# Is the set of zero sets of rational coefficient polynomials countable?

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

• Well, we know each polynomial has at most $d$ roots where $d$ is the degree. That's good enough. – lulu May 31 at 14:31
• @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. – hampster May 31 at 14:34
• 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. – lulu May 31 at 14:39
• This is false. Every real number is a zero of the zero polynomial. – Shalop May 31 at 22:29
• @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. – 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 $$\deg(q)=\deg(p)-1$$.