Can a field with cardinality $3$ be an ordered field?

Question

This is the question I'm confused about:

Consider a field $F = \{0,1,a\}$ with exactly 3 elements. Can it be an ordered field? Explain your answer in all details.

Well obviously, if $F$ is a field, then as $a$ is an element of $F$, $a + a$ also has to be in the field. So I thought that $2a = 1$, since if $2a = 0$ or $2a = a$, then $a = 0$, which is contradictory to the statement that the field has exactly 3 elements.

Then it's concluded that $a = 1/2$, but if that's true, $1 + 1/2 = 3/2$ also has to be in this field, but it's not. Therefore I concluded that it can't be an ordered field.

But something seems so wrong about my solution. I would appreciate any advice on what I was wrong about, and any correct ways to solve this problem.

Answer 1

In general, a finite field can never be an ordered field.

Let $a$ be some positive element -- that is, $0<a$. By repeatedly adding $a$ on both sides of this, we find $$ 0 < a < (a+a) < (a+a+a) < (a+a+a+a) < \ldots $$ but since the field is finite, there must be two of the expressions in this sequence that have the same value. If that value is $t$, then by transitivity we have $t<t$, which contradicts $t=t$.

Answer 2

Our $F$ is isomorphic with $\Bbb Z_3$, so we can regard $a$ as $2$. Observe that $2+2=1$ in $\Bbb Z_3$. Please check the axioms of the ordered field.

Of course we could directly work with our $a$ by knowing that $a+a=1$.

Your mistake was working with rational numbers with ordinary arithmetic operations.

Answer 3

Without loss of generality, we can say that the elements of the field are $\{0,1,2\}$, making it the field $\mathbb{Z}/3\mathbb{Z}$, the set of elements modulo $3$.

If it is an ordered set, then we have either $0 < 1$ or $1 < 0$. Suppose $0 < 1$. Since addition preserves inequality, we would that $2 < 0$, by adding $2$ to both sides (and because of $2 + 1 = 0$ in this field). Because of transitivty, we have $2 < 0 < 1$ and hence $2 < 1$. Adding $1$ to both sides, we find that $0 < 2$, but we already had $2 < 0$. This gives us a contradiction. The same trick works in the case that $1 < 0$.