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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

YoYo
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    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
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    Thanks. This question on another SE site http://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
  • Related: https://math.stackexchange.com/questions/3353809. In short, the answer in the MO question linked above is "The fundamental theorem of algebra tells us that the number of solutions to $f(z) = a$ is the maximum of the degrees of the numerator and denominator." – Watson Feb 08 '21 at 15:20

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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.

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

Magdiragdag
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  • 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. – YoYo Aug 29 '16 at 19:43
  • @Alain24 Can you then clarify what your question actually is? Which quantities do you already know to be equal and which degree do you mean where? – Magdiragdag Aug 29 '16 at 19:56
  • 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 : https://en.wikipedia.org/wiki/Degree_of_a_continuous_mapping . Thank you. – YoYo Aug 29 '16 at 20:02