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

Question

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 $ ?

Thanks in advance for your help.

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

Answer 1

One quick and dirty way to see the degree of a continuous map is to pick a generic point in the target and count the number of its preimages. (If the point is not generic then you'd have to worry about multiplicities and other things.)

A solution to $f(X) = c$ is a root of $P(X) - cQ(X)$, which generically has $\max\{\deg P , \deg Q\}$ roots.

Answer 2

Because the minimum polynomial of $X$ over ${\mathbb C}(f)$ is $Q(T) f(X) - P(T)$, which has degree $\max(\deg P,\deg Q)$.