A finite field cannot be an ordered field.

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.

-
Do you know that in any ordered field the only possibility is $$\sum_{i=1}^n a_i^2=0\iff a_i=0\;\;?$$ –  DonAntonio Dec 23 '13 at 20:21

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.

-
Clear, short, accurate. +1 –  DonAntonio Dec 23 '13 at 20:44
@Bill: It seems to me that your answer might be useful to someone tackling this from an algebraic point of view; not from the point of view of Baby Rudin. –  Asaf Karagila Dec 23 '13 at 21:03
@Bill: Yes, I agree. But "Linearly ordered groups are torsion free" can be outright confusing to someone unfamiliar with these terms. –  Asaf Karagila Dec 23 '13 at 21:17
@Asaf Agreed, if one knows no group theory. Probably I should have said "positives are closed under addition" and later segued into the more general group-theoretical view, to avoid scaring away readers. –  Bill Dubuque Dec 23 '13 at 21:24
I found both answers useful, Asaf was more straight to the point and Bill was more profound, The only reason I accepted Asaf and not Bill's is it took me more time to understand what Bill said. –  dREaM Dec 23 '13 at 23:11

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.

-