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 $\mathbb{C}$...
Here are some hints:
For (ii), you are meant to show the field axioms are satisfied. This basically says that you can do arithmetic as freely as with, say the rational numbers.
Explicitly, for every $x,y,z \in \mathbb{Z}/p$:
- $x+y = y+x$
- $x \cdot y = y \cdot x$
- $(x+y)+z = x+(y+z)$
- $(x \cdot y) \cdot z = x \cdot (y \cdot z)$
- $x + [0] = x$
- $x \cdot [1] = x$
- for every $x$, there is a $-x$ such that $x+(-x) = [0]$
- for every $x \not = 0$, there is a $x^{-1}$ such that $x \cdot x^{-1} = [1]$
- $x \cdot (y+z) = (x \cdot y) + (x \cdot z)$
Can you show these facts?
For (iii), recall an Ordered Field is a field (so it satisfies the above) equipped with a relation $<$ satisfying properties which say $<$ isn't a bad choice of symbol:
- $x < y$ and $y < z$ implies $x < z$
- exactly one of $x = y$, $x < y$, $y < x$ holds for any $x,y$.
We also want to know that $<$ is compatible with the field structure, namely the following (again, obvious) equation should hold:
- $x < y$ implies $x + z < y + z$
- $0 < x$ and $0 < y$ implies $[0] < x \cdot y$
Can you see how to use $x < y$ implies $x + z < y + z$, coupled with $$\underbrace{x + x + x + \ldots + x}_{p \text{ times}} = 0$$ to show that no such order can exist?
Feel free to comment on this answer if you have more specific issues.
I hope this helps ^_^
Best Answer
"multiplication of numbers" is actually a relative concept, depending on the algebraic structure which you are working with. For your question, the most important example is $ \frac{\mathbb{Z}}{p\mathbb{Z}} $ which is a finite field (with mod $p$ summation and multiplication).For example when $p = 5$ we have this equalities (in $ \frac{\mathbb{Z}}{5\mathbb{Z}} $): $$ 1+1+1+1+1 = 5 = 5*1 = 0 (mod \text{ }5) $$ so you see that $5*1 = 0$ in this context.
Now, I will go through (i), (ii) and (iii):
(i): Defining axioms of an ordered field (F, +, ., <) imply that for any $0 \ne x \in F$ we have $x^2 > 0$ because if $x > 0$ we get $x*x = x^2 > 0$ and if $x < 0$ we have $x + (-x) < 0 + -x$ so $0 < -x$ and then again we have $x^2 = (-x)*(-x) > 0$ and we are done. so in a special case, we get $1 > 0$ in any ordered field (F, +, ., <). Now if we have some positive $m$ which $m*1 = 0$, we will get a contradiction, because: $$ 1>0 , 1+1>0, 1+1+1>0, ..., 0 = 1+1+...+1(m\text{ times}) > 0 $$
(ii): Assuming $F$ is a finite field with cardinality $n$, if we add up $1$: $$ 1, 1+1, ..., 1+1+...+1(n+1 \text{ times}) $$ we have to get two equal numbers in this sequence like $m*1 = p*1$ and $m \ne p$ so we get $(m-p)*1 = 0$.
(iii): This part is straight. if we look at vector space axioms, we see from that (i) and (ii) its not possible to define scalar product for a finite field $F$ over $\mathbb{R}$ and $\mathbb{C}$ as usual, like: $1_{\mathbb{R}}*1_{F} = 1_{F}$ because then we will have some $m \in \mathbb{Z}\text{: } m*1_F = 0_F$ and: $$ 0_F = \frac{1}{m} * 0 = \frac{1}{m}(m*1) = 1_{\mathbb{R}}*1_{F} = 1_{F} $$ which is absurd.