TITLE: How do extension fields implement $>, <$ comparisons?
QUESTION [10 upvotes]: I'm taking an abstract algebra course, and we just hit extension fields - for example, you define $\sqrt{2}$ by starting with the field $\mathbb{Q}$ and defining $\sqrt{2}$ as a solution to the irreducible (over $\mathbb{Q}$) polynomial $x^2 - 2$.
This is unintuitive to me, because I know something additional about $\sqrt{2}$: intuitively I want to be able to make statements like $1 < \sqrt{2} < 2$, but I'm not sure how $<$ is even defined when $\sqrt{2}$ is fabricated like so.
So, my question: using the extension-field definition of $\sqrt{2}$, what additional construction allows us to make comparisons such as $1 < \sqrt{2} < 2$?  And why do these constructions not allow us to make comparisons on $i$, which is similarly defined via extension fields (since $1 < i < 2$ is a nonsensical statement)?

REPLY [2 votes]: The answers so far have already explained the whole issue, but here is an additional remark:
Although we cannot distinguish $\sqrt{2}$ from $-\sqrt{2}$ in $\mathbb{Q}(\sqrt{2})$ in the language of fields, we can do so in $\mathbb{R}$ (in fact already in $\mathbb{Q}(\sqrt[4]{2})$ with the same argument). In fact, the order relation on $\mathbb{R}$ can be defined by means of the field structure since $x \geq 0 \Leftrightarrow \exists y (x=y^2)$.
This observation also implies that the only ring endomorphism of $\mathbb{R}$ is the identity, which is not the case for $\mathbb{Q}(\sqrt{2})$. Every endomorphism of a field can be used to twist an arbitrary ordering to get a potentially new one, but for $\mathbb{R}$ the ordering is unique.