### Sunday, September 26, 2010

## JSH: How complicated can it be?

I'm getting replies from posters questioning what I see as a trivial argument, so how hard can it be? Like, once again, as I try to use the simplest expression I've found to explain what I think is trivial, try in the ring of algebraic integers to use:

9(g_1(x) + 1)(g_2(x) + 2) = (f_1(x) + 9)(f_2(x) + 9) = 9*P(x)

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

You can find cases in the ring of algebraic integers where NEITHER of the f's can have 9 as a factor, for certain values of x, so you wander off the to the complex plane—because you've a VERY SMART math student—so you can see exactly what is happening.

There you note that:

(9g_1(x) + 9)(g_2(x) + 2) = (f_1(x) + 9)(f_2(x) + 9) = 9*P(x)

follows trivially by the distributive property so FOR THAT CASE the ring of algebraic integers presumably would have 9 as a factor for all x, but can be shown to NOT have 9 as a factor as noted above.

Being an EXTREMELY INTELLIGENT math student you find this puzzling, so you consider it until it makes sense.

Now then, how can you multiply by 9 in the field, but not be allowed to have 9 as a factor in the ring of algebraic integers?

I think the answer is easy, but am I wrong? Can any of you solve the puzzle?

If you get completely lost, just go to Google.

Search on: algebraic integers vs complex numbers

I put it all in a paper, which is on Scribd.

9(g_1(x) + 1)(g_2(x) + 2) = (f_1(x) + 9)(f_2(x) + 9) = 9*P(x)

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

You can find cases in the ring of algebraic integers where NEITHER of the f's can have 9 as a factor, for certain values of x, so you wander off the to the complex plane—because you've a VERY SMART math student—so you can see exactly what is happening.

There you note that:

(9g_1(x) + 9)(g_2(x) + 2) = (f_1(x) + 9)(f_2(x) + 9) = 9*P(x)

follows trivially by the distributive property so FOR THAT CASE the ring of algebraic integers presumably would have 9 as a factor for all x, but can be shown to NOT have 9 as a factor as noted above.

Being an EXTREMELY INTELLIGENT math student you find this puzzling, so you consider it until it makes sense.

Now then, how can you multiply by 9 in the field, but not be allowed to have 9 as a factor in the ring of algebraic integers?

I think the answer is easy, but am I wrong? Can any of you solve the puzzle?

If you get completely lost, just go to Google.

Search on: algebraic integers vs complex numbers

I put it all in a paper, which is on Scribd.