Can an ordered field be finite? I came across this question in a calculus book. 

Is it possible to prove that an ordered field must be infinite? Also - does this mean that there is only one such field? 

Thanks
 A: An ordered field must be infinite.  Notice that each field has a subset of numbers that behave like the natural numbers, with $0<1<1+1<1+1+1\dots$
However, not every ordered field is isomorphic to all other ordered fields.  Notice that both the rational numbers and real numbers are ordered fields.
A: Hint $\ $ 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).
A: Recall that in an ordered field we have:


*

*$0<1$;

*$a<b\implies a+c<b+c$.


Suppose that $F$ is an ordered field of characteristic $p$, then we have in $F$ that $$\underbrace{1+\ldots+1}_{p\text{ times}} = 0$$
Therefore: $$0<1<1+1<\ldots<\underbrace{1+\ldots+1}_{p\text{ times}} = 0$$
Contradiction! Therefore the characteristic of $F$ is $0$ and therefore it is infinite, since it contains a copy of $\mathbb Q$.

Few fun facts on the characteristic of a field:
Definition: The characteristic of a field $F$ is the least number $n$ such that $\underbrace{1+\ldots+1}_{n\text{ times}}=0$ if it exists, and $0$ otherwise.
Exercises:


*

*If a field has a positive characteristic $n$ then $n$ is a prime number.

*If $F$ is a finite field then its characteristic is non-zero (Hint: the function $x\mapsto x+1$ is injective, start with $0$ and iterate it $|F|$ many times and you necessarily got $0$ again.)

*If $F$ is finite and $p$ is its characteristic then $p$ divides $|F|$.

