# Proving the countability of algebraic numbers

I am trying to prove that algebraic numbers are countably infinite, and I have a hint to use: after fixing the degree of the polynomial, consider summing the absolute values of its integer coefficients, and setting the sum less than or equal to $m$, for each $m \in \mathbb{N}$. I am also allowed to use the fact that every polynomial has a finite number of roots. I am not sure where to begin... Perhaps after fixing the degree of the polynomials, I could try enumerating them, and after proving countability of the set that contains the roots of all polynomials of a certain degree, I could state that the union of infinitely many ($\bigcup_{n=1}^{\infty}$, $n \in \mathbb{N}$) countably infinite sets is countable, and thus the algebraic numbers are countably infinite, too, but I am not sure if this is a good approach/how to enumerate the polynomials. Thanks!

• +1 for showing your thoughts about the problem - this is unfortunately quite rare. – Zev Chonoles Jul 10 '11 at 17:28
• An alternative approach to showing that polynomials with integer coefficients are countable: consider the bijection $\phi: \mathbb{Z}[x] \rightarrow \mathbb{N}$ that sends the polynomial $a_0 + a_1x + a_2x^2 + \ldots + a_nx^n$ to the natural number $2^{b_0}3^{b_1}5^{b_2}\cdots p_n^{b_n}$ (notation: $p_i$ is the $i$-th prime number and $b_i$ is the image of $a_i$ under any bijection from the integers to the natural numbers). – Elliott Jul 10 '11 at 18:05

## 3 Answers

The hint pretty much tells you what to do.

First, consider the polynomials of degree $1$ in which the sum of the absolute value of the coefficients is $1$; there's only finitely many of them, each with finitely many roots. (Think of this collection as corresponding to the pair $(1,1)$, the first $1$ giving the degree, the second $1$ giving the sum of the absolute value of the coefficients).

Then, consider the polynomials of degree $2$ whose sum of the absolute value of the coefficients is $1$; again, only finitely many, each with finitely many roots; make this set correspond to $(2,1)$.

Then, the polynomials of degree $1$ whose sum of the absolute value of the coefficients is $2$; only finitely many, each with finitely many roots. The set corresponds to the pair $(1,2)$.

Then consider the pair $(3,1)$; then the pair $(2,2)$; then the pair $(1,3)$. Then move on to the pair $(4,1)$, followed by $(3,2)$, followed by $(2,3)$, followed by $(1,4)$. Etc.

This will ensure that you cover each and every polynomial of positive degree with integer coefficients after finitely many steps; many algebraic numbers will occur more than once, but that's not a problem to showing they are at most countable (that they are infinite should be trivial).

• You wrote: First, consider the polynomials of degree 1 in which the sum of the absolute value of the coefficients is 1; there's only finitely many of them, each with finitely many roots. My question is why are there only finitely many of them? – Kara Nov 6 '14 at 19:45
• Because there are only so many ways to add up to 1, to 2, to 3, etc using positive numbers. I've got it now.... – Kara Nov 21 '14 at 19:21

That's exactly the right approach! The hint provided to you is about how might you go about enumerating the polynomials over $\mathbb{Z}$. Can you show there are finitely many polynomials over $\mathbb{Z}$ of degree $n$ and with sum-of-absolute-values-of-coefficients less than $m$, for any $m$? How many polynomials of degree $n$ are there over $\mathbb{Z}$, then? How many polynomials over $\mathbb{Z}$ are there, then? How many roots of polynomials over $\mathbb{Z}$ (i.e. algebraic numbers) are there, then?

HINT $\$ Enumerate the polynomials by "height" := max(degree, max |coefficient|)

• Maybe I am missing something but this problem seems really easy. For each positive integer $n$ , there are countably many polynomials of degree $n$ with integer coefficients. Each such polynomial has finitely many roots. There are countably many integers. Is that good enough? – Stefan Smith Mar 2 '13 at 1:45
• My comment above uses that fact that countable unions of countable sets are countable. I think this requires some form of the Axiom of Choice. Are the proofs above intended to avoid AC? – Stefan Smith Mar 3 '13 at 16:20