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)?

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    $\begingroup$ I like the smiley in the title. $\endgroup$ – Martin Brandenburg Mar 8 '13 at 23:04

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.

  • $\begingroup$ This makes sense, but it sounds like we need to define $\mathbb{R}$ before we can get a satisfactory comparison-allowing definition of $\sqrt{2}$. But we need to define $\sqrt{2}$ in order to define $\mathbb{R}$, don't we? $\endgroup$ – GMB Mar 8 '13 at 22:53
  • $\begingroup$ @Lev: you can define an order on $\mathbb{Q}(\sqrt{2})$ without talking about $\mathbb{R}$, it's just not unique and the easiest way to say what this order actually is is to say it's the induced order coming from a real embedding. $\endgroup$ – Qiaochu Yuan Mar 8 '13 at 23:04
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    $\begingroup$ @LevDub: We can define $\Bbb R$ to be the completion of $\Bbb Q$ under the absolute difference metric $d(x,y)=|x-y|$ ($x,y\in\Bbb Q$)--or any of a number of other ways. There's no need to define $\sqrt{2}$ first, but we'll certainly want to show that there is some positive number in $\Bbb R$ whose square is $2$--otherwise, we can't embed into $\Bbb R$ as we'd like to. $\endgroup$ – Cameron Buie Mar 8 '13 at 23:06
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    $\begingroup$ @Qiaochu, are you sure that it’s unfortunate that there are two different embeddings of $\mathbb Q(\sqrt2)$ into $\mathbb R$? $\endgroup$ – Lubin Mar 8 '13 at 23:32

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.

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

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.


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 $f(a)<f(b)$.


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.


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