I am trying to prove that the set of algebraic numbers is countable. I have managed to prove that the set of rational coefficient polynomials is countable. My idea is to prove that the set of zero-sets of rational coefficient polynomials is countable and argue the algebraic numbers are a subset of that set. Here is my reasoning:

If we assume each polynomial of positive degree k has at most $k \in \mathbb{Z}_+$ roots (we assume this without proof), then for each $p \in \mathbb{Q}[X]$ the set $P_0 = \{ x \in \mathbb{Q}\, |\, p(x) = 0 \}$ is countable (since the set is finite). If we let $\mathcal{P}$ be the collection of zero-sets, then the set of all zero-sets is $\bigcup_{P_0 \in \mathcal{P}} P_0$. This is a countable union, since each zero set corresponds to a unique rational coefficient polynomial (the set of which is proven to be a countable set). Thus the set is a countable union of countable sets i.e countable. Is my argument correct?

NOTE: This is not for homework. This is for self studying. EDIT 2: polynomial of positive degree k

  • $\begingroup$ Well, we know each polynomial has at most $d$ roots where $d$ is the degree. That's good enough. $\endgroup$ – lulu May 31 at 14:31
  • $\begingroup$ @lulu But don't we need the fact that the union is specifically a countable union? At least the book I'm studying (Munkres Topology) states that a countable union of countable sets is countable. $\endgroup$ – hampster May 31 at 14:34
  • $\begingroup$ Sure. There are countably many polynomials of degree $d$. Each of them has at most $d$ roots. Thus there are countably many algebraic numbers of degree $d$. But every algebraic number has a degree. $\endgroup$ – lulu May 31 at 14:39
  • $\begingroup$ This is false. Every real number is a zero of the zero polynomial. $\endgroup$ – Shalop May 31 at 22:29
  • $\begingroup$ @Shalop I've added the clarification, though it was implicit in the text that we consider positive degree polynomials, since the motivation for the proof regards algebraic numbers. $\endgroup$ – hampster Jun 1 at 5:15

The proof is correct. Actually the number of roots of a polynomial in $\Bbb Q[X]$ is not greater than its degree. Note that if $\alpha$ is a root of $p(X)$ then $p(X)=q(X)(x-\alpha)$ for some polynomial $q$ and $\deg(q)=\deg(p)-1$.

  • $\begingroup$ Oops, didn't realise I missed the "less than degree" part. Thanks very much! $\endgroup$ – hampster May 31 at 14:36

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