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.
Can we make $\mathbb{R}^{2}$ an ordered field?
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4Well, it has the right cardinality that we an give it the same structure as $\mathbb{R}$. But if we require that the structure as a real vector space is kept, then no, there is no way to do so. – 2012-10-19
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2You're imposing no restraints on the field structure? Addition and multiplication can both be completely arbitrary? – 2012-10-19
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2What field operation do you envisage? You can define a bijection with $\mathbb R$ which gives an ordered field structure. – 2012-10-19
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0I don't assume anything, I just want to know whether we can make it as an ordered field. Field as in $(\mathbb{R}, +)$ is an abelian group and $(\mathbb{R}^{*}, \cdot)$ is an abelian group. With simple order $<$ with [$a < b$ implies $a + c < b + c$] and [$0 < a, b$ implies $a < b$] for all $a, b, c \in \mathbb{R}^{2}$. If you have suggestion, let me know or any reference will be appreciated as well. – 2012-10-19
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0@ChrisEagle Yes, you can define any operators as you want. – 2012-10-19
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0Do you want the underlying abelian group structure of $\mathbb{R}^2$ to be preserved? – 2012-10-19
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0@Conrad I haven't thought about it, but that's a good question! – 2012-10-19
1 Answers
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 $f(a)>0$.
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0Thank you so much for your comprehensive answer! – 2012-10-19