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.
Best Answer
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.