The set $\{\alpha \in \mathbb{C} \mid \alpha $ is algebraic over $ \mathbb{Q}\}$ is countable 
The set $\{\alpha \in \mathbb{C} \mid \alpha $ is algebraic over $ \mathbb{Q}\}$ is countable.

Would you give me a direction towards solving this one? The definitions are clear and although the countability of $\mathbb{Q}[X]$ isn't very intuitive to me this is something the exercise gives me as a hint.
 A: Hint: If $\alpha$ is algebraic, then it is the root of some non-constant polynomial. Each polynomial has only finitely many roots, and how many possible polynomials are there?
Edit: It seems you've already been made aware of the countability of $\Bbb Q[X]\setminus \Bbb Q$. Short proof of this countability:

For each possible degree $n$, the set of polynomials of degree $n$ is a subset of $\Bbb Q^{n+1}$, and therefore countable. Now list all polynomials the following way: The first degree-one polynomial. Then the first degree-two polynomial and the second degree-one polynomial. Then the first degree-three polynomial, the second degree-two polynomial and the third degree-one polynomial, and so on. Each polynomial will eventually be included this way, so the set is countable.

Now for counting the algebraic numbers, do it like this: Number all polynomials in some way. Take the first polynomial according to this numbering, and list all its roots. Then take the second polynomial and list all its roots. Then the third polynomial and all its roots. And so on. For conciseness, you could at each step only include roots you haven't previously listed. This makes a bijection from $\Bbb N$ to the set of algebraic numbers.
A: Take a polynomial $\displaystyle p(x)= \frac{a_n}{b_n} x^n + \cdots + \frac{a_1}{b_1} x + \frac{a_0}{b_0}$ with $a_i, b_i \in \mathbb Z$.
Define the height of $p$ as $n+|a_n|+\cdots+|a_0|+|b_n|+\cdots+|b_0|$.
Define the height of an algebraic number as the height of its minimal polynomial.
Then there are only finitely many polynomials of a given height and so only finitely many algebraic numbers of a given height.
Therefore, the set of all algebraic numbers is a countable union of finite sets and so is countable.
