Integral Domains, Fractional Linear Transforms

Polynomials as Functions

Let R be a ufd and let F be the fraction field of R. Recall that R[x] is the ring of polynomials over R, which is also a ufd. Let Q be the fraction field of this ring. In other words, Q consists of quotients of polynomials, reduced to lowest terms. We won't write x2-1 over x-1, since that is the same as x+1. don't forget to cancel any common factors that come from R.

An element of Q can act as a function from R into F, or from F into F. Substitute for x and plug and chug.

Let's see which elements of Q produce the zero function. Any polynomial p(x) has finitely many roots over F, so if R is infinite, every polynomial (other than 0) produces a function that is nonzero most of the time. As a corollary, different polynomials produce different functions, since there difference is a polynomial that cannot produce 0 for every x.

Extend this to Q. If a/b and c/d produce the same function, except perhaps for a few poles where c and d become 0, cross multiply and write ad = bc. These produce the same functions, and are in fact the same polynomial. The fractions are reduced, so a and b are relatively prime, meaning a divides c. Since c and d are relatively prime, c divides a. Thus a = c and b = d, and the fractions are identical. Distinct elements of Q determine unique functions from R into F, or from F into F.

Of course there may be functions from R into F that cannot be implemented as quotients of polynomials, such as Ex in the reals.

If you want a polynomial that implements the zero function, let R be the finite field Z7 and consider x7-x.

If R is finite, as shown by the above example, let S = R[w], where w is an indeterminant. Thus S is an infinite ufd that contains R. The functions of Q are quotients of polynomials taken from R[x], but they are also quotients of polynomials drawn from S[x]. With respect to S, which is an infinite domain, distinct elements of Q implement distinct functions from S into its fraction field. It is sometimes convenient to apply this trick when the base ring is finite.

Next, we want to look at function composition, and that means we need three functions, three elements of Q, called g h and j. As you recall, these are reduced quotients of polynomials from R[x]. Remember that g h and j also represent functions from F into F.

The function g can be composed with the function h, and then the function j. This composite might be writtne j(h(g(x))). Substitute and expand, and substitute and expand, to get a rational function in Q.

Wait a minute! How do you know it's a function in Q? Let the first function be 5, or 5/1 if you prefer, and let the secont be 1 over x-5. Fold the first into the second and get 1/0, which is not a valid member of Q.

Oops. You caught me. Let Q+ be the rational functions of degree 1 or greater. There is an x upstairs, or downstairs, or both. Let g be such a function. Finitely many values of x map to 0 in g, namely the roots of the polynomial in the numerator. What is the preimage of the constant a/b under g? The numerator of g, times the constant b, has to equal the denominator of g times a. Move everything to one side and we are looking for the roots of a polynomial in R[x]. Like 0, every element of F has a finite preimage under g, as long as g belongs to Q+.

Consider the application of g, and then h. The denominator of g is a polynomial with finitely many roots. These roots are poles, and g is not defined there. The same goes for h. If h is not defined for some x, take the preimage under g to find finitely many values that lead to poles in h. Combine these with the poles of g and we have a finite number of values where h(g) is not defined. Remember, R is infinite, and if it's not, we can embed R into R[w] as we did before; so think of R as infinite. The composite h(g(x)) is now well defined on most of R. It is not something over 0.

Suppose h(g(x)) is a constant c in F. Pull c back through h to find a finite preimage, and pull these points back through g to find a finite preimage. Yet R is infinite, so this is a contradiction. The composition of functions in Q+ gives another function in Q+. Function composition is a well defined binary operator on Q+.

Composition is not commutative. Let g(x) = 2x and h(x) = x+1. Now h(g(x)) = 2x+1, while g(h(x)) = 2x+2.

Although composition is not commutative, it is associative. Think of g h and j as functions, not polynomials. Function composition through arbitrary sets is associative. Take any x in F and map it through g, through h, and through j, and find another element of F. It doesn't matter if g compose h is applied first, then j, or if g is applied, then h compose j. Either way x maps to j(h(g(x))). Since the two functions are the same, and since S is infinite, the formulas that implement those functions are identical. Composing elements of Q+ is associative. Combine this with the identity function x/1, and Q+ becomes a monoid.

In general, you cannot compose h with something to get x/1 back again. Let h be x2 over the reals. Substitute h in any other polynomial, and all terms have even degree. The result is not x/1, hence h is not invertible. But if we restrict attension to a smaller subset of Q+, we might find invertible transformations, and a group.

Fractional Linear Transforms

Let Q1 be the functions in Q+ that have degree 1. There are no terms beyond x1, and x appears upstairs, or downstairs, or both. These are called fractional linear transforms. After all, each is a quotient of linear transformations - shifted linear transformations if you want to be technical. Examples include 1/x, (x+3)/(x-3), (x+5)/2, but not 0/x or 3/5, and certainly not x2/7.

These transforms can be composed, as described above. Although it is not immediately obvious, the result is another fractional linear transform. We know the composition lies in Q+, we only need show the result lies in Q1, i.e. its degree is 1. Let g(x) be a fractional linear transform and replace x with h(y), where y is another quotient of monomials. In fact, let h(y) have denominator jy+k. When x ← h(y) in the numerator, the numerator becomes a fraction with denominator jy+k. Make the same substitution downstairs and it two becomes a fraction with denominator jy+k. We now have a quotient of fractions, where each fraction has denominator jy+k. This denominator cancels, leaving a quotient of monomials, which is another fractional linear transform.

If g does not have an x above and below, g(h) clearly belongs to Q1.

Fractional linear transforms are closed under composition, and composition is associative, with x/1 acting as the identity element. Therefore Q1 is a monoid inside the larger monoid Q+, inside Q. But is Q1 a group? Let's find the inverse of a fractional linear transform.

Let g(x) = (ax+b)/(cx+d), where the coefficients a b c d are arbitrary elements of R, or perhaps F. Set g(x) = y and solve for x, giving h(y) = (dy-b)/(a-cy). Verify g(h(y)) = y/1 and h(g(x)) = x/1. We only get in trouble, i.e. a composite function of 0/0, if ad = bc, also written a/b = c/d. But then g(x) and h(y) reduce to constants in F, and are not in Q1 in the first place. Thus Q1 is a group under function composition.

Let g(x) be the general transform (ax+b)/(cx+d), with inverse (bx-d)/(a-cx) as described above. Now g(x) is 1-1 and onto, except when cx+d = 0, or when a-cx = 0. Create a mythical point at infinity and let g(-d/c) = ∞. At the same time, map ∞ to a/c, a point that is otherwise inaccessible. Now g maps the augmented field onto itself, and is invertible. Combine with the earlier results, and Q1 has become a group of invertible functions mapping the augmented field F+∞ onto intself.

Three Points Determine A Transform

For every distinct a, b, and c in F, a fractional linear transform maps a to 0, b to 1, and c to ∞.

g(x) = (b-c)/(b-a) × (x-a)/(x-c)

If a b or c = ∞, the following transforms do the trick.

   a = ∞: (b-c) over (x-c)
   b = ∞: (x-a) over (x-c)
   c = ∞: (x-a) over (b-a)

Carry three points onto 0 1 and ∞, then on to three other points, and there is a transform that carries any three distinct points in F+∞ onto any other three distinct points in F+∞.

Suppose g(x) and h(x) both carry three points to the same image points. Subtract the functions to find a transform that maps three different points to 0. Yet every fractional linear transform is invertible, so the difference g-h must be a constant. In fact it must be 0, hence g = h. Given three points in F+∞, their image, i.e. any three distinct points in F+∞, corresponds uniquely with a fractional linear transform.

This makes sence if you think about it. Start with (ax+b)/(cx+d) and clear c from the numerator and denominator, leaving (ax+b)/x+d). Three coefficients determine the transform.