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Let $n$ a positive number, and let $A_n$ 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 $A_n$ is countable.

Hint: For each positive number $m$, consider polynomials

$\sum_{i=0}^n a_i x^i$ that satisfy

$\sum | a_i | \le m$.

I'm having difficulty grasping the concepts and method to write the proof. Can someone please explain in simple terms?

Thank you.

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You have a fairly low acceptance rate; it's been nine days since you asked this question and you've gotten two good answers. You may want to think about accepting one of them. –  Rick Decker Oct 11 '12 at 1:05
    
Sorry, don't know fully how this website works yet. –  Alti Oct 12 '12 at 1:40
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2 Answers

up vote 1 down vote accepted

Hint: Here is a hint along the same lines as in the post, but with a slight twist that makes things easier.

For any polynomial $P(x)$ with integer coefficients, let $C(P)$, the complexity of $P$, be the sum of the degree of $P$ and the sum of the absolute values of the coefficients of $P$. Instead of sum of the absolute values of the coefficients, one can just use the maximum absolute value of the coefficients, but we definitely want the degree as a component of the complexity.

For any $k$, there are only finitely many polynomials of complexity $k$, and these produce only finitely many algebraic numbers.

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Thank you, but why is it countable and not finite? That's what I'm confused about. –  Alti Oct 1 '12 at 4:46
    
@Alti: Technically, finite sets are countable. But the number of algebraic numbers is actually countably infinite. For note that any natural number $n$ is algebraic, since it is a root of the polynomial equation $x-n=0$. So there certainly are infinitely many algebraic numbers. –  André Nicolas Oct 1 '12 at 4:50
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There are countably many polynomials with rational coefficients and each of them has finite number of roots, so there has to countably many roots of such polynomials.

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