Announcement

I'm going to stop writing the super-real fields series for now. It's still interesting me, but there are some other subjects which have been bumped up my priorities list recently that I want to learn first. I may well make some posts about them later.

Super-real fields 3: Artin-Schreier theory of ordered fields

I was originally going to use this post to talk about gaps in ordered sets, but I thought I'd detour and talk about Artin-Schreier theory instead. I'm going to need it at some point so I may as well include it now.

First of all a note: I made a mistake in my first post on how to specify an order by specifying the positive elements. There's a missing condition. If P is our proposed set of positive elements we also need to have that P does not contain 0 and that the ring is P u {0} u -P. (Note that because P is closed under addition and does not contain 0, P is disjoint from -P). One can easily verify that this condition is satisfied wherever I've used it.

Also, this approximately completely fails to follow Martin Isaac's Algebra: A Graduate Course. At any rate, this is the main reference text I used in writing this, and at least one of the proofs has been lifted out of it.

Let K be any ordered field. Note that any field homomorphism f : R -> K is automatically order preserving. Why? Because every positive element of R is a square, so if x > 0 then x is a square, so f(x) is a square and so positive.

Q has the same property - any field homomorphism from Q is order preserving - but there are rather a lot of elements of Q which are not squares.

So, which fields have this property?

Another way to look at it is to ask when there are multiple orderings on a field which are compatible with the field structure. (These are clearly the same question - if f is a non-order preserving homomorphism then x <~ y iff f(x) < f(y) is a different ordering on the field. If there is a different ordering on the field then the identity map is a non-order preserving homomorphism).

It turns out that the answer is that those elements which are determined to be positive are precisely those which are sums of squares. Note that m/n = mn (1/n)^2, so every positive rational is a sum of squares, and thus the ordering is fixed (though we knew that anyway). It essentially comes down to proving the following statement:

Let K be a field which admits an ordering and let x \in K be such that x is not a sum of squares. Then there is an ordering on K which makes it into an ordered field and has x < 0. Thus if neither x nor -x are sums of squares there are orderings in which x is negative and orderings in which x is positive.

We'll prove this in a bit.

Anyway, all this motivates the following definition:

Let K be a field. K is formally real if -1 is not a sum of squares.

It will turn out that K is formally real iff there is an ordering on K which makes K into an ordered field.

R has the property that no algebraic extension of it is formally real (this is trivial, because the only algebraic extensions of it are R and C). We call this being real closed.

We will be very interested in studying field extensions of R which are real closed, so it's time to develop some theory for them.

Proposition 1:

A field K is formally real iff 0 is not a sum of non-zero squares.

Proof:

First suppose that K is formally real. Suppose x_1^2 + ... + x_n^2 = 0. Then (x_2/x_1)^2 + .. + (x_n/x_1)^2 = -1.

Conversely, if x_1^2 + ... + x_n^2 = -1 then x_1^2 + ... + x_n^2 + 1 = 0.

QED

Corollary: Formally real fields have characteristic 0.

Proposition 2:

Let K be formally real and x an element of K such that -x is not a sum of squares. Then K( sqrt(x) ) is formally real.

Proof:

Set t = sqrt(x). Suppose sum ( a_n + b_n t)^2 = 0 with not all the a_n, b_n equal to 0.

sum(a_n + b_n t)^2 = sum (a_n^2 + b_n^2 t) + 2t sum a_n b_n. So, using the uniqueness of the representation, sum a_n^2 + t sum b_n^2 = 0.

Now, because K is formally real we must have some of the b_n non-zero (as else it would just be \sum a_n^2 , which we know is non-zero). Further, sum b_n^2 != 0 (because K is formally real). Thus -t = sum a_n^2 / sum b_n^2

Now note that (sum b_n^2)^{-1} = sum (b_n / sum b_m^2)^2. So (sum b_n^2)^{-1} is a sum of squares. Thus -t is a product of sums of squares, and thus a sum of squares, contradicting the hypothesis.

QED

Proposition 3:

Let K be real closed. For every x != 0 precisely one of x and -x has a square root.

Proof:

At most one of x and -x can be a sum of squares, because if not then 0 would be a sum of non-zero squares. Without loss of generality we may say -x is not a sum of squares. Then K(sqrt(x)) is formally real, so by maximality of K we have K(sqrt(x)) = K, and x has a square root.

QED

Proposition 4:

Let K be real closed. The set of squares is closed under addition.

Proof:

Essentially the same as above. If x is a sum of squares then -x cannot be, so we may adjoin sqrt(x) to the field, and by maximality it's already there.

QED

Proposition 5:

Every real closed field has a unique order on it which makes it into an ordered field.

Proof:

If we denote the set of squares by S, we have shown that S is closed under addition. It is obviously closed under multiplication. Further because for every x != 0 exactly one of x or -x has a square root, we have that K = S u {0} u -S. So S works as a set of positive elements for the order. Further, in any order all squares have to be positive, so it is the only set that does.

QED

Proposition 6:

Every formally real field has an algebraic extension which is real closed.

Proof:

Embed K in its algebraic closure, say L, and apply zorn to the set of formally real subfields of L containing K.

QED

We call such an extension a real closure.

Corollary:

Every formally real field has an order which makes it into an ordered field.

However, unlike the algebraic closure there is not a unique real closure. The following will shed some light on this:

Proposition 7:

Let K be an ordered field. There is a real closure RK such that the canonical ordering on RK extends that on K.

Proof:

Use zorn to adjoin a square root of every positive element and then take the real closure.

QED

This apparently *is* unique. I will need that, but I currently have no idea how to prove the uniqueness statement. However, we don't need that to show that real closures are not unique. Take two different orderings on Q(x) - say one which is archimedean and one which is not. Then the real closures extending these orderings can't be isomorphic.

We'll now explore some more of the algebraic properties of being real closed and find an internal characterisation.

Proposition 8:

Let K be formally real and L a field extension of K with [L:K] odd. Then L is formally real.

Proof:

We'll prove this by induction on n = [L:K]. If n = 1 then there's nothing to prove, so assume n > 1.

Pick t in L \ K. If L != K(t) then [K(t):K] and [L:K(t)] are both odd and less than n, so by applying the inductive hypothesis K is formally real. So assume L = K(t).

Now, if L is not formally real then there exists a_k with \sum a_k^2 = -1. We may write a_k = g_k(t) where g_k is a polynomial of degree < n with coefficients in K.

Now, let h(X) = 1 + \sum g_k(X)^2. We have h(t) = 0, so the minimial polynomial of t (say f) divides h.

Let m be the maximum degree of the g_i. Note that m < n. Then deg(h) <= 2m. The coefficient of X^2m is a sum of squares, so because K is formally real must be non-zero. Thus deg(h) = 2m, and deg(h/f) = 2m - n < n.

Take an odd degree irreducible factor of h/k, say l, and adjoin a root s of l to K. Then [K(s):K] = deg(l) <= 2m - n < n, so by the inductive hypothesis K(s) is formally real. But h(s) = 0, so -1 = sum g_i(s)^2, contradicting that K(s) is formally real.

QED

Corollary: Let K be real closed. Every odd degree polynomial in K has a root.

In fact this lets us give an equivalent internal definition of real closed.

Proposition 9:

Let K be a field such that:

a) The set of non-zero squares is closed under addition.
b) For every x either x or -x has a square root.
c) Every odd degree polynomial has a root.

Then K is real closed.

I'm not going to prove this now - the proofs I've seen all need too much Galois theory, and I don't really need the result. The basic idea of the proof is to show that every irreducible polynomial in K has degree 1 or 2 and so that any proper algebraic extension of K contains a square root of -1.

Note this is a first order characterisation (if you don't speak logician this means that it only quantifies over elements of K. e.g. 'forall x \exists y y^2 = x or y^2 = -x' is first order but 'forall n nx != 0' is not because it quantifies over n). This means that because ultrapowers are elementarily equivalent to R (satisfy the same first order statements) that all ultrapowers are real closed. So we already have an important example of real closed field extensions of R.

This has all got a bit longer than I expected. I'll probably tidy up this post and LaTeX it. Also I'll try and figure out how one proves that uniqueness statement, as I'm going to need it.

Super-real fields 2: Hyperreals and Ultrapowers

This article has more of an analysis and set theory flavour than the last one. I'm trying to keep it fairly light, but it neccesarily creeps in. This is a common theme in this subject - it bounces back and forth between analysis, set theory and algebra. This is one of the reasons I like it.

The hyperreals, R*, were originally studied for model theoretic reasons. They're a particular example of what's called an ultrapower, which have the nice property that they're elementarily equivalent to the original structure (that is to say they satisfy precisely the same first order statements). Note that 'x is an infinitesimal' is not first order, because it requires you to quantify over the naturals. Because of this they provide an alternate approach to real analysis and calculus which some people quite like, as it allows one to make sense of a lot of formal manipulations with infinitesimal and infinite quantities. A good example of this is Keisler's "Elementary Calculus: An Approach Using Infinitesimals".

I'm not going to go into this in depth here. This is just for background. I'm quite interested in learning more about ultrapowers, so I may at some later date write an article on them as I do so, but it's really tangential to this subject.

Note: I've performed a slight dodge in talking about 'the' hyperreals. They actually need not be unique up to isomorphism and can depend on the choice of ultrafilter. This depends on the underlying set theory. I really don't want to have to worry about this at the moment, so I'm instead going to assume they aren't unique and refer to them as ultrapowers instead. Later on once we've developed a lot of theory which doesn't depend on the underlying set theory (at least past ZFC. I don't intend to quibble about the axiom of choice at all) we'll be able to give an easy condition which guarantees uniqueness.

Alright, enough logic and set theory. I've probably scared off all the algebraists. :)

The reals are in some sense quite small. They have a countable dense subset, and they inherit a lot of behaviour from this. In fact, every subset of R has a countable dense subset.

Ultrapowers are on the other hand a lot larger. We have the following result:

Proposition 1:

Let K be an ultrapower and A, B be countable subsets of K such that for every a in A and b in B we have a < b. Then there exists x such that for every a in A and b in B we have a < x < b.

I'm temporarily blanking on the proof of this. It should just be a clever diagonalisation argument. I'll edit it in later.

We'll need to consider such things quite a lot, so we'll introduce the following notation: A << B if for every a in A, b in B we have a < b. We will usually write A << {x} as A << x.

This will give us an awful lot of infinitesimals, because it means we can pick any decreasing sequence of positive numbers and there will be a smaller one. e.g. there is some x_0 such that 0 < x < 1/n for every n. Then there is some x_1 such that 0 < x_1 < x_0^n for every n, x_2, etc. then there is y with 0 < y < x_n for every n, etc. You get the idea. There are a lot of infinitesimals.

Given a bit of faffing we can turn the above argument into the following:

Proposition 2:

There is a strictly decreasing sequence of positive infinitesimals of length aleph_1.

I think this will be important later, but I'm not sure.

Also, note that proposition 1 means that there are no convergent sequences in the order topology of K except those which are eventually constant. Proof:

Suppose we have a sequence x_n -> x such that for all n, x_n != x. Let A = { x_n : x_n < x} and B = { x_n : x_n > x}. Then we can find s with A << s < x and t with x < t << B. Then the sequence x_n never enters the neighbourhood (s, t) of x, so does not converge to x. Contradiction.


We can restate proposition 1 as:

Proposition 1':

Let K be an ultrapower and A, B \subseteq K be countable with A << B. Then there exists x with A << x << B.

This motivates the following definitions:

Let X be an ordered set and A, B be subsets which are not both empty. Then (A, B) forms a pregap if A << B.

If A << x << B then we say x interpolates the pregap (A, B).

A pregap is a gap if it has no interpolating element. It is countable if both A and B are countable.


So, one more restatement of the proposition:

Proposition 1'':

Let K be an ultrapower. K has no countable gaps.

We'll turn this into a definition. An ordered set is an eta_1 set if it has no countable gaps.

Also, an ordered set is an alpha_1 set if every subset of it has a countable order dense subset. (A <= X is order dense if for every x, y \in X there is an a in A with x < a < y). e.g. R is an alpha_1 set.

The intuition here being that alpha_1 sets are small, eta_1 sets are large. We will mostly be considered with ordered fields which are eta_1 as ordered sets. Ultrapowers were our first example of such, but we'll see a lot of other examples later.

It will apparently turn out to be very important to study what kinds of gaps our ordered field extensions have. I'll include more on that next time.

Super-real fields 1: Basic background material

This is just to establish some background. It doesn't contain an of the hard stuff.

The basic idea is that we want to study ordered field extensions of R. Why? Why not? There turn out to be various interesting reasons for studying them, but for now I'm just going to do it for its own sake. I'll try and justify some of this once I understand the justifications.

Recall that the reals, denoted by R, are the unique complete ordered field, where here complete means that every subset which is bounded above has a least upper bound.

The reals have an important property. They are Archimedean. That is to say that for every real x there is some natural number n such that n > x. Proof: Suppose x where an upper bound for N (the natural numbers). Then so is x - 1. Thus there can be no least upper bound, so by completeness the natural numbers are not bounded above.

This is equivalent to saying that for any x > 0 there is some n with 1/n < x. So, in short, R contains no infinitely large numbers or infinitely small non-zero numbers.

We'll turn this into a definition: Let K be an ordered field. If |x| < 1/n for every natural number n then x is an infinitesimal. If |x| > n for every natural number n then x is infinitely large. If x is not infinitely large then it is finite.

Note that a field contains a non-zero infinitesimal iff it contains an infinitely large number, as if x is infinitely large then x^{-1} is infinitesimal and vice versa.

Now, we have the following important observation:

Proposition 1: Any ordered field properly containing R contains an infinitely large number.

Proof: Let K be such an ordered field and x \in K \setminus R. If x is infinitely large then we're done. Else { t \in R : t < x } is bounded above. Let y = sup { t \in R : t < x }. Then x - y is a non-zero infinitesimal.

QED

So, in order to study ordered field extensions of R, we're really going to have to study infinitely large and infinitesimal additions to R. It turns out to be more tractable to study the infinitesimals.

We make the following definitions:

Let K be an ordered field. K^# is the set of finite elements of K. K^0 is the set of infinitesimal elements of K.

Proposition 2:

• These are both subrings of K (don't start with me about rings needing to have a 1).


• K^0 is an ideal of K^#.


• Every element of K^# \setminus K^0 is invertible. So K^0 is the unique maximal ideal of K^#, and K^# is a local ring.

Looking at our proof of proposition, we've actually proved something more.

Proposition 3:

Let K be an ordered field containing R. For every x \in K^# there is a real number y such that x - y \in K^0. Further, this number is unique. We may thus define a map st : K^# -> R by letting st(x) be the unique real that is infinitesimally close to x. Then st(x) is an order preserving ring homomorphism with kernel K^0. Thus K^# / K^0 is isomorphic to R.

We'll now recall some useful tricks for defining total orders on rings.

Recall that specifying a total order on a ring is equivalent to specifying a set of positive elements which is closed under addition and multiplication. We may then define x > y iff x - y is positive.

An easy example. Let K be an ordered field. We can turn K[X] into an ordered ring by letting the positive elements be the ones whose highest power coefficient is positive.

Now, suppose we have a totally ordered integral domain, R. We can turn its field of fractions into a totally ordered field by declaring the positive elements to be the ones of the form x/y, for x, y positive. In particular we can turn K(X) into an ordered field.

Note that having done this we have that 1/X < y for any y \in K with 0 < y. So we've added an infinitesimal which is smaller than all the positive infinitesimals of K.

In particular this gives us our first lot of examples of ordered field extensions of R. It's a bit trivial, but it works for starters. Note that R(X)^# is the set of f/g with deg(f) <= deg(g) and R(X)^0 is the set of f/g with deg(f) < deg(g).

For our first genuinely non-trivial example of an ordered field extension of R, we turn to the hyperreals.

Let U be a free ultrafilter on R. Consider the direct product R^N and define an ideal I_U = { f : { x : f(x) = 0} \in U }. This may easily be verified to be maximal, so R^* = R^N / I_U is a field. Define [f] \leq [g] iff f \leq g. This may easily be checked to be well defined and turn R^* into an ordered field.

But more on these later.

A Mathematician's Scratchpad

I use my site for posting serious articles about mathematics. I'll use this to put up general musings on the subject, sketch notes of things, etc. I'm going to start with using it to put up some articles based on me trying to understand Dales and Woodin's "Super-Real Fields" and see where it goes from there.

David