How do extension fields implement $>, <$ comparisons?

Question

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)?

Answer 1

They don't. You need to pick an embedding of $\mathbb{Q}(\sqrt{2})$ into $\mathbb{R}$, which is an ordered field, and you can make such comparisons in ordered fields. ($\mathbb{Q}(i)$ has no such embeddings.) Unfortunately, there are two embeddings: in the other embedding, $\sqrt{2}$ is sent to $-\sqrt{2}$, and then it's not true that $1 < - \sqrt{2} < 2$. In other words, knowing that $\sqrt{2}$ is a root of $x^2 - 2$ is not enough information for you to conclude that it's positive.

More generally, a number field has both real and complex embeddings, and moving to a different real embedding doesn't preserve the truth value of statements involving orderings.

Answer 2

I will call $\sqrt{2}$ by $\alpha$, instead, to distinguish it from the positive real number whose square is $2$.

Amusingly, there are two distinct ways that we can make $\Bbb Q(\alpha)$ an ordered field with the usual operations, under an order relation whose restriction to $\Bbb Q$ is just the standard order relation on $\Bbb Q$. Consider the functions $\varphi_1,\varphi_2:\Bbb Q(\alpha)\to\Bbb R$ given by $$\varphi_1(p+q\alpha)=p+q\sqrt{2}\quad\text{and}\quad\varphi_2(p+q\alpha)=p-q\sqrt{2},$$ where $p,q$ range over $\Bbb Q$. These are injective field homomorphisms, as you can check.

We can induce order relations $<_1$ and $<_2$ on $\Bbb Q(\alpha)$ from the standard order $<$ on $\Bbb R$, by saying $$x<_jy\Leftrightarrow\varphi_j(x)<\varphi_j(y)$$ for $j=1,2.$ Now, $\Bbb Q(\alpha)$ is an ordered field with the standard operations under both $<_1$ and $<_2$, and these orders are not the same. However, for any rational $p,q$, we have $p<_jq$ if and only if $p<q$ ($j=1,2$), so these orders both "act the same" on the rationals.

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Upshot: You can put either $1<\alpha<2$ or $1<-\alpha<2$. Take your pick.

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As for $\Bbb Q(i)$, we can't embed it into $\Bbb R$ in this way, since there isn't any real number whose square is $-1$. We can certainly order $\Bbb Q(i)$ in any number of ways, but such an ordering will never make $\Bbb Q(i)$ into an ordered field with the usual operations.

Answer 3

I think there can be multiple appropriate responses (the obvious, but probably not helpful, one being simply "an ordering of the field"). To give you a more satisfactory answer, I would respond that the additional construction you need to make "comparison statements" in a field $L$ is: a specified ordered field $K$ together with a specified map $f:L\to K$.

If the ordering of $K$ is denoted with "$<$", you would then define an ordering "$\prec$" on $L$ by declaring that, given elements $\alpha,\beta\in L$, we have $\alpha\prec\beta$ if and only if $f(a)<f(b)$.

Answer 4

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.