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Suppose $F = \mathbb{R}^2$ is a field equipped with the following operations:

Addition: Defined entrywise i.e.

$$(x_1, x_2) + (y_1, y_2) = (x_1 + y_1, x_2 + y_2)$$

Multiplication:

$$(x_1, x_2) \cdot (y_1, y_2) = (x_1y_1 - x_2y_2, x_1y_2 + x_2y_1)$$

Provided with the operations above, and knowing that $F$ is a field, show that no ordered relation makes $F$ into an ordered field.

I'm attempting to solve the problem above. I started with "Assume that < is an ordered relation on $F$ and $(x_1, x_2), (y_1, y_2), (z_1, z_2) \in F$. We see that: " in an attempt to find a contradiction but I haven't been able to find any. I'm not exactly sure how I can proceed by using the axioms though I'm sure it has something to do with multiplication since it isn't defined entrywise. Also, I'm not sure if I can use this but I found the additive identity to be $(0, 0)$ and the multiplicative identity to be $(1, 0)$. Any hints or assistance are much appreciated.

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Hint: $(0,1)^2=(-1,0)=-(1,0)$. In an ordered field, every nonzero square is positive…

The exercise however should also tell you that the set is $F=K^2$, where $K$ is at least a commutative ring, so the operations are well defined.

When is $F$ a field? Precisely if and only if $K$ is a field and, for all $x,y\in K$, $x^2+y^2=0$ implies $x=y=0$.

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  • $\begingroup$ Thanks for the reply. So then, would I need to define some order relation or assume one exists in order to show that $-(1, 0)$ is, in fact, negative? Also, I forgot to mention that $F = \mathbb{R}^2$; I've edited it accordingly! $\endgroup$
    – SupremePickle
    Commented Jan 14, 2021 at 11:58
  • $\begingroup$ @TRONIIX You just assume it exists and draw a contradiction. You want to prove it can't be an ordered field, so you cannot define a suitable order relation to begin with. Oh, by the way, you $F$ is nothing else than the complex field. $\endgroup$
    – egreg
    Commented Jan 14, 2021 at 12:01

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