Re: Some math, algebraic integers

From: Nora Baron (norabaron_at_hotmail.com)
Date: 09/18/04 

Date: 18 Sep 2004 06:40:52 -0700

jstevh@msn.com (James Harris) wrote in message news:<3c65f87.0409171539.1773e0d9@posting.google.com>...
> Luckily for me the math that explains my work is actually quite simple
> in many ways. All that is required is that you accept algebra. That
> might sound weird, but pay attention and you'll understand by the end.
>
> Now with rationals you can have a simple case like
>
> 2x^2 + 5x - 3 = 0
>
> where the roots are x=1/2 and x=-3, if I did my algebra correctly.
>
> Now notice you have an integer paired with a fraction, which, of
> course, is NOT an integer.
>
> Now consider some numbers q_1 and q_2, which I won't say a lot about
> now, though in a bit I'll put them in a particular ring, where you get
> a polynomial with
>
> x = q_1/2 and x=-3q_2 as the roots.
>
> That polynomial looks like
>
> 2x^2 + (6q_2 - q_1)x - 3q_1 q_2 = 0
>
> and let q_1 q_2 = 1, and 6q_2 - q_1 be an integer.
>
> It turns out that q_2 can't be an algebraic integer.
>

  Let's see, you're saying q_1/2 and -3*q_2 are the roots of

       2*x^2 + 5*x - 3.

  Of course the roots are known. They are 1/2 and -3.
You just said that above.

  Therefore q_1/2 = 1/2, or q_1 = 1.

  And -3*q_2 = -3, so q_2 = 1.

  Gee, those both look like algebraic integers to me.

  Let's check the other conditions.

        q_1*q_2 = 1 * 1 = 1.

  Yes, that looks OK.

        6*q_2 - q_1 = 6*1 - 1 = 5.

  Yes, that is an integer, as you required.

  What in the world were you thinking when you
said above

  "It turns out that q_2 can't be an algebraic integer."

  ???

  Nora B.

[snip]

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