Is A_n Countable for Each Fixed n in Algebraic Number Theory?

[rbzima]

Homework Statement

Fix $$n \in$$ N, 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. (For each $$m \in$$ N, consider the polynomials $$a_nx^n + a_n_-_1x^n^-^1 + ... + a_1x + a_0$$ that satisfy $$\left|a_n\right| + \left|a_n_-_1\right| + ... + \left|a_1\right| + \left|a_0\right| \leq m$$.)

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

Homework Equations

The Attempt at a Solution

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 $$A_n$$. If anyone has any suggestions, this would be most helpful!

 

[Mathdope]

Show...? You forgot to put in the statement.

 

[rbzima]

Show...? You forgot to put in the statement.

My bad, I just fixed it. I want to show it's countable!

 

[Mystic998]

This is from Rudin, right? I personally found the hint rather unhelpful.

I remember that the way I did this was to fix n and show that the roots of all nth degree polynomials with integer coefficients forms a countable set. Then to get the roots of all polynomials of finite degree with integer coefficients, you just take a countable union of those sets.