# Rational functions in one indeterminate - useful concept?

to mr. tashi, re post #21: it is true such "substitutions" seem odd, but maybe one can make sense of it by abstracting the ordinarily intuitive process of composition, or substitution.

So saying one can "substitute" an element a of R for the symbol π, is just the fact that one can define a ring homomorphism from Z[π]-->R that sends π to a. I.e. "substituting a for π" just means sending π to a by a ring map. Does this have any (at least logical) appeal?

Yes, it does. Can each homomorphism ##\mathbb{Z}[x] \rightarrow \mathbb{R}## be realized as such a map?

As I understand the ring of polynomials in one "indeterminate" and similar mathematical constructs, they are defined essentially as strings of symbols together with rules for deriving strings from given strings. If done properly, this would be phrased in the same style as is used to define computer languages and in the modeling of formal logic as strings of symbols. I've never seen a algebra text that adopts that style of presentation. In algebra texts it is taken for granted that the reader understands how "substitution" can be used to map a polynomial in one indeterminate to an element in another ring. The concept of "substitution" is so intuitive and universal in mathematics that it isn't formalized. Using the concept of "substitution", we can implement the idea of "composition" of polynomials.

What make me intellectually uneasy is not understanding what, if any, limitations are imposed on the concept of "ring" by assuming it must have a structure that is a string of symbols whose operations are described by manipulating the strings. There is a viewpoint in Group Theory where groups are described in such a manner. I haven't seen this view extended to representing rings.

WWGD
Gold Member
A few remarks: if f(x,y) is an irreducible polynomial with complex coefficients, then the locus f=0 defines a plane curve N in C^2, whose ring of polynomial functions equals the integral domain C[x,y]/(f). I.e. this is the ring of polynomial functions on the plane restricted to N, so that those which equal zero on N are set equal to zero. The field of fractions of this domain is called the field of rational functions on N. This field is isomorphic to the field of meromorphic functions on the Riemann surface of N, i.e. the compactified desingularization of N. It is a theorem that a compact non singular connected one dimensional complex manifold is determined up to isomorphism by its field of meromorphic functions.

The special case where f = y, gives the field C(x), rational functions of one variable. the associated Riemann surface is the Riemann sphere, hence a Riemann surface is isomorphic to the Riemann sphere if and only if its field of meromorphic functions is isomorphic to the field C(x). An irreducible plane curve N is called rational if its field of rational functions is isomorphic to C(x), if and only if its Riemann surface is isomorphic to the Riemann sphere.

So fields of rational functions are a fundamental invariant of Riemann surfaces, and the simplest example is C(x). Notice that for almost any function f as above, (except for f = x-a), the field C(x) is a subfield of the fraction field of C[x,y]/(f), corresponding to the fact that we can project a plane curve (other than a vertical line) surjectively onto the x axis. Thus it is natural to study any Riemann surface via such a projection, making the surface a finite cover of the Riemann sphere (the compactified x axis). Correspondingly, any rational function field of any plane curve is a finite algebraic extension of the basic field C(x). In this way the field C(x) is a very basic tool for studying all Riemann surfaces. Riemann himself for example proved that if a Riemann surface S has topological genus g, and if d is the smallest integer d ≥ (g/2)+1, then S is a finite cover of the Riemann sphere of degree d. Equivalently, the rational function field of S is an algebraic extension of C(x) of degree d.

E.g. if S is a torus of genus one, then S is a double cover of the Riemann sphere, (branched at 4 points). Equivalently a rational function field of genus one is a quadratic extension oif the field C(x). E.g. the elliptic curve y^2 = x(x^2-1), has rational function field generated by x and y, where y is quadratic over C(x), by virtue of this defining equation. But quadratic rational function fields can have any genus g, by considering curves with equations of form y^2 = x(x-1)(x-2).....(x-(2g+1)).

Remark: Thus some curves of genus g are 2-1 covers of the sphere, but the curve x^4+y^4 = 1 of genus 3, is a 3-1 cover but not a 2-1 cover of the Riemann sphere. I.e. the curves of genus g listed 2 or 3 lines above, are all double covers of the sphere, hence are not general curves of genus g, but special ones.
Isn't a cover a local homeomorphism? If so, how can a 1-d curve be a cover for a sphere? I guess I am missing something obvious. Is it just that the topological 2-sphere is a 1d Complex manifold?

Infrared
Gold Member
A (nonsingular) complex curve is a 2D as a real manifold. And here he means a branched cover (https://en.wikipedia.org/wiki/Branched_covering), which is not everywhere a local homeomorphism.

WWGD
Gold Member
A (nonsingular) complex curve is a 2D as a real manifold. And here he means a branched cover (https://en.wikipedia.org/wiki/Branched_covering), which is not everywhere a local homeomorphism.
Yes, thanks, wanted to verify it was a complex curve so that dimensions matched up.

mathwonk
Homework Helper
2020 Award
my bad, mixing my metaphors. If I am going to speak of the projective line as the [riemann] sphere, then i should speak of a [complex curve] as a riemann surface. thanks for this clarification.

• WWGD
WWGD
Gold Member
my bad, mixing my metaphors. If I am going to speak of the projective line as the [riemann] sphere, then i should speak of a [complex curve] as a riemann surface. thanks for this clarification.
Kind of strange, to me at least, to see how Complex lines become spheres through projectivization.

mathwonk
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