TITLE: A finite field cannot be an ordered field.
QUESTION [17 upvotes]: 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.

REPLY [35 votes]: 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.