I should be able to figure this out, but I’m just not making any progress. I don’t want the answer outright, but I’d greatly appreciate some hints.
Any ordered field F has a subset P with the following properties:
[ul]
[li]For any non-zero a [symbol]Î[/symbol] F, exactly one of a and -a is in P.[/li][li]a, b [symbol]Î[/symbol] P implies that a + b, ab [symbol]Î[/symbol] P.[/li][/ul]
It’s very easy to show that P exists, but I’m stumped on showing that P is unique. It certainly seems like it ought to be (but of course I don’t know that for sure). Any hints on how to obtain this result?
If there are two distinct subsets of F–call them P and Q–with the properties outlined above, then there is at least one x [symbol]Î[/symbol] F with x [symbol]Î[/symbol] P and -x [symbol]Î[/symbol] Q. Since both P and Q are closed under multiplication, x[sup]2[/sup] is an element of both P and Q.
That seems like an important insight, but I can’t see where to go with it.
Strictly speaking, P is obviously not unique; 0 may be in or not in P, so there are at least two such sets. So let’s say 0 is in P.
I haven’t completely checked this, so I may have left something important out, but: Have you considered a field such as Q[sqrt(2)]? This can be made an ordered field by considering it as a subfield of R, with the set of positive elements determined accordingly…… but there are two square roots of 2 in R.
Yes, I should have been clear. Since P is meant to be the set of positive elements, I’d say assume that 0 is not in P, but it probably doesn’t make any difference.
I’m afraid I don’t really see why this might be relevant.
Good point. I think it’s obvious that any ordered field is infinite, and as such, Q is a subfield. Since 1 [symbol]Î[/symbol] P, Z[sup]+[/sup] [symbol]Í[/symbol] P.
Furthermore, P is closed under multiplicative inverses. If a [symbol]¹[/symbol] 0, 1/a exists and is non-zero, so either 1/a or -1/a is in P. If a and -1/a are both in P, then 1/a[sup]2[/sup] is in P, and a * 1/a[sup]2[/sup] = 1/a is also in P. From there, we can conclude that Q[sup]+[/sup] [symbol]Í[/symbol] P. But again, I don’t see where to go with this.
My point is that I think this gives two different orderings of the same field, and so two different sets satisfying your axioms. (Spoilers, assuming I’m not being stupid, below.)
Order Q[sqrt(2)] by the ordering induced from R (i.e., say (p,q) in Q[sqrt(2)] is positive iff p+q*sqrt(2)>0); this gives you a set P of positive numbers.
But now order Q[sqrt(2)] by the “conjugated” ordering: say (p,q) is positive iff p-q*sqrt(2)>0 (note the negative sign), and use this to get a new set P’ of positive numbers. Because the formal field axioms work the same whichever square root of 2 you choose, these both (I think) are consistent orderings, but they’re different.
Sure, but so what? The two requirements in the definition of P make no reference to the ordering relation, so its existence can only depend on the base field and not on what < we choose. Both P and P’ satisfy the requirements in any ordering of Q[sqrt(2)].
No, it’s not unique. The trick is to view both P = { (p, q) | p + qsqrt(2) > 0 } and P’ = {(p, q) | p - qsqrt(2) > 0 } as subsets of Q[sqrt(2)] without regard for the ordering. I think that’s what you were getting at, but the phrasing threw me off.
That then raises the question of when P is unique. I think we’ve shown that P is unique in Q. Are there any elements of R - Q that you can add to Q and retain that property? Probably not–the counterexample above seems like it would generalize.
I’m not sure I understand; let me see if I’ve got this straight. Do we agree that the set “P” (as defined in the OP) need not be unique for an ordered field?
And now, is the question whether there exists a bijection between orderings (F,<) of a base field F, and sets P satisfying the OP’s criteria?
Yeah, that’s what I was getting at. Sorry for the oblique phrasing; I’m pretty bad at giving hints.
I agree that P is unique in Q. I don’t know enough about the classification of (ordered) subfields of R to know whether there are other examples, but all of the fields Q [sqrt(a),sqrt(b),…] (with square-free natural numbers a,b,…) should have the same nonuniqueness issues.
Or any field that contains such a field. It’s hard to believe that any wouldn’t.
The interesting thing about P = { (p, q) | p - q*sqrt(2) } in Q[sqrt(2)] with the standard ordering is that, while it has the properties of the positive numbers, not all of its elements are positive (e.g., (1, -1)). It’d be nice if those properties did specify a subset of the positive numbers, but it’s clear that they don’t.
Since one of each pair {a,-a} has to be in P, this can’t ever happen.
Given an ordering <, it’s easy to see that P={a in F:0<a} works. Similarly, given P satisfying the above, define < by x<y iff y-x in P; it’s easy to check that this satisfies the ordering axioms. So I think P is unique per ordering.
Wait, didn’t we just establish that such a scheme won’t work? Is there an error in the following argument?
Take F = Q[sqrt(2)] with (p, q) > (p’, q’) iff p + qsqrt(2) > p’ + q’sqrt(2) in R. Consider P = { (p, q) | p + qsqrt(2) > 0 } and P’ = { (p, q) | p - qsqrt(2) > 0 }.
For any non-zero (p, q) in F, either p + qsqrt(2) > 0 or -p - qsqrt(2) > 0, so P satisfies the first condition. If a + bsqrt(2) > 0 and c + dsqrt(2) > 0, then a + c + (b + d)*sqrt(2) > 0 and ac + 2bd + (ad + bc)*sqrt(2) > 0, so P satisfies the second condition.
A similar argument holds for P’. If p - qsqrt(2) < 0, -p + qsqrt(2) > 0. If a - bsqrt(2) > 0 and c - dsqrt(2) > 0, then a + c - (b + d)*sqrt(2) > 0 and ac + 2bd - (ad + bc)*sqrt(2) > 0, so we have both conditions for P’.
I think that, given a set P like we’re talking about, x < y iff y - x is in P induces an alternative (partial) ordering on F. For the right choice of P, you get your original ordering back, but I don’t think you can take that to mean that P is unique for a given ordering.
Sorry, bad wording on my part. What I was trying to say is that there is a bijection between total orderings on F and sets P satisfying your criteria: given an ordering < you can find a P, and given a P you can find an ordering.
I agree that a given ordered field can have multiple such sets (as my earlier example Q[sqrt(2)] showed).
I thought about this a little more and generalized my quadratic example:
Suppose x[sub]i[/sub] are the real roots of a polynomial p (in Q), of degree n, which is irreducible over Q. Then Q[x[sub]i[/sub]] is an algebraic extension of Q, a vector space of dimension n over Q. The Q[x[sub]i[/sub]] are isomorphic as fields, since they are all Q/p(x) (the ring of polynomials with rational coefficients, modulo the congruence p(x)==0).
We can impose a total order < on Q/p(x), by the order induced from R. Specifying an order is equivalent (as I said above) to defining the set of positive values (the elements y which satisfy y > 0). Specifying an order is also equivalent to defining a real root x[sub]i[/sub] of p(x): Writing
a = (a[sub]0[/sub],a[sub]1[/sub],…,a[sub]n-1[/sub]) = a[sub]0[/sub] + a[sub]1[/sub]x + … + a[sub]n-1[/sub]x[sup]n-1[/sup]
for an element a of Q/p(x) (a[sub]i[/sub] in Q), the total order can be used to find the subset of Q={(q,0,…,0)} containing all values less than (0,1,0,…,0)=x; but this is just a Dedekind cut and so defines the real number x.
So for an algebraic extension, the number of possible orderings (and hence the number of sets P satisfying the OP’s criteria) should equal the number of real roots of p. For a quadratic field, this is either 0 (e.g., Q[sqrt(-2)]) or 2, but fields with any desired number of orderings should thus exist.
Does this seem right to you? … I still need to think about transcendental extensions.