Can we make $\mathbb{R}^{2}$ an ordered field? We know that if we give $\mathbb{R}^{2}$ the complex field structure, we cannot make it an ordered field. Is there any field structure that we can put on $\mathbb{R}^{2}$ that makes this ordered field? I don't think there is, but I don't know how to start my argument.
 A: Whenever we say the term "ordered field", this means the order on the field has something to do with the field structure. Otherwise as a set you can put an order anyway.
In $\Bbb{R}^2$ we have the well-known dictionary ordering which becomes total ordering. But what I was talking about is, the total order has to be compatible with the field operations in order to be an ordered field. A total order "$<$" on a field $F$ is said to be compatible with $F$ if the following holds true for all $a,b,c\in F$.


*

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

*$a\leq b$ and $c>0$ implies $ac\leq bc$.


Next we see how the fact $i^2=-1$ in $\Bbb{C}$ ensures that there is no total order on $\Bbb{C}$ which is compatible with the field operations of $\Bbb{C}$.

Let $<$ be any arbitrary total ordering on $\Bbb{C}$. Then $i\neq 0$ gives, either $i<0$ or $i>0$. But we will show none of them holds true.


If $i>0$ then from the condition
$(2)$ we get, $i\cdot i>i
\cdot0\implies -1>0$. Now some may think that, we have arrived at a contradiction but unfortunately no. Since $<$ is an arbitrary ordering so this may happen. But apply condition $(2)$ again and we get, $(-1)\cdot i>0\cdot i\implies -i>0$. Now using condition $(1)$, $i>0$ and $-i>0\implies i+(-i)>0+0\implies 0>0$. Which is a contradiction.


Similarly if we put $i<0$ then from condition $(2)$ we get, $i\cdot i>0\cdot i\implies -1>0$. Then again apply condition $(2)$ on $i<0$ to get, $i\cdot (-1)<0\cdot (-1)\implies -i<0$. Again using condition $(1)$, $i>0$ and $-i>0$ $\implies i+(-i)>0+0\implies 0>0$. Which is a contradiction.

Hence $i$ and $0$ are not comparable, so there is no total order on $\Bbb{C}$ which makes it an ordered field.
A: As mentioned in one of the comments, if you want to keep the vector space structure of $\mathbb{R}^2$, then the answer is no. The reason is that $\mathbb{R}^2$ then necessarily would have to be an algebraic field extension of $\mathbb{R}$ of degree $2$, so it would be of the form $\mathbb{R}(j) =\{a+bj:a,b\in\mathbb{R}\}$, where $j$ would be the root of a quadratic polynomial without real roots. This automatically makes $\mathbb{R}(j)$ isomorphic to $\mathbb{C}$ (which is the only nontrivial algebraic extension of $\mathbb{R}$), which cannot be ordered because $i^2 = -1 < 0$.
Note that the transcendental extension of $\mathbb{R}$ of degree $1$, i.e., the rational functions with real coefficients, can be ordered.
Also, if you don't impose any conditions at all, $\mathbb{R}^2$ can be trivially (and very uselessly) ordered by finding a bijection $f:\mathbb{R}^2 \to \mathbb{R}$ and defining operations on $\mathbb{R}^2$ by $a+b = f^{-1}(f(a)+f(b))$ and $ab = f^{-1}(f(a)f(b))$, as well as defining $a$ to be positive iff $f(a)>0$.
