Construct an order on a field $\mathbb{Q}^2$ Let $\mathbb{Q}$ be the field of rational numbers. Let $\mathbb{Q}^2=\{(a,b):a,b\in\mathbb{Q}\}$ and define addition and multiplication as follows:
$$
(a,b)+(c,d)=(a+c,b+d)\\
(a,b)\cdot(c,d)=(ac+2bd, ad+bc)
$$
Then $(\mathbb{Q}^2, +, \cdot)$ is a field, and $(0,0)$ is its zero element while $(1,0)$ is its unit element. The inverse of $(a,b)$ is $(\frac{a}{a^2-2b^2},-\frac{b}{a^2-2b^2})$. I'm asked to construct an order on it such that it becomes an ordered field, namely invariant under addition and multiplication with a positive element.
What I know is $(1,0)>(0,0)$, so $(n,0)>(0,0)$ for any natural number $n$. I don't quite know how to proceed. Is there any rules to follow to construct an order on a field?
Thank you for any help!
 A: There aren't any general rules, but in this case your field is actually (isomorphic to) a sub-field of $\mathbb{R}$. If you calculate what $$(0,1) \cdot (0,1)$$ is it may suggest how to map your field into $\mathbb{R}$. Then you can "pull back" the order on $\mathbb{R}$ onto your field.
BTW, @Toby's observation that such an order is not unique supports the idea that there are not general rules. If there were a set of rules they would likely lead to a unique answer, but we can see that there isn't a unique answer. And while we're talking more generally it's also worth noting that sometimes there is no such order, e.g. there never is for any finite field.

Follow up: If my comments about uniqueness don't make sense to you right now feel free to ignore them; they're just there to give a wider more general view.
Yes, that's the isomorphism I was thinking of. Because $$(1,0) \cdot (a,b) = (a,b)$$ for any $(a,b)$ we can see that $(1,0)$ acts like $1 \in \mathbb{R}$, and because $$(0,1)^2 = (0,1)\cdot(0,1) = (2,0)$$ we can see that $(0,1)$ acts like $\sqrt{2} \in \mathbb{R}$ (or like $-\sqrt{2}$ if you prefer - you can choose either but then must stick with it,)
So it makes sense to define $\phi : \mathbb{Q}^2 \rightarrow \mathbb{R}$ by $\phi( (a,b) ) = a+b\sqrt{2}$. Now you must explicitly show that $$ \phi((a,b)+(c,d)) = \phi((a,b)) + \phi((c,d))$$ and $$ \phi((a,b)\cdot(c,d)) = \phi((a,b)) \cdot \phi((c,d))$$ (As you write out the proof for multiplication you might see why we had to choose $\pm\sqrt{2}$ for $(0,1)$.
After you've done that you can "pull back" the ordering on $\mathbb{R}$, which means you define that "$(a,b) \lt_{\mathbb{Q}^2} (c,d) \iff \phi((a,b)) \lt_\mathbb{R} \phi((c,d))$", where the $\lt_\mathbb{R}$ on the right-hand side is the usual order on $\mathbb{R}$. You should then be able to show that  $\lt_{\mathbb{Q}^2}$ is invariant under addition and multiplication by using the invariance of $\lt_\mathbb{R}$. Note that this is something that you have to write out explictly; it's not enough to observe it. If you're just getting started in abstract algebra you might find this a little hard but really you're just applying $\phi$ and $\phi^{-1}$ and "pushing symbols around".
Notice that this method assumes the existence of both $\mathbb{R}$ and its invariant ordering $\lt_\mathbb{R}$ (and also the existence of $\sqrt{2})$. Since this will make the problem much easier I'd go ahead and assume we know they exist. If we were not allowed to assume them we'd have to find a way to express "$a+b\sqrt{2} \lt c+d\sqrt{2}$" without using the concept of $\sqrt{2}$, which gets a lot more complicated.
A: The quick and easy answer is to notice that this field is isomorphic to $\mathbb Q[\sqrt{2}]$ by sending $(a,b)$ to $a+b\sqrt 2$, so $(a,b)<(c,d)$ if $a+b\sqrt 2<c+d\sqrt 2$.
