Countability of Algebraic Numbers as Roots of Polynomials

[DPMachine]

Homework Statement

A real number $$x \in R$$ is called algebraic if there exist integers $$a_{0},a_{1},a_{2}...,a_{n}$$, not all zero, such that

$$a_{n}x^{n} + a_{n}_{1}x^{n-1} + ... + a_{1}x + a_{0} = 0$$

Said another way, a real number is algebraic if it is the root of a polynomial with integer coefficients...

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 $$m \in M$$, consider polynomials $$a_{n}x^{n} + a_{n}_{1}x^{n-1} + ... + a_{1}x + a_{0}$$ that satisfy $$\left|a_{n}\right| + \left|a_{n-1}\right| + ... + \left|a_{1}\right| + \left|a_{0}\right| \leq m$$.)

Homework Equations

The Attempt at a Solution

I'm not sure how to explain this coherently... here is what I have. I feel like there are some holes.

-Every polynomial has a finite # of roots (therefore it is countable)

-$$m \in N$$ is the sum of all integer coefficients for the roots of polynomials.

-Let $$C_{m}$$ be a set containing all possible polynomials whose integer coefficients add up to $$m$$ for a fixed $$n$$. Since there are finite ways to express $$m$$ as a sum of integers, each $$C_{m}$$ is countable.

-Every $$A_{n}$$ is made up of $$C_{m}$$, so $$A_{n}$$ is countable (since union of a countable # of countable sets is countable).

 

[Dick]

I don't see anything wrong with that. As you said, a union of countable sets is countable. That's the point, right?

 

[DPMachine]

I don't see anything wrong with that. As you said, a union of countable sets is countable. That's the point, right?

Yeah, what I came up with at the end is probably correct... I guess I'm more concerned about whether each step makes sense or if there's anything I should clarify more.