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I read a question about ordering of complex numbers, and saw an answer showing that there cannot exist an ordering of the complex numbers because regardless of how $i$ is placed in that order, it would imply that $i^2 = -1$ would be positive.

This proof can obviously be generalized to prove that for any non-zero $a \in Q$ it is impossible for $ai$ to be an element of an ordered field. But there is no obvious generalization of the proof showing that the same hold for all imaginary numbers.

Does there exist a sub-field of the complex numbers containing at least one imaginary number, which can be assigned a consistent ordering?

Or stated in different terms, does there exist a subset of the complex numbers, which is a formally real field and is not a subset of the reals?

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kasperd
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    The reals are a sub-field of the complex numbers. By the way, the question in the end and the title are not the same. – Ofir Schnabel Jan 23 '15 at 11:59
  • If such a $P$ exists, would $i$ or $-i$ be in it? – Autolatry Jan 23 '15 at 12:13
  • @OfirSchnabel I updated the question to explicitly state that it must contain at least one imaginary number. – kasperd Jan 23 '15 at 12:45
  • @Autolatry If such a field exists it cannot contain $i$ or $-i$ as proven by the linked answer. – kasperd Jan 23 '15 at 12:47
  • So if it contain imaginery it must contain also reals and (just multiply the imaginary by itself), then it must contain at least $\mathbb{Q}$ so you don't have an order there. – Ofir Schnabel Jan 23 '15 at 12:50
  • I have one idea as to how a solution could possibly constructed. Given a transcendental real number $x$, there exist a smallest field containing $x$. This is an ordered field as it is a subset of the reals. If instead of a real $x$ a transcendental imaginary number $z$ was chosen, there would also exist a smallest field containing that $z$. That field would not have an obvious ordering, but they might exist a homomorphic mapping between those two fields, which could be used to assign an ordering. – kasperd Jan 23 '15 at 12:55
  • @OfirSchnabel Any ordered field must contain Q. That is true regardless of whether you try to include any complex numbers or not. If you take the smallest field containing one chosen complex number, you will get a countably infinite set containing Q as a subset, it is however not a given, that the set will contain any real numbers outside of Q. – kasperd Jan 23 '15 at 13:02
  • But there is no order on $\mathbb{Q}[i]$ also so such ordered field do not exist. – Ofir Schnabel Jan 23 '15 at 13:06
  • @OfirSchnabel I already mentioned twice, that the solution cannot contain i. However the proof doesn't apply to for example Q[e+i], so there might exist an ordering on Q[e+i]. – kasperd Jan 23 '15 at 13:09
  • See formally real field. – Bill Dubuque Jan 23 '15 at 16:31
  • @BillDubuque Yes, that definition sounds like what I am asking about. – kasperd Jan 23 '15 at 22:36
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    Consider an irreducible polynomial $f \in \mathbb{Q}[X]$ with real and non-real zeros. If $\alpha,\beta$ are zeros of $f$, then $\mathbb{Q}(\alpha) \cong \mathbb{Q}(\beta)$. If $\alpha \in \mathbb{R}$ then $\mathbb{Q}(\alpha)$ is an ordered field (a subfield of $\mathbb{R}$), hence $\mathbb{Q}(\beta)$ can be made an ordered field. Take for example $f(X) = X^4 - 2$ and $\mathbb{Q}(i\sqrt[4]{2})$. – Daniel Fischer Jan 23 '15 at 22:51

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This is an old question but since it is unanswered I thought I would provide an interesting example of such a field (technically it is answered: Daniel Fischer's answer suffices as far as I can tell). Another reason I wanted to provide this answer as opposed to the more elegant one given in Daniel Fischer's comment is to highlight the fact that $\mathbb{C}$ contains some pretty strange subfields.

Let $C(\mathbb{R})$ be the ring of continuous real-valued functions defined on $\mathbb{R}$ and let $I' \subseteq C(\mathbb{R})$ be the ideal consisting of functions with compact support, with $I$ the corresponding maximal ideal. Consider the residue field $C(\mathbb{R})/I$. It is proven here (rather famously) that this is a real-closed field containing a proper subfield isomorphic to $\mathbb{R}$, so there is a unique total order $\leq$ we can define on this field. We now note that $F=C(\mathbb{R})/I$ has cardinality $\mathfrak{c}$:

Proof: It is obvious that $|F| \geq \mathfrak{c}$ since $F$ contains a subfield isomorphic to $\mathbb{R}$. To prove the other inequality, note that $F$ is actually a collection of equivalence classes, i.e. $F= \left\{ [f]: f \in C(\mathbb{R}) \right\}$, for some equivalence relation. By the axiom of choice, there is a function $g:F \rightarrow C(\mathbb{R})$ such that $g([f]) \in [f]$ for each $f$. Such a map is necessarily an injection, so $|F| \leq |C(\mathbb{R})|=\mathfrak{c}$. Schroeder-Bernstein then gives our result. $\Box$

So $F$ has the cardinality of the continuum, and it is also easy to see that $F$ has characteristic $0$. Since any field of characteristic zero and cardinality at most continuum is isomorphic to a subfield of $\mathbb{C}$ (stated here and I've also seen this invoked in other places as well), we have $F \cong K \subseteq \mathbb{C}$ for some field $K$. Give $K$ the order induced by the isomorphism with $F$. Since $F$ contains a proper subfield isomorphic to $\mathbb{R}$ and $\mathbb{R}$ contains no such thing, we know $K$ is not contained in $\mathbb{R}$, so $K$ is a field satisfying your desired properties.

Now, if you are familiar with the paper I linked above by Hewitt, you'll recognize that this will actually work for any non-compact, completely regular topological space $X$ with a maximal ideal $I$ such that $C(X)/I$ has cardinality $\mathfrak{c}$ and has the so-called eta set property. We can find infinitely many such topological spaces and ideals, thus it is tempting to believe that we can actually produce infinitely many non-isomorphic examples of such fields in the way we did above. A word of caution about this: it is completely consistent with ZFC that all fields produced in this manner are isomorphic (see here, it is an implication of the continuum hypothesis). Under CH, the unique (up to isomorphism) field we have constructed is a "hyperreal number field" and happens to be the most common one used in non-standard analysis.

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