Let G be a group that is linearly ordered. This means group operations respect order. Add u to v (in either order) and you get something bigger than v, assuming u is bigger than e, where e is the group identity. If u < e then uv and vu are less than v.
You can think of e as 0, the center of the "values" of G, but I don't want to call it 0, because we're going to take linear combinations of elements of G wiht coefficients in R. So c*e could be a nonzero element in our ring, and it would be rather confusing to call this c*0. So c*e it is.
Let S be the set of linear combinations of elements of G with coefficients in R, such that the elements of G form a well ordered set, relative to the linear ordering on G. This does not mean G itself is well ordered. Look again at the laurent example. The group G is Z, the integers, which is not well ordered. Yet every laurent series is bounded below. It has a least term, and every subcollection of the terms in a laurent series has a least term. When G = Z, S is the laurent series over R.
By convention, coefficients are placed on the left.
Addition and subtraction occur term by term, as with any series. Since the union of two well ordered sets is well ordered, the terms of the sum are well ordered, and the sum lies in S.
Multiplication requires a twist, which is defined by a group homomorphism σ from G into the ring automorphisms of R. When defining the group of automorphisms, use the reverse convention, so that the composition of two automorphisms is the second followed by the first. You'll see why this is necessary later on.
When G = Z, a cyclic group, σ is determined by σ on the generator x. The twist on x2 has to be the twist of x invoked twice, and so on. This is the twisted laurent example. Groups that are not cyclic can support more complicated maps into the automorphisms of R.
With the twist in hand, consider the product of two individual terms cu * dv, where c and d lie in R, and u and v lie in G. The result is cσu(d)uv. When d is pulled past u, it is run through the automorphism associated with u. Do this for all terms in the first series cross all terms in the second, and regroup to find the product of two series.
To prove that a regrouping of terms is possible, we need a lemma. A countable sequence of elements of G can always be packed into the closed interval [0,1]. Place the first point at 1/2, and put the second point at 1/4 or 3/4, depending on whether it is smaller or larger than the first. The third point can be placed on a rational number with denominator 8, according to its relationship to the first two points. Continue this forever, until the entire sequence is embedded, in a manner that respects the order of G. If the sequence is infinite it implies a cluster point that I will call l. A subsequence now approaches l from above, or below.
Let h be an element of G, and suppose an infinite number of products of the form ui*vi contribute to h. Restrict atention to a countable subset of these products. The group elements drawn from the first series, u1 u2 u3 etc, are linearly ordered, and can be embedded in the unit interval. The group elements drawn from the second series, v1 v2 v3 etc, are linearly ordered, and can be embedded in a second copy of the unit interval. In each case we have a limit point, and a subsequence that converges to said limit. Looking at u, the subsequence cannot converge from above, else the set of group elements in the first series would not be well ordered. So it converges to l from below. Restrict u to this subsequence. Now as each ui gets larger, the corresponding vi has to get smaller, so that their product remains fixed at h. This violates the well ordering of the terms of the second series. Therefore finitely many products from the first series cross the second contribute to h, and the coefficient on h is well defined. Apply this across the board, and the product of two series in S is well defined.
Does the product lie in S? Suppose there is an infinite sequence h of descending terms in the product. Start at the top and find u0 and v0 from the two series such that u0v0 = h0. Step down to h1 and find u1*v1 = h1, and so on. Embed u and v in their closed intervals, and note that a subsequence of u approaches a limit from below. Since the points of h decrease, the corresponding points of v must also decrease, to compensate for the fact that u is increasing. This builds a descending chain in v, which is impossible. Therefore the terms of the product are well ordered, and the product belongs to S.
Next we need to prove S is a ring. This is an exercise for you. Here are two hints.
1. The distributive law works because each twist is a ring automorphism. We can pull a sum past an element of G, or pull each summand past G individually; the result is the same.
2. Multiplication is associative because σ is a group homomorphism on G. The twist of G1G2 is the twist of G2 followed by the twist of G1. Expand c1G1 * c2G2 * c3G3 in two different ways and you'll see what I mean.
In summary, the twisted series of G by R is a ring. The additive identity is 0, and the multiplicative identity is 1e. The group ring of G by R, with the aforementioned twist, embeds in S. These look like polynomials, having finitely many terms. If R is commutative, and σ is trivial, then S is commutative.
Assume the product of two series is 1, or 1e if you prefer. This is the least term of the product, derived from the least terms of the two operands. One of the two coefficients is run through an automorphism, but that moves units to units, and causes no trouble. Therefore the first coefficient of each operand is a unit in R.
The converse is not trivial, though the proof is a natural extension of the laurent case. Let w be a series that begins with a unit in R, which is also a unit in S. Every member of G is a unit, so nothing changes if we multiply w on the right by something in G. Choose a member of G that is the inverse of the least group element in w. This shifts w so that the least term is e. This is analogous to shifting a laurent series so that it becomes a power series with a nonzero constant term.
If e has a coefficient of c, multiply by c inverse on the left. We can do this because the first coefficient is a unit in R, which is a unit in S. Now w begins with 1. For convenience, change the notation, so that our series is 1-w. Thus w holds the positive terms of the invertible series.
Let z be the infinite sum 1+w+w2+w3+… We need to prove that regrouping is possible (whence z is well defined), and the terms of z are well ordered (whence z belongs to S). Oddly enough, the second implies the first. Let's prove this lemma now.
Let w be as above, or any positive series for that matter, and let z be the collection of elements of G that is produced by all the powers of w. Assume z is well ordered. Select an h ∈ G that belongs to z, and consider all the tuples that contribute to h. Sort the terms in each tuple according to the order of G. For instance, if w4 contributes u1u22u3, we have arrange the factors in the product so that u1 < u2 < u3. There is no trouble sorting each tuple, since the tuples are finite.
Find the first tuple, lexicographically. It begins with u1, where u1 is the least group element in all the tuples that produce h. The next factor is u2, or perhaps another copy of u1, based on the least group element that appears in a tuple containing u1, that produces h. Continue in this manner, until the process stops. If it continues forever, making longer and longer tuples, the "rest" of the tuple, that takes up the slack, and produces the element h, belongs to z, and these remainders form a descending sequence in z. This means z is not well ordered, which is a contradiction. So there is indeed a least tuple, lexicographically, that equals h.
If this tuple has length n, there can be no other tuple of length n, for some of its elements would be larger, and that produces something bigger than h. Similarly, tuples longer than n go beyond h. So we are restricted to tuples of length n or less, which come from the lower powers of w. Since wi is well defined, tuples do not pile up at h infinitely often. If z is well ordered, z is well defined, and z belongs to S.
For any h ∈ G that is present in z, h is created by some wn. Specifically, h is the product of an n-tuple of terms drawn from w. The max of these terms, raised to the n, is at least h. So it is meaningful to define j(h) as the least u in w such that some power of u is greater than or equal to h.
Suppose the terms of z are not well ordered. There is at least one descending sequence in z. If h begins such a sequence, i.e. h is at the top, let j(h) be the corresponding element of w, as described above. These values of j form a well defined subset of the terms of w. Select an h that maps to the smallest possible j(h). I will simply call this j from now on.
Separate w into two pieces, the part before j, and the part at and beyond j. Call these α and β respectively. Realize that α could be empty, as it is in the laurent case.
The tuples produced by the powers of α build terms that are well ordered. Otherwise we could select a smaller h, and a smaller j inside α. Invoke the earlier lemma, and the tuples produced by the powers of α don't pile up at any particular term infinitely often. Therefore the sum zα = α+α2+α3+… is a well defined member of S.
Now let's go back to h, or h0 if you prefer, which is the top of a descending sequence hi in z. A tuple that lands on hi is a tuple from zα, times a tuple from zβ. The second tuple cannot be longer than n, since jn ≥ h0.
We already proved the terms of zα are well ordered. Since zβ can be restricted to a finite sum of products, its terms are also well ordered. Yet they combine to build a descending sequence. We already proved this is impossible. This is our long sought contradiction, hence the members of z are well ordered. Apply the earlier lemma, and z is well defined. Therefore the ring S contains the following member.
z = 1 + w + w2 + w3 + w4 + …
Now we have to prove (1-w)*z = 1. Since S is a ring, (1-w)*z is y for some series y. Note that y begins with 1, from 1*1 in 1-w cross z.
A positive term h in y has finitely many contributions from 1-w cross z. These can be grouped into pairs: -u cross v in -w cross wn, and 1 cross uv in 1 cross wn+1. These pairs cancel, and the result is 0. All the positive terms are 0, and (1-w)*z = 1.
In summary, a series that begins with a unit in R is a unit in S. Conversely, a unit in S begins with a unit in R if R is an integral domain, or if S is restricted to nonnegative terms, such as the formal power series.
If R is a division ring then so is S. This follows directly from the above.
Let σ be trivial, the identity map on R, and let R be a division ring. Let S be the twisted series of G by R, which is a division ring. Within this ring lives a subring, the free R algebra on A B and C.
In general, the free R algebra on any set of well ordered indeterminants embeds in a division ring. At first this seems like a trivial result. After all, every integral domain embeds in its fraction field. But things are different in the noncommutative world. There are domains that don't embed in any division ring. such a domain has to be infinite, for a finite domain is a field. I don't have an example handy, but these pathological domains do exist.