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.
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Is Rudin's hint meant to help us prove (i) or (ii)?
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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?
Best Answer
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 a countable union of finite sets. Since each such polynomial, other than the null polynomial, has only finitely many roots, it follows that there are only countably many algebraic numbers.