Clarification on Baby Rudin Chapter 2 Exercise 2 I have a basic question about Baby Rudin Chapter 2 Exercise 2. There are a number of solutions online and on StackExchange, but I'm still left with some questions.

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

Some solutions note that (i) polynomials of degree $n$ have at most $n$ different solutions, and (ii) since there are countably many $n$-th degree polynomials with integer coefficients, the set of algebraic numbers is a union of countable sets and hence it is countable.

*

*Is Rudin's hint meant to help us prove (i) or (ii)?


*Is the following proof for (ii) correct?

Fix $z$. Map the polynomial $a_0z^n + a_1z^{n-1} + a_n = 0$ to the list $(n, a_0, a_1, \ldots, a_n)$. The set of lists $\{(n, a_0, \ldots, a_n): n\in \mathbb{N}, a_0 \in \mathbb{Z}, \ldots a_n \in \mathbb{Z}\}$ is countable, since the Cartesian product of a finite $n$-tuple of integers $(a_0, \ldots, a_n)$ is countable and the Cartesian product of the two countable sets $\mathbb{N}$ and $\mathbb{Z}^n$ is countable. Hence, for each $z$, the set of polynomials with integer coefficients is at-most countable.


*

*I can find proofs for (i) online, but would anyone mind showing me how Rudin's hint can be used to prove (i)? Or (ii), if that's what the hint is for?

 A: The idea of proving that, for each $N\in\Bbb Z_+$, there are only finitely many polynomials $a_0z^n+a_1z^{n-1}+\cdots+a_{n-1}z+a_n$, with $n=|a_0|+|a_1|+\cdots+|a_n|=N$ is that then the set of all polynomials with integer coefficients is countable, since it is an union of finite sites. Since each such polynomial, other than the null polynomial, has only finitely many roots, it follows that there are only countably many algebraic numbers.
A: There are a few answers out there for this question, but as a beginner, I had a hard time understanding many of them. Here's my approach, which may be overly simple/wordy but (if correct) could help beginners searching for a more "obvious" answer.
Recall that an algebraic number $z \in Z$ satisfies
$$a_0 z^n + a_1 z^{n-1} + \ldots + a_n = 0$$
where $a_0, a_1, \ldots, a_n$ are all integers.
The set $Z$ is infinite: when $n = 1$, algebraic numbers are those satisfying $a_1 z + a_0 = 0$ or $z = -a_1/a_0$. So rational numbers are a subset of algebraic numbers, and since rational numbers are infinite, so are algebraic numbers.
$Z$ is at most countable We use Rudin's hint that, for any natural number $N$, there are only finitely many equations with
$$N = n + |a_0| + |a_1| + \ldots + |a_n|$$
For each $N$, we have a set of coefficients satisfying the equation above. We can associate each of these with an expression. In the table below, I've considered the case where $N = 5$ and listed just some of the coefficients satisfying the equation above, associated expressions, and the values $z$ where the expressions equal zero.




$n$
$a_0$
$a_1$
$a_2$
$a_3$
$a_4$
Expression
Zeros




4
1
0
0
0
0
$z^4$
$z=0$


3
1
0
0
1
0
$z^3 + 1$
$z = -1$


2
1
-2
0
0
0
$z^2 - 2z$
$z =2$ or $z = 0$





*

*Let $A_N$ be the set of expressions associated with $N$ (e.g., $A_5$ includes the expressions $z^4$ and $z^3 + 1$ among others). Rudin tells us this set is finite. (See edit at end).


*The number of roots associated with each expression in $A_N$ is also finite, so the roots represented by the expressions in each $A_N$ are also finite. I'll call the set of roots $B_N$.
Most answers say that Rudin assumes we know that polynomials of degree $n$ have, at most, $n$ roots. But, I think we just need to be convinced that each polynomial has a finite number of roots.

*

*Since $N$ is countable and each $B_N$ is finite, the union $\bigcup_{N=1}^\infty B_N$ is countable. The set of algebraic numbers is only a subset of this union, since some roots are repeated (e.g., $z = 0$ in the first and third rows of the table). So, the algebraic numbers are at most countable.

Since the algebraic numbers are infinite and at-most countable, they are countable.
Edit: As @fleablood commented to my initial question, Rudin's hint prevents the reader from making incorrect argument that the set of polynomials is countable, since we can enumerate the coefficients (in the countable set $\mathbb{Z}$) and their degrees (in the countable set $\mathbb{N}$).
