NOTE: I know that countability of algebraic numbers has been proven on this site before, but I'm concerned about this specific hint they give and I don't know how to prove it using that advice.

I am solving an exercise in Rudin to prove that the set of algebraic numbers is countable. In particular,

A complex number is said to be algebraic if there are integers $a_0, \dots, a_n$, not all $0$, such that $a_0 z^n + a_1 z^{n-1} + \dots + a_n = 0$. Prove that the set of all algebraic numbers is countable.

Hint: For every positive integer $N$ there are only finitely many equations with $n + |a_0| + |a_1| + \dots + |a_n| = N$

I have proven the statement using essentially the method in this post, but I'm curious as to how one can use the hint to prove the statement.


2 Answers 2


I believe that this hint helps you to establish that $\mathbb Z[X]$ is countable.

For a polynomial $f(X)=a_nX^n+\cdots +a_1X+a_0\in\mathbb Z[X]$, $a_n\neq 0$, define the number $h(f)$ by: $$h(f)= n+|a_n|+\cdots+|a_1|+|a_0|.$$ The function $h:\mathbb Z[X]\longrightarrow \mathbb N$ measures some kind of a complexity of a polynomial.

It is fairly obvious that the set $P_N=\{f(X)\in\mathbb Z[X]\mid h(f)= N\}$ is finite, and that $$\mathbb Z[X]= \bigcup_{N\geq 0}P_N.$$ Hence $\mathbb Z[X]$ is countable union of finite sets, hence countable. (Let me say, very nice trick!)

Now, the usual argument finishes the proof: the set of algebraic numbers is $$\bigcup_{f\in\mathbb Z[X]}\operatorname{roots}(f),$$ and it is countable union of finite sets, hence countable.

  • 1
    $\begingroup$ What exactly is $\mathbb{Z}[X]$? $\endgroup$ Mar 25, 2021 at 6:33
  • $\begingroup$ I think i understand now: $\mathbb{Z}[X]$ is the set of all finite complex polynomials with integer coefficients? $\endgroup$ Mar 25, 2021 at 19:56

Maybe we can understand the answer in this way:

  1. $a_i,i=0,...,n$ are all integers, the hint indicates that there are only finitely many combinations of $\{a_i\}_{i=0}^{n}$ for each given positive integer N;
  2. for each combination of $\{a_i\}_{i=0}^{n}$, the ${n^{th}}$ order equation implies that there are only finitely many solutions.

So for all of the integers, each corresponds to a finitely many options of $\{a_i\}_{i=0}^{n}$, each with a finitely many solutions.

Now we can arrive at the conclusion that the set of all solutions has to be countable.


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