Total orderings on $\mathbb{Q}(\sqrt2)$

Question

One of the first exercises from part II of Garling's A Course in Mathematical Analysis asks us to define two different total orderings over $\mathbb{Q}(\sqrt2)$. Now, one of them is the obvious lexicographical ordering, i.e. $r+s\sqrt2 < r'+s'\sqrt2$ iff $r < r'$ or $r=r'$ and $s<s'$. I'm not too sure about the other one, though; I thought about using the order inherited from the reals, i.e. if $p, q \in \mathbb{Q}(\sqrt2)$, then $p \leq q$ iff there is an $x \in \mathbb{Q}(\sqrt2)$ such that $q = p + x^2$. This is a partial order, but I'm not entirely sure it is a total order; moreover, if it is a total order, I'm not sure if it's not the same as the lexicographical one. Any hints here?

Also, does this exercise generalize, i.e. is there in general just two ways of defining total orders over quadratic fields?

EDIT: I just noticed the exercise asked for a total order which made $\mathbb{Q}(\sqrt2)$ into an ordered field, so the above won't work. Fortunately, other posters have already replied with orderings that do respect the ring structure of $\mathbb{Q}(\sqrt2)$; in light of this, I'd like to ask two more questions:

(i) are the total orderings on a quadratic rational field always given by the embeddings of the field into the reals? More generally, are the total orderings on quadratic fields always given by embeddings into a field extension?

(ii) is there any way of determining the number of total orderings (irrespective of the ring structure) on a field of given cardinality?

Answer 1

We can give two total orders that also respects the ring structure, i.e. satisfying $a,b\ge 0\implies a+b,\ a\cdot b\ge 0$:

1. the one inherited from the reals, that is, obtained by pulling back $\le$ along the embedding $\Bbb Q(\sqrt 2)\hookrightarrow\Bbb R$
2. the one obtained by the other embedding $a+b\sqrt2\mapsto a-b\sqrt2$

Besides these, there are a lot total orderings on the countable set $\Bbb Q(\sqrt 2)$.

Answer 2

Your order defined as $p \leq q$ iff there is an $x \in \mathbb{Q}(\sqrt2)$ such that $q = p + x^2$ is not total: consider $q = \sqrt{2}$, $p = 0$. Then there's no $x \in \mathbb{Q}(\sqrt{2})$ such that $\sqrt{2} = 0 + x^2$, neither there's $y \in \mathbb{Q}(\sqrt{2})$ such that $0 = \sqrt{2} + x^2$.

There's no need to try to emulate the definition from the reals, if you can just inherit it directly. There's an inclusion $i: \mathbb{Q}(\sqrt{2}) \hookrightarrow \mathbb{R}$, $i(p) = p$, and it's easy enough to define an order on $\mathbb{Q}(\sqrt{2})$ as $p \leq q \iff i(p) \leq i(q)$, where second $\leq$ comes from the ordering in $\mathbb{R}$. Clearly it's a total ordering which makes $\mathbb{Q}(\sqrt{2})$ an ordered field.