Construct all rational functions f (quotients of polynomials) in C such that (f)=6[0]−4[5], where (f) is the divisor of the function f. My answer/attempt: From my understanding of divisors, there seems to be exactly one function, namely, f(z)=z6(z−5)4. 1.Are there any other functions with this divisor? If not, why? 2.If we restrict ourselves to divisors of rational functions only, then if we are given a divisor of a △ on C, does there exists a unique rational function f with (f)=△?

Question

Construct all rational functions f (quotients of polynomials) in C such that (f)=6[0]−4[5], where (f) is the divisor of the function f.    My answer/attempt: From my understanding of divisors, there seems to be exactly one function, namely, f(z)=z6(z−5)4.    1.Are there any other functions with this divisor? If not, why?    2.If we restrict ourselves to divisors of rational functions only, then if we are given a divisor of a △ on C, does there exists a unique rational function f with (f)=△?

jorgegalar0xk · Accepted Answer

The answer to your question is yes, but one must invoke some machinery.    On a smooth projective curve, all functions have (f) a degree zero divisor. By adding a point at infinity we see that if △ is a divisor on ∀{C}1, then for any fϵC(z) we have that (f) is equal to △ if and only if when extended to PP1 we have (f)=△−deg(△)[∞].    By the Riemann-Roch Theorem, since PP{C}1 has genus 0, we have that dimC⁡H0(PP{C}1,O(D))=deg(D)+1 (whenever deg(D)≥−1). Now take the divisor D=−(△−deg(△)[∞]), which has degree 0. Then dimC⁡H0(PP{C}1,O(D))=1.    We have then proved that there is a 1-dimensional C vector space of functions f∈C(z) such that (f)≥−D, but since D has degree 0, equality must hold. Thus (up to multiplication by a scalar) there is a unique function f∈C(z) such that the divisor of f is △ on ∀{C}1.

Construct all rational functions f (quotients of polynomials) in CC such that (f)=6[0]-4[5]

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