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I'm wondering if it's possible to order a field extension $\mathbb{Q}[\sqrt{x}]$ for $0<x\in\mathbb{Q}$ such that it is an ordered field with an ordered subfield isomorphic to $\mathbb{Q}$.

It seemed that the ordering $$\leq_\sqrt{x}=\{(a+b\sqrt{x},c+d\sqrt{x}):a^2+c^2-2ac\leq x(b^2+d^2-2bd)\}$$ was promising, due to the algebraic manipulations $$a+b\sqrt{x}\leq c+d\sqrt{x}\iff a-c\leq\sqrt{x}(d-b)$$ $$\iff (a-c)^2\leq x(b-d)^2\iff a^2+c^2-2ac\leq x(b^2+d^2-2bd).$$ Despite this, if we use the canonical product structure $$(a+b\sqrt{x})(b+d\sqrt{x})=ab+bdx+(ad+bc)\sqrt{x}$$ it is not evident to me via algebra-crunching that orders are well-preserved under multiplication. Specifically, it isn't clear that for $a,b,c,d\in\mathbb{Q}[\sqrt{x}]$ we have that $a<b$ and $c<d$ imply $ac<bd$. Is this the case, and if not is there an ordereing we can explicitly define that makes a field extension like this ordered with $\mathbb{Q}$ as a subfield?

Edit

More specifically, suppose we have (a-priori) the ordered field $\mathbb{Q}=\langle\mathbb{Q},+,-,\times,\div,\leq\rangle$, and we wish to totally order the field extension $\mathbb{Q}[\sqrt{2}]=\langle\mathbb{Q},+_{\sqrt{2}},-_{\sqrt{2}},\times_{\sqrt{2}},\div_{\sqrt{2}},\sqrt{2}\rangle$ with a total ordering $\leq_\sqrt{2}$ expressed in the language of ordered fields such that the subfield $\{a+b\sqrt{2}:b=0\}\subsetneq\mathbb{Q}[\sqrt{2}]$ is isomorphic to $\mathbb{Q}$ as an ordered field under the projection $\pi_0$ of $\mathbb{Q}[\sqrt{2}]$ onto its first factor, $$\pi_0(a+b\sqrt{2})=a.$$ What is an explicit formula for $\leq_\sqrt{2}$? Does the above formula for $\leq_\sqrt{x}$ work in the case that $x=2$?

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    $\begingroup$ I'm confused when you say $x \in \mathbb Q$. Then $x$ is not an indeterminate, but a rational numbr? So for example your question could be about $\mathbb Q[\sqrt{2}\;]$ ... why not use the ordinary real-number order on those? $\endgroup$ – GEdgar Oct 29 '17 at 13:12
  • $\begingroup$ Maybe you want the ordering to be definable in the language of rings? $\endgroup$ – Adayah Oct 29 '17 at 15:46
  • $\begingroup$ @GEdgar I am, in reality, trying to totally order a splitting field extension of a nonstandard model of $\mathbb{Q}$ for some research I'm working on. I thought this question wasn't really research level, and I can extrapolate from the standard model case to the nonstandard model I'm working with on my own pretty easily. I would also like to do this without an a-priori construction of a nonstandard model of the $\mathbb{R}$ to work with, as the connection between those two things is what I'm trying to understand more deeply. $\endgroup$ – Alec Rhea Oct 30 '17 at 0:45
  • $\begingroup$ @Adayah I believe that should be sufficient, or even in the language of ordered fields. $\endgroup$ – Alec Rhea Oct 30 '17 at 2:28
  • $\begingroup$ That looks suspicious. If ${\leqslant} \in L$ is interpreted in $\mathbb{Q}[\sqrt{2}]$ as the usual order on $\mathbb{R}$, why not just take $\leqslant$ as the order? Otherwise, how is $\leqslant$ interpreted in $\mathbb{Q}[\sqrt{2}]$? Or perhaps we somehow can compare rational numbers but can't directly compare elements of $\mathbb{Q}[\sqrt{2}]$ - in that case, how is the statement formalized? $\endgroup$ – Adayah Oct 30 '17 at 7:01
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Note: I'll do the specific case of $\sqrt{2}$.

I think it's best to do it through the set of positive numbers. Remember that in an ordered field $K$, for any $a\in K\setminus\{0\}$ either $a>0$ or $-a>0$. Moreover if $a>0$ and $b>0$ then $a+b>0$ and $ab>0$.

So we have the field $\mathbb Q[\sqrt{2}]$ with elements of the form $a+b\sqrt{2}$ with $a,b\in\mathbb Q$. Now for elements of $\mathbb Q$ we know what numbers are positive (and indeed, it's not hard to show that we don't have a choice there). But what is the sign of $\sqrt{2}$? We have two choices: Either $\sqrt{2}>0$, then $-\sqrt{2}<0$, or $\sqrt{2}<0$, then $-\sqrt{2}>0$. But both choices are actually equivalent, as there's a field isomorphism exchanging $\sqrt{2}$ and $-\sqrt{2}$, and thus the choice just amounts to which of the two solutions of $x^2-2=0$ we call $\sqrt{2}$. Going with the usual convention, we choose $\sqrt{2}$ to be positive.

Now consider a general element $a + b\sqrt{2}$. We can concentrate on the case $b>0$ since for $b<0$ we simply get the opposite sign of $-a-b\sqrt{2}$ which again has a positive value of $b$. And for $b=0$ we just have a rational number, whose sign we already know.

Now we use the fact that with $b>0$ we also have $b^{-1}>0$, and multiply the above number with $b^{-1}$ (remember, multiplying with a positive number does not change the sign). Thus we get the condition $ab^{-1}+\sqrt{2}>0$.

If we write $c=ab^{-1}$, we therefore have to check for which values of $c$ we get $c+\sqrt{2}>0$.

Now if $c\ge 0$, then this is obviously positive. So we need to consider the case $c<0$. To more easily see the sign, let's define $d=-c$, so that $d>0$.

Now we can rewrite the condition as $$0 < {\sqrt{2}-d} = \frac{(\sqrt{2}-d)(\sqrt{2}+d)}{\sqrt{2}+d} = \frac{2-d^2}{\sqrt{2}+d}\quad.$$ The denominator is, again, clearly positive, so we get a positive number iff $d^2<2$.

Substituting everything back and combining the cases, we get for the case $b>0$ the condition: $$a\ge 0 \lor a^2 < 2b^2$$ For $b<0$ we need the opposite condition for $-a$ and $-b$: $$\lnot(-a\ge 0 \lor (-a)^2 < 2(-b)^2))$$ which can be rewritten as $$a>0 \land a^2\ge 2b^2$$ The three cases ($b=0$, $b>0$ and $b<0$) can now of course be combined into a big expression, but I don't see an obvious way to simplify that.

Of course, as usual, you have $u>v$ iff $u-v>0$, which you can insert in the above conditions to make it even more complicated.

Your condition $$a+b\sqrt{2} \le c+d\sqrt{2}\iff a^2+c^2-2ac\leq 2(b^2+d^2-2bd)$$ does not work, as it is symmetric under exchange of the two numbers. A concrete example that fails is $a=b=1, c=d=0$, which gives $$1^2+9^2-2\cdot 1\cdot 0 = 1 \le 2 = 2(1^2+0^2-2\cdot 1 \cdot 0$$ so you'd get $1+\sqrt{2}\le 0$.

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  • $\begingroup$ This is excellent, thank you. $\endgroup$ – Alec Rhea Nov 15 '17 at 22:54
  • $\begingroup$ @AlecRhea: You're welcome. $\endgroup$ – celtschk Nov 15 '17 at 22:58

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