You should be able to solve this.

True or false: Are algebraic numbers countable?
You can't use axiom of choice.

I'd say you can probably get round the axiom of choice problem because you can define a well ordering on the set of all integer valued polynomials.

I've not put a lot of thought into it, but a nice well-ordering normally solves the problem.


How about explicitly doing it then.


For ease of, well everything, I'm just going to take positive co-efficients, to include negative ones we'd have to inject the integers into the naturals (easily done without AC) to use the method I'm gonna do.


So we're going to assign each algebraic number 3 integers.

To each irreducible polynomial we assosciate it's degree.

Within each degree of polynomial we can inject into the natural numbers as there are only a countable number of each degree (A finite union is still countable without AC), so we assosciate to each polynomial a second number, it's order in this ordering.


Then to each solution of that polynomial we order them by modulus and then argument in the complex plane, and assign it a third number.


Each algebraic number is uniquely defined by these three numbers (Because we used irreducible polynomials), so let's do an injection:


r |-> (p_a^p_b)^p_c

Where a,b,c are the numbers assosciated with r, and p_i is the ith prime.


That's an injective function into the natural numbers, hence they're countable?


I can't see much choice here, I'm defining countable as "can inject into w (that's an omega)", possibly you might need choice to extend to a bijection, although from what I remember the proof of the Cantor–Bernstein–Schroeder theorem doesn't require AC, (Looked up the proof, it didn't).

Hope I win an internets