# Degree theory of rational fractions over $\mathbb{C}$.

Could someone explain to me please, why, if : $f = \dfrac{P}{Q} \in \mathbb{C} (X)$, is a rational fraction with coefficients in $\mathbb{C}$, with $P$ and $Q$ are two non zeros coprime polynomials, then : $\mathrm{deg} f = [ \mathbb{C} (X) : \mathbb{C} (f) ] = \mathrm{max} \{ \ \mathrm{deg} P , \mathrm{deg} Q \ \}$, where : $[ \mathbb{C} (X) : \mathbb{C} ( f) ]$ is the degree extension of $\mathbb{C} (X)$ over the subfield $\mathbb{C} (f)$ generated by $f$ ?

edit : I mean by degree of a map $f$ as it is defined here : https://en.wikipedia.org/wiki/Degree_of_a_continuous_mapping . What is the connexion between the three members of the equalities above ? Thank you

• Thanks for the edit. Could you specific what the manifold in question is? The Riemannsphere or something else? – quid Aug 29 '16 at 19:56
• $f \in \mathbb{C} (X)$ is a rational fraction with coefficients in $\mathbb{C}$, and $Z$ the finite set of its poles. We identify $\mathbb{C}$ with the domain of an affin chart of $\mathbb{P}_1 ( \mathbb{C} )$, and we denote $\infty$ its complement point. – YoYo Aug 29 '16 at 19:58
• Thanks. This question on another SE site mathoverflow.net/questions/198648/degree-of-a-rational-function might answer you question. It's a bit terse though. – quid Aug 29 '16 at 20:01
• No problem. I also learned something via this post, so that's great. – quid Aug 29 '16 at 20:10

A solution to $f(X) = c$ is a root of $P(X) - cQ(X)$, which generically has $\max\{\deg P , \deg Q\}$ roots.
Because the minimum polynomial of $X$ over ${\mathbb C}(f)$ is $Q(T) f(X) - P(T)$, which has degree $\max(\deg P,\deg Q)$.
• I mean by degree of a map $f$ as it is defined here : en.wikipedia.org/wiki/Degree_of_a_continuous_mapping . What is the connexion between the three members of the equalities above ? Thank you. – YoYo Aug 29 '16 at 19:43
• The quantities i already know is $[ \mathbb{C} ( X ) : \mathbb{C} ( f ) ] = \mathrm{max} \{ \mathrm{deg} P , \mathrm{deg} Q \}$, what i would like to know is why is $\mathrm{deg} f = [ \mathbb{C} ( X ) : \mathbb{C} ( f ) ]$ such that the degre notion that i use is as it is defined here : en.wikipedia.org/wiki/Degree_of_a_continuous_mapping . Thank you. – YoYo Aug 29 '16 at 20:02