Baby Rudin Chapter 2, Exercise 2 proof check

Question

I am currently self-studying Baby Rudin and I have written what I think is a solution to Exercise 2 of Chapter 2. It reads:

$\textbf{Exercise 2}:$ A complex number $z$ is said to be algebraic if there are integers $a_0,a_1,\dots,a_n$, not all zero, such that $$a_nz^n+a_{n-1}z^{n-1}+\dots+a_1z+a_0 = 0.$$ Prove that the set of all algebraic numbers is countable. Hint: For every positive integer $N$, there are finitely many equations satisfying $$n+|a_0|+|a_1|+\dots+|a_n| = N.$$

My proof does not use this hint, hence I am wondering if it is valid. It uses Theorem 2.12 from Rudin which states that a countable union of countable sets is countable. Here's the proof:

Proof. Consider the set (where $n \in \mathbb{N}$) $$E_n = \{p \in \mathbb{Z}[x] : \deg(p) = n\}.$$ Since $E_n$ is in bijection with $\mathbb{Z}^n$ (the map $(p = a_nx^n+\dots+a_0)\mapsto(a_n,\dots,a_0)$ is the obvious one), $E_n$ is countable. Now, consider $$S_n = \bigcup_{p \in E_n} \{z \in \mathbb{C} : p(z) = 0\}.$$ Each polynomial $p(x)$ in $E_n$ has at most $n$ roots in $\mathbb{C}$, so $\{z \in \mathbb{C} : p(z) = 0\}$ is finite. Since $E_n$ is countable, by Theorem 2.12, $S_n$ is countable. (It can be noted that, in abstract algebra language, $S_n$ contains each algebraic complex number whose minimal polynomial degree is $n$). Now, consider $$A = \bigcup_{n=1}^{\infty} S_n$$

By Theorem 2.12, $A$ is countable, and it is clear that every algebraic number is contained in this set. Hence, the algebraic numbers are countable. $\blacksquare$

Any help is appreciated. Thank you!

Answer 1

This might be an approach to consider in outline form, that utilizes the hint.

Hint: For every positive integer $N$ there are only finitely many equations with

$$ n+ \left|a_{0} \right| + \left|a_{1} \right| + \cdots + \left|a_{n}\right| = N. $$

Suppose $N=1.$ We can then consider the case $n=1.$
For $n=1$, one has

$$ 1+|a_{0}|+|a_{1}|=1 $$

or

$$ |a_{0}|+|a_{1}|=0 $$

Obviously, the only solutions are $a_{0}=a_{1}=0$ which are not acceptable.

Suppose $N=2$. We can consider $n=1$ and $n=2.$

For $n=1$:

$$ 1 + \left| a_{0} \right| + \left|a_{1} \right|=2 $$ or $$ \left| a_{0} \right| + \left|a_{1} \right|=1 $$

There are only $4$ possible corresponding equations of $z$ with:

$a_{0}=0, a_{1}=-1$, $a_{0}=0, a_{1}=1$,
$a_{0}=-1, a_{1}=0$, $a_{0}=1, a_{1}=0$

Each of these has $n=1$ solutions, so the possible number of algebraic numbers $z$ is $4$, a finite number.

For $n=2$, one has:

$$ 2 + \left| a_{0} \right| + \left|a_{1} \right|+ \left|a_{2} \right|=2 $$

which again leads to $a_{0}=a_{1}=a_{2}=0$, not allowed.

Continuing on like this, for each $N$ and $n$, one can see that the number of equations will be finite for each combination. Therefore, the number of solutions (with each equation having $n$ solutions) can be put into a correspondence with the natural numbers and thus implies the algebraic numbers (i.e. the solutions to these equations) are countable.