Consider $\mathbb{Q}$ and then consider "adding" a transcendental $\zeta$ to it, while still retaining the field axioms (i.e. $\mathbb{Q}(\zeta)$).
We could add another transcendental in an obvious extension to make yet another subfield of $\mathbb{R}$ – my question is this:
Does adding an uncountable number of transcendentals to $\mathbb{Q}$ necessarily make it into $\mathbb{R}$? Or could we add some uncountable subset of the transcendentals, such that the resultant field is not $\mathbb{R}$?
Best Answer
You can make (a field isomorphic to) $\mathbb C$ from $\mathbb Q[i]$ by adjoining $2^{\aleph_0}$ independent transcendentals and taking the algebraic closure. Easy-peasy.
$\mathbb R$ from $\mathbb Q$ is not so simple a matter.
If you just take $\mathbb Q(\zeta)$ you don't get any $\sqrt[3]\zeta$, even though all reals have cube roots. You will need to do something additional to make it exist. Adjoining more independent transcendentals will not do, because if they can create $\sqrt[3]\zeta$, they were not algebraically independent from $\zeta$ in the first place.
Be sure not to do anything as crude as requesting an algebraic closure while doing this, because you don't want to get any $\sqrt{-1}$.
Because positive reals can be characterized algebraically as "all the squares of nonzero reals", simply knowing the field operations of $\mathbb R$ is enough to reconstruct the ordering of $\mathbb R$. This means that if you succeed in constructing $\mathbb R$ by somehow "adding" transcendentals, the resulting structure will determine exactly how each of your transcendentals compares to each rational -- that is, you will end up having determined which real it is. And that means that you need to make a lot of decisions along the way that reach farther than "here is another transcendental, same as all the other ones".
Conversely: By Zorn's lemma, $\mathbb R$ has a maximal algebraically independent subset -- a transcendence basis. Every real number is root in a polynomial whose coefficients are finite rational polynomials of the basis elements, in an "essentially unique" way.
If you take any proper subset of the transcendence basis -- which may well be $2^{\aleph_0}$-sized -- and take all real roots of polynomials with coefficients that are rational polynomials of the selected elements, then you get a field that probably matches your intuitive idea of "adding" $2^{\aleph_0}$ transcendentals. It will not be all of $\mathbb R$, and not even isomorphic it $\mathbb R$. Its field structure alone will encode exactly which of the reals are missing.
On the other hand, you can also (in one sense) "go too far" and add too many transcendentals, getting a real closed field that properly extends $\mathbb R$. (For example, adjoin an infinitesimal element using a model-theoretic compactness argument, possibly followed by repeating the above construction to select exactly how many supernumerary independent transcendentals you want). Or even one that is neither a subfield nor an extension of $\mathbb R$.