I am reading baby Rudin and it says all ordered fields with supremum property are isomorphic to $\mathbb R$. Since all ordered finite fields would have supremum property that must mean none exist. Could someone please show me a proof of this?
Thank you very much, Regards.
HINT: Suppose that $(F,0,1,+,\cdot,<)$ is an ordered field which is finite of characteristic $p$. Then $0<1<1+1<\ldots$, conclude a contradiction.
Hint $\ $ In an ordered ring, positives are closed under addition (so a sum of positives is $\ne 0$).
Remark $\ $ More generally, note that linearly ordered groups are torsion free: $\rm\: 0\ne n\in \mathbb N,$ $\rm\:g>0 \:\Rightarrow\: n\cdot g = g +\cdots + g > 0,\:$ since positives are closed under addition. Conversely, a torsion-free commutative group can be linearly ordered (Levi, $1942$).
Hint: any finite field must have a non-zero characteristic.