Can an ordered field be finite?

Recall that in an ordered field we have:

1. $0<1$;
2. $a<b\implies a+c<b+c$.

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 $\mathbb Q$.

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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:

1. If a field has a positive characteristic $n$ then $n$ is a prime number.
2. 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.)
3. If $F$ is finite and $p$ is its characteristic then $p$ divides $|F|$.

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.

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).