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.