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.