TITLE: $\mathbb{R}$ is not isomorphic to a proper subfield of itself
QUESTION [8 upvotes]: let $\mathbb{R}$ be the field of real numbers. I found stated in this pretty work On Groups that Are Isomorphic to a Proper Subgroup, that there is no proper subfield $K$ of $\mathbb{R}$ which is isomorphic to $\mathbb{R}$ itself.
Does someone have a proof of this fact?
Thank you very much for your help in advance.
NOTE1. Contrast this situation with the case of the field $\mathbb{C}$ of complex numbers, for which there exist proper subfields isomorphic to $\mathbb{C}$ itself: see e.g. Automorphisms of the Complex Numbers, Concluding Remark 2.
NOTE2. This issue arouse in my post Proper Subgroup of O_2(R) Isomorphic to O_2(R) about whether the orthogonal group $O_2(\mathbb{R})$ is co-Hopfian or not.

REPLY [3 votes]: Hint: This has to do with the fact that the ordering in $\mathbb R$ can be expressed purely as a logical statement expressed in the language of rings/fields: $x\le y\equiv (\exists z)y=x+z^2$. Now, let $\phi:\mathbb R\to\mathbb R$ be a field monomorphism. Using the previous, you can prove that $\phi$ preserves ordering (i.e. is monotonically increasing). As you can (trivially) prove that $\phi$ maps rational numbers to themselves, i.e.$\phi(q)=q, q\in\mathbb Q$, then (due to preserving the ordering, and the fact that irrational numbers can be seen as Dedekind cuts in $\mathbb Q$), you can conclude that $\phi$ also maps every real number to itself