And why is $f(y)=g(y^p)$?

Thank you for the answer! Could you tell me what is $K^{1/p^\infty}$?

@Random You're right!

@PeterTaylor First sentence in Comtet's article: "Soit y une fonction de la variable complexe x...". A part from this, in equation (4) from the proof, he divides by the derivative wrt y, which could be zero in positive characteristic.

Can we have a Vitali set without choice?

By "the real world" I wanted to say the standard model of PA.

Sorry I was wrong. Now I realized that if CH+ZFC implies Goldbach then Goldbach has to be true in every model of ZFC, so that we can always use completeness theorem.