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.

KermitI saw this article on other page. It had identical

sense but in a completely different words, they use advanced article

rewriter, you should read about it, just type in google:

Niachight’s rewriter