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Question: Explain the construction below (taken directly from Counter Examples in Analysis):

An ordered field is a field $F$ that contains a subset $P$ such that $P$ is closed with respect to addition and multiplication and exactly one of the following are true: $$x \in P; x = 0; -x \in P$$

The set $F$ of all numbers of the form $r + s\sqrt{2}$, where $r$ and $s$ are rational and the operations of addition and multiplication are those of the real number system R of which $F$ is a subset, is an ordered field. Let $P$, defined above, be the set of all members of $F$ that are positive members of R. A second way in which $F$ is an ordered field is provided by the subset $B$ defined: $$ r+s\sqrt{2} \in B \iff r - s\sqrt{2} \in P.$$

Based on how I've read it these few questions in particular stand out to me.

Q1. Isn't the inverse of addition destroyed by selecting only the positive members of $P$?

Q2. Can we say that this construction is effectively showing that there exists two automorphisms for the field extension ${\mathbf Q}[\sqrt{2}], {\mathbf Q}\rightarrow {\mathbf R}$? (${\mathbf Q}$ being the rationals)

Q3. What are the practical consequences for having these two distinct ways? If there are many, can you sum up their theme? (Read: why is this a counter example except to the rationals and irrationals, or why is this counter example important?)

Note: I'm new to fields, if this is an easy/poorly asked question let me know. I'll delete it and ask it again after I've read more, although I would be thankful for a hint in the right direction.

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Remember that we only care that P is a subset, closed under addition & multiplication, and for any x $\in$ F, $x\in P$ or $-x\in P$ so the ordered field is not worried about additive inverses. Note: I think you're on the right track with automorphisms. I think that these subsets show some consequences of defining automorphisms on a field extension, having these two distinct automorphisms of F is a base step for Galois theory which is monumental in math and helps with proving that there is no general solution by radicals of polynomials with degree greater than 4 –  Mr.Guy Jan 12 '13 at 5:11
    
@Mr.Guy Very nice, I'll have to think on this a while. –  J C Jan 12 '13 at 7:35
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Basically, if $F$ is any field and you can write down a field embedding $f \colon F \rightarrow {\mathbf R}$, you can pull back the usual ordering on ${\mathbf R}$ to define an ordering on $F$: for $a$ and $b$ in $F$, declare $a <_f b$ iff $f(a) < f(b)$ in ${\mathbf R}$. The two different embeddings of ${\mathbf Q}[\sqrt{2}]$ into ${\mathbf R}$, based on sending $\sqrt{2}$ to $\sqrt{2}$ or to $-\sqrt{2}$, provide the two embeddings in the example. It is a theorem that when $F$ is a finite extension of ${\mathbf Q}$, all orderings on $F$ arise from embeddings of $F$ into the real numbers. –  KCd Jan 12 '13 at 7:46
    
@KCd, Thanks! I have more to read now :) –  J C Jan 12 '13 at 7:50
    
Not all orderable fields can be embedded in $\mathbb{R}$ (e.g. $\mathbb{R}(x)$). However, there is a larger class of fields called real closed fields that can play this role. –  Hurkyl Jan 12 '13 at 8:39
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2 Answers

up vote 6 down vote accepted

Q1. Isn't the inverse of addition destroyed by selecting only the positive members of P?

If you mean that the set $P$ is not closed under taking additive inverses, yes, that's certainly true. For instance, $P$ is not a subgroup of the additive group of the field $F$. Your language ("destroyed") makes this sound like a bad thing, but it isn't. If it helps, the subset $P$ is not the "ordered field" itself, it's the extra structure on the field which allows you to view it as an ordered field, namely $x < y \iff y-x \in P$. Often $P$ is called the positive cone of the ordering, and this language is more suggestive since in both the frozen dairy industry and higher mathematics, cones are not required to be closed under inversion.

Q2. Can we say that this construction is effectively showing that there exists two automorphisms for the field extension $\mathbb{Q}(\sqrt{2})/\mathbb{Q}$?

I would say that the construction is effectively using the nontrivial automorphism $a+ b \sqrt{2} \mapsto a- b\sqrt{2}$ of $\mathbb{Q}(\sqrt{2})$: the logic of it goes in the other direction. Namely, whenever you have an ordering $<$ on a field $K$ and an automorphism $\sigma$ of $K$, you can pull back the ordering by the automorphism to get an ordering $<_{\sigma}$ defined by $a <_{\sigma} b \iff \sigma(a) < \sigma(b)$. In the case when $K = \mathbb{Q}(\sqrt{2})$, this is exactly how we get the second ordering from the first (which itself comes from the ``standard embedding'' into $\mathbb{R}$: in other words, the second ordering $<_\sigma$ on $\mathbb{Q}(\sqrt{2})$ is characterized by $a+b\sqrt{2} >_\sigma 0$ if $a-b\sqrt{2} > 0$.

It is easy to check that in full generality, $<_{\sigma}$ is an ordering on the field $K$. We did not check in general that $<_{\sigma}$ is a different ordering from $<$. In the above example this was clear: e.g. $\sqrt{2}$ is regarded as positive in exactly one of the two orderings. However it is possible for $<_{\sigma}$ and $<$ to coincide: this happens (rather tautologically) exactly when $\sigma$ is not just an automorphism of the field $K$ but of the ordered field $(K,<)$. Here is what I think is the simplest example:

Let $K = \mathbb{Q}(t)$ be the rational function field in one variable. Then the automorphism group of $K$ is $\operatorname{PGL}_2(\mathbb{Q}) = \{t \mapsto \frac{at+b}{ct+d} \ | a,b,c,d \in \mathbb{Q}, \ ad \neq bc\}$. Then it makes a nice exercise to show:

(i) There is exactly one ordering $<$ on $\mathbb{Q}(t)$ in which $n < t$ for all $n \in \mathbb{Z}$: one says that the element $t$ is infinitely large.
(ii) The automorphism $t \mapsto t+1$ preserves this ordering.

In general the set $X(K)$ of all orderings on a field $K$ can be naturally endowed with the structure of a topological space: see e.g. $\S 15.8$ of these notes. This is already an interesting structure, and as we saw it carries a natural -- but not necessarily free -- action under the group $\operatorname{Aut}(K)$. (It is easy to see that the action is free when $K$ is a number field. Beyond that I don't know much and would be interested to learn more.)

Q3. What are the practical consequences for having these two distinct ways? If there are many, can you sum up their theme? (Read: why is this a counter example except to the rationals and irrationals, or why is this counter example important?)

In real analysis I don't know of any real consequences of the above considerations, and the above counterexample does seem rather peripheral to me: I was slightly surprised to see it in Gelbaum and Olstead's book.

However, the example is an important one for general culture and perspective. In general, when two objects are conjugate under a group action, it is often awkward to prefer one over the other: that entails a certain symmetry-breaking. An effect of this is to make algebraists view fields more abstractly: it is not fully helpful to view $\mathbb{Q}(\sqrt{2})$ as a subfield of $\mathbb{R}$ because it breaks the symmetry between $\sqrt{2}$ and $-\sqrt{2}$. A more sophisticated (and ultimately more useful) perspective is to start with some more intrinsic definition of $\mathbb{Q}(\sqrt{2}$) -- e.g. as $\mathbb{Q}[t]/(t^2-2)$ -- and then realize that it has exactly two embeddings into $\mathbb{R}$. Carrying this idea further, a number field $K$ is a finite degree field extension of $\mathbb{Q}$. It is not hard to show that any number field can be embedded into the complex numbers $\mathbb{C}$, so that one could define a number field as a subfield of $\mathbb{C}$ with finite dimension over $\mathbb{Q}$. But this definition is subtly unnatural in number theory: it is much better to keep track of the set of embeddings of $K$ into $\mathbb{C}$...

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Wow, this response is great. I wish I had checked back earlier. I really like your answer of Q3, I sort of felt like I understood Q1 and Q2 after the comments on the original post but this really nails it down for me. Thanks again Pete! –  J C Mar 29 '13 at 18:30
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Another answer to Q3: What practical consequences are there to having these two ways.

The real numbers have exactly one order which respects the field operations (i.e., one subset P as in the definition you guys are using). The complex numbers have NO orders which respect the field operations (i.e., you can not order the complex numbers in such a way that

  1. $a>b \implies a+c\geq b+c$,
  2. $a>0 $ and $ b>c \implies ab\geq ac$

for all $a,b,c$ complex numbers, see 'ordered field' wiki or google 'real field').

So: the field $Q(\sqrt{2})$ described above, where elements look like $a+b\sqrt{2}$, has (at least!) two real orderings, the reals numbers have exactly one, and the complex numbers have none. It turns out that fields that have this second property are 'very nice' for studying abstractly, and these kinds of fields are called real closed.

By showing that $Q(\sqrt{2})$ has at least two orders, we see that it can't be real closed.

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