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My book says yes, but I'm not so convinced. There is:

$$ F:=\{a+b\sqrt2: a,b \in \mathbb Q\} $$

Sum and product are defined in $F$ as follow.

$$ (a+b\sqrt2)+(c+d\sqrt2) := a+c+(b+d)\sqrt2\\ (a+b\sqrt2)·(c+d\sqrt2) := ac+2bd+(ad+bc)\sqrt2 $$

The order relation to be considered is: $$ a+b\sqrt2\ \lesssim c+d\sqrt2 \iff c-a+(b-d)\sqrt2 \geq 0 $$

I'm not so convinced due to the multiplication property of ordered fields: $$ \forall a,b,c,\ c\gt0,\ a\lesssim b\implies a·c\lesssim b·c $$

So I should demonstrate that $$ (a+b\sqrt2)·(e+f\sqrt2)\lesssim (c+d\sqrt2)·(e+f\sqrt2);\\ \text{with } e+f\sqrt2 \gt0 $$ From the product definition this is equivalent to: $$ (ae + 2bf) + (af+be)\sqrt2 \lesssim (ce+2df) + (cf+de)\sqrt2 $$ From the relation operation definition: $$ (ce+2df)-(ae+2bf) + [(af+be)-(cf+de)]\sqrt2 \geq0\\ (c-a)(e-f\sqrt2)+(b-d)(e-f\sqrt2)\sqrt2 \geq0\\ (e-f\sqrt2)[c-a+(b-d)\sqrt2] \geq0 $$ I cannot say if $e-f\sqrt2 \geq0$, so I cannot say that $F$ is an ordered field. Am I right or is my book right?

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  • $\begingroup$ What book would that be? And what is your background? $\endgroup$
    – Will Jagy
    Sep 16, 2013 at 19:09
  • $\begingroup$ @WillJagy My background: high school. The book: an italian analysis book ( ISBN 9788843052899 ) $\endgroup$
    – Fra H
    Sep 17, 2013 at 12:08

4 Answers 4

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$F \subseteq \mathbb R$ with the field operations and ordering induced from $\mathbb R$. So your book is right, $F$ is ordered.

Instead of looking at formulas, just think of the elements of $F$ as real numbers. If you have two real numbers $a \leq b$ and a third $c > 0$ then, because they are just real numbers, you have $ac \leq bc$.

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    $\begingroup$ The ordering is not the one induced from $\mathbb{R}$, unless $\sqrt{2}$ denotes here the negative square root of $2$. $\endgroup$
    – Daniel Fischer
    Sep 16, 2013 at 18:04
  • $\begingroup$ $\lesssim$ could be different from $\leq$. I shouldn't consider $\leq$ here. $\endgroup$
    – Fra H
    Sep 16, 2013 at 18:21
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    $\begingroup$ @DanielFischer: Good catch, but we do still get an order-preserving embedding into $\mathbb R$ by mapping $a + b\sqrt2 \in F$ to $a - b\sqrt2 \in \mathbb R$, so it's still (isomorphic to) a subfield of $\mathbb R$. $\endgroup$
    – Ilmari Karonen
    Sep 16, 2013 at 19:32
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Note that $\mathbb Q(\sqrt 2)$ is a subfield of $\mathbb R$, which is an ordered field, and "expected ordering" is the one derived from $\mathbb R$, and since it is compatible with the field operations on $\mathbb R$ we can say that $\mathbb Q(\sqrt 2)$ with the "expected order" is an ordered field.

However there is not a canonical ordering on $\mathbb Q(\sqrt 2)$ as an extension field of $\mathbb Q$ - the extension does not come with an order attached.

You can see that if you explore the order defined in the question effectively by $a+b\sqrt 2 \gt 0$ in $\mathbb Q(\sqrt 2)$ if and only if $a-b\sqrt 2 \gt 0$ in the usual order on $\mathbb R$. This just reverses the sign of $\sqrt 2$ wherever it occurs, and arises because $\mathbb Q(\sqrt 2)$ has a non-trivial automorphism which fixes $\mathbb Q$. All the computations are compatible with the order.

So there is a choice of possible orderings, and in this sense we can say that $\mathbb Q(\sqrt 2)$ can be ordered as a field in either way.

Note I would say that it is wrong to say that $\mathbb Q(\sqrt 2)$ "is" an ordered field.

Note also that in any ordered field the squares are positive, and this defines a unique order on $\mathbb R$, for example.

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  • $\begingroup$ Nice and complete academical answer, but I cannot figure out why in $\mathbb R$ $a-b\sqrt2 > 0 \iff a+b\sqrt2 >0$ in $\mathbb Q(\sqrt2)$ $\endgroup$
    – Fra H
    Sep 16, 2013 at 18:56
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    $\begingroup$ @F.Haymard'Ettory In $\mathbb R$ we have the normal ordering. The alternative order in $\mathbb Q$ is defined this way. It works because we have no algebraic way in $\mathbb Q$ of telling whether $\sqrt 2$ is positive or negative - so we can consistently choose the "wrong" sign. I wrote my answer mainly to try to put this example in a wider context, rather than to fill out the details - on its own it is a curiosity, but it is a window onto some bigger and significant ideas. $\endgroup$
    – Mark Bennet
    Sep 16, 2013 at 19:09
  • $\begingroup$ All clear, thank you. $\endgroup$
    – Fra H
    Sep 16, 2013 at 20:19
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You should notice that $a+b\sqrt 2\in F$ is positive iff $a-b\sqrt 2\in\mathbb R$ is positive. So since $\mathbb R$ is ordered, $F$ ought to turn out ordered as well.

And in fact you arriuved just at that point, but fell for a typo: In what you should demonstrate you wrote as condition "with $e+f\sqrt 2>0$", but the correct condition would be "with $0\lesssim e+f\sqrt 2$". In fact you had the same typo already in the quoted multiplication property for general ordered fields.

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QUOTE: The order relation to be considered is: $$ \color{magenta}{a+b\sqrt2\ \lesssim c+d\sqrt2 \iff c-a+(b-d)\sqrt2 \geq 0} $$

In particular, taking $a=0, \; \; b=0,$ we have $$ 0 \lesssim c+d\sqrt2 \iff c-d\sqrt2 \geq 0 $$

There is nothing special about the letters, they are all just rational numbers. So, in the line just above, I could take substitutions $c \mapsto a, \; \; d \mapsto b$ and arrive at

$$ 0 \lesssim a+b\sqrt2 \iff a-b\sqrt2 \geq 0 $$

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