blow an expression out of the ring of algebraic integers for certain cases? Answer: symmetry breaking

If one considers those expressions carefully you DO find that in BOTH cases the f's can be algebraic integer functions. And it turns out that one of the g's can always be an algebraic integer function as well, but one cannot in the second example, while BOTH can in the first example.

What is happening is that the rules of the ring of algebraic integers do not allow it to allow the second example in certain particular circumstances, so it blocks it. It doesn't allow those particular expressions to exist as given.

And all it takes is changing a 1 to a 2.

One way to look at the issue is that the ring of algebraic integers has symmetry requirements that allow it to specify to algebra itself that certain algebraic expressions are not allowed within that ring, under highly particular circumstances, unlike any other ring known.