Let $C$ be a curve and $P \in C$ a smooth point. Then the local ring $k[C]_P$ of $C$ at $P$ is a discret valuation ring.

Question

First of all, $k$ is an algebraically closed field, and by "curve" I mean a projective variety of dimension one.

I should also mention that I have defined discret valuation rings as rings that look like $$\{ a \in k | ν(a) \geq 0 \},$$ where $ν: k-\{0\} \rightarrow \mathbb{Z}$ is a discret valuation.

Any help would be greatly appreciated.

Answer 1

I'll use criterion 5 in the Wikipedia article: R is an integrally closed Noetherian local ring with Krull dimension one.

Integrally closed: this is because $p$ is smooth. The Auslander-Buchsbaum theorem implies that $O_{C, p}$ is a UFD. (see chapter 19 of Eisenbud's Commutative Algebra with a View Toward Algebraic Geometry for a proof)

Noetherian : This is because $C$ is a variety, so it's Noetherian. Choose any affine open set $U = Spec(R)$ containing $p$. Therefore, $O_{C, p} = R_p$ which is also Noetherian since it's the localization of a Noetherian ring. See here for a proof.

Local : the stalk at a point is a local ring.

Krull dimension 1 : $C$ is a curve.