Let $C$ be an affine smooth curve over a field $k$. I want to prove that the coordinate ring $k[C]$ is a Dedekind domain, using the following proposition:
Let $A$ be a Dedekind domain with fraction field $K$. Let $L$ be a finite separable field extension of $K$ and denote by $B$ the integral closure of $A$ in $L$. Then $B$ is a Dedekind domain.
The following is my attempt of solution. Suppose that we have a finite morphism $f: C\to \mathbb{A}^1_k$, i.e. the pullback $f^*:k[\mathbb{A}^1_k]=k[t]\hookrightarrow k[C]$ is injective and $k[t]\subseteq k[C]$ is an integral extension of rings. Then, since $C$ is a smooth curve, $k[C]$ is integrally closed, so the integral closure of $k[t]$ in $k(C)$ is $k[C]$. Now, if $k(C)/k(t)$ is separable, I can apply the previous proposition with $A=k[t]$.
So, my questions are:
Is it true that a finite morphism $f: C\to \mathbb{A}^1_k$ always exists?
Is it true that $k(C)/k(t)$ is a separable extension of fields?
- Do we need to assume that $k$ is algebraically closed?
Thank you.
Edit: Maybe, 1. is a particular case of Noether normalization lemma, right?