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I'm recently trying to study basics about algebraic curves. However, apparently I'm still quite unfamiliar with the subject as there have occured 2 questions that seem quite basic to me, yet I don't directly see a way to make them clear to myself:

1.) For an (edit: smooth) algebraic curve $C/K$, why does $K(C)$ contain uniformizers for any point $P$ of $C$?

My - so long - first and only idea was to take a minimum degree set of generators of the maximal ideal $M_P$ at $P$ and try find a contradiction if all of them were also contained in $M^2_P$. However, notation of degree in this situation is not entirely clear to me and apparently there are lots of things that can go wrong, so I have quite the feeling this is the wrong path.

2.) Why is the ramification index $e_\Phi(P) := ord_P(\Phi^{\ast} t_{\Phi}(P))$ for a uniformizer $ t_{\Phi}(P)$ at $P$ well defined and doesn't depend on the choice of the uniformizer?

Thanks a lot in advance.

Louis
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1 Answers1

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  1. You have to assume that $C$ is smooth. Then every stalk $\mathcal{O}_{C,P}$ is regular, noetherian and of dimension $1$, hence a discrete valuation ring.

  2. You should say what is $\Phi$ is. I assume that it is finite morphism of curves. Then you can check that $e_{\Phi}(P)$ is the order of $\Phi^*(\pi_P)$ for every uniformizer $\pi_P$ at $P$. The reason is simply that uniformizes only differ by units, and $\Phi^*$ (as a ring homomorphism) preserves this property.

Martin Brandenburg
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