Can we make $\mathbb{R}^{2}$ an ordered field?

Question

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.

Answer 1

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 $f(a)>0$.