I was thinking about the field of computable numbers, and it's intuitively pretty obvious that they'll be isomorphic to the characteristic $0$ algebraically closed field, but I realised I don't know a good proof of this.
The sticking point is that I don't know how to find countably many computable numbers which are algebraically independent of each other. With the complex numbers, you can use a cardinality argument to find countably many algebraically independent numbers, but that doesn't work with the computable numbers.
My instinctive guess was that taking countably many numbers $a_0, a_1, \dots$ which are linearly independent over $\mathbb{Q}$ would give $e^{a_0},e^{a_1}, \dots$ which are algebraically independent. With a bit of searching, I found this is indeed true, and is in fact the Weierstrass Lindemann Theorem, with the restriction that the $a_i$ must be algebraic. So this provides a proof. But it also seems to be quite a heavy piece of machinery, and I don't at all understand it.
As such, it doesn't fully answer my curiosity, as I am interested in how to prove the isomorphism of fields, not the fact that it is true. And it seems like an obvious fact should have a much simpler proof than this. So I would like to know if there is an simpler way to find countably many algebraically independent computable numbers, which is comprehensible to someone with less background.
So is there a simpler way of showing that the computable number field is isomorphic to the algebraically closed field of countable transcendence degree?