Is there a classification of ordered fields? I was reading an analysis textbook, where the author defines $\mathbb{R}$ as an ordered field with axiom of completeness, and I'm wondering why this definition is okay. (I mean why there is only one such field). I also wondered if there is a classification for ordered fields, but that's a slightly different question (I am interested in both questions )
For consistency of axioms I can prove that every ordered field must have zero characteristic, so it must contain the $\mathbb{Q}$ as a subfield. And since we have the axiom of completeness, each such field must contain $\mathbb{R}$ as a subfield, but why it can't be bigger than $\mathbb{R}$? 
For classification I have no idea. I found that there are even some non-Archimedean fields as an example for ordered fields (so we have something else than just $\mathbb{R}$ and all it subfields as an example for ordered fields), but I didn't find any related work on classification even on arxiv.
 A: Suppose $\Bbb R$ is a subfield of the ordered field $F,$ with the same order $<$ on both, and suppose $x\in F\setminus \Bbb R.$ Then $0<|x|\not\in \Bbb R.$
Let $r$ be the $lub$ in $\Bbb R$ of $S=\Bbb R\cap (-|x|,|x|).$ Then $r\ne |x|.$ Now
Case $(i).$ If $r>|x|$ then $\Bbb R\cap (|x|,r)=\emptyset$ because if $r'\in \Bbb R\cap (|x|,r)$ then  $r'$ would be an upper bound in $\Bbb R$ for $S$ with $r'<r$, contrary to the def'n of $r.$
Case $(ii).$ If $r<|x|$ then $ \Bbb R\cap (r,|x|)=\emptyset$ because if $r'\in \Bbb R\cap (r,|x|)$ then $r< r'\in S,$ contrary to the def'n of $r.$
In either Case, let $y=|\,(|x|-r)\,|.$ Then $y>0$. And $\Bbb R\cap (0,y)=\emptyset.$ Because if $r''\in \Bbb R\cap (0,y)$ then in Case $(i)$ we would have $r'=r-r''\in \Bbb R\cap (|x|,r),$ or in Case $(ii)$ we would have $r'=r+r''\in \Bbb R\cap (r,|x|).$
We can now show that $\Bbb R^+$ has NO $glb$ in $F.$  Suppose by contradiction that $z=glb_F(\Bbb R^+).$ Then $z>0$ because  $y$ is a positive lower bound in $F$ for $\Bbb R^+,$ so $z\ge y>0.$ But then $z<2z$ and $$\forall t\in \Bbb R^+\,(z\le t/2)\implies \forall t\in \Bbb R^+\,(2z\le t)$$ so $2z$ is a larger lower bound for $\Bbb R^+$ than $z=glb_F(\Bbb R^+),$ a contradiction.
So $F$ is not order-complete.
