TITLE: Is $\mathbb{R}$ an algebraic extension of some proper subfield?
QUESTION [7 upvotes]: Is there a (proper) subfield $K$ of $\mathbb{R}$ such that $\mathbb{R}$ is an algebraic extension of $K$?
From this question, Is there a proper subfield $K\subset \mathbb R$ such that $[\mathbb R:K]$ is finite?, it is clear that for such an $K$, $[\mathbb{R}:K]=\infty$.
This question seems to be non-trivial, and I suspect any existence result will be non-constructive. Any references will be appreciated.

REPLY [8 votes]: Just to have at least one actual answer to the question, I'll convert the comment by PVAL to an answer.
Using Zorn's lemma there exists a transcendence basis $B$ of $\Bbb R$ over $\Bbb Q$: a maximal algebraically independent (over $\Bbb Q$) set of real numbers. Then $\Bbb Q(B)$ is a subfield of $\Bbb R$, and by maximality every element of $\Bbb R$ is algebraic over $\Bbb Q(B)$. So $\Bbb R/\Bbb Q(B)$ is an algebraic extension.