Let $B$ be a $\Bbb Q(\sqrt 2)$-basis of $\Bbb R$ and let $E$ be a $\Bbb Q(e)$-basis of $\Bbb R$.
What are the degrees $[\Bbb R : \Bbb Q(B)]$ and $[\Bbb R : \Bbb Q(E)]$ ? Are they countable, uncountable?
According to this question (which motivated mine), these degrees can't be finite. We know that $[\Bbb R : \Bbb Q] = [\Bbb R : \Bbb Q(B)][\Bbb Q(B) : \Bbb Q] = 2^{\aleph_0}$. If we have $[\Bbb Q(B) : \Bbb Q] = \aleph_0$, then $[\Bbb R : \Bbb Q(B)]= 2^{\aleph_0}$. But I believe that $[\Bbb Q(B) : \Bbb Q] = 2^{\aleph_0}$, which doesn't tell us anything about $[\Bbb R : \Bbb Q(B)]$. I haven't seen any question asking whether $\Bbb R$ could an extension of some subfield, with degree $\aleph_0$ (I only found that germane question... which doesn't quite help because of this one).
I think that these degrees are independent from the choice of the basis. However the fact that any two $\Bbb Q(\sqrt 2)$-basis of $\Bbb R$ have the same cardinality is not sufficient to prove it.
Anyway, I tried to think whether I could prove that $\sqrt 2 \in \Bbb Q(B)$ which implies $\Bbb Q(B) = \Bbb R$ (because a basis of $\Bbb R$ over $\Bbb Q$ is the products of elements from $B$ and from $\{1,\sqrt 2\}$).
Thank you for your comments!
