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Algebraic Countability

  1. Feb 6, 2008 #1
    I posted this in the Homework/Coursework section, but I really don't consider it that at all because I'm working through this text on my own, and I'm a little stuck on this problem.

    Fix [tex]n \in[/tex] N, and let [tex]A_n[/tex] be the algebraic numbers obtained as roots of polynomials with integer coefficients that have degree n. Using the fact that every polynomial has a finite number of roots, show that [tex]A_n[/tex] is countable. (For each [tex]m \in[/tex] N, consider the polynomials [tex]a_nx^n + a_n_-_1x^n^-^1 + ... + a_1x + a_0[/tex] that satisfy [tex]\left|a_n\right| + \left|a_n_-_1\right| + ... + \left|a_1\right| + \left|a_0\right| \leq m[/tex].)

    By the way, this only deals with real roots. Complex roots are simply negligible.

    So, I know a few things, but bringing the big picture together is really messing me up here. For example, I know that the sum of the absolute value of the coefficients for quadratic equations only has a certain number of solutions. So, whatever I elect m to be, there will always be a finite number of solutions. Also, the number of quadratics with coefficients is less than or equal to m: this is also finite. When we multiply this fact times the number of roots, we have the number of roots of a quadratic whose absolute value sums to some value less than or equal to m.

    The big problem I have is trying to generalize this statement for all [tex]A_n[/tex]. If anyone has any suggestions, this would be most helpful!
     
  2. jcsd
  3. Feb 6, 2008 #2

    mathman

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    Your description for the quadratic essentially applies for any fixed m, for the specified n. There are only n+1 integer coeficients, so there are only a finite number of possibilites for the sum of abs. values bounded by m. Each polynomial has at most n roots - you can include the complex. Therefore for fixed m the total number of roots for all equations is finite.

    You now have a countable sum of finite numbers when you let m -> inf.
     
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