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According to the lecture,

[Observation] Let $C$ be the projective plane curve. Then a local ring at a regular point on the curve $C$ is a discrete valuation ring(DVR),

Based on the equivalence relation of DVR, I tried to prove the above statement.

Proof. When taking a regular point $p$ at $C$, the dimension of tangent plane is $1$. and since the local ring , we can find a unique maximal ideal $\mathfrak{m}$. For convenience, let $\mathcal{O}_{X,p} $ be the local ring at the regular point $p$) Then,

$$1= \operatorname{dim}_{k} (T_pC)=\operatorname{dim}_{k} ((T_pC)^{*})= \operatorname{dim}_{k}(\mathfrak{m}/\mathfrak{m}^2)$$

(where $(T_pC)^{*}$ is a dual space of $T_pC$ and $k=\mathcal{O}_{X,p}/\mathfrak{m} $ )

Then, $\mathcal{O}_{X,p} $ becomes a DVR by deifntion.

But my problem is here : When observing the TFAE...

Proposition. Let $R$ be a Noetherian local domain the following are equivalent.

$(1)$ $R$ is a discrete valuation ring

$(2)$ $\operatorname{dim}_{k}(\mathfrak{m}/\mathfrak{m}^2)=1$

$(3),(4),(5)...$... mumble,mumble...

In order to apply this theorem for the given observation, the given ring $R$ must satisfy Noetherian local domain. Clearly, the given ring $\mathcal{O}_{X,p} $ is local ring and integral domain as well. But I am now stuck why $\mathcal{O}_{X,p} $ should be Noetherian. (such doubt seems to be very fundamental result, but it is not easy for me)

Anyway, the following two ways are my trial :

  1. In my thought, the most plausible reason why $\mathcal{O}_{X,p} $ is a Noetherian is Hilbert's Basis Theorem, because the projective plane curve consists of a homogenous polynomial $f\in k[x,y,z]$. However, the quotient of polynomial is not in an element $k[x,y,z]$ again. Thus, such I cannot get the desired result.

  2. Meanwhile, I try to apply the following theorem : If $R$ is a principle ideal domain(PID),then $R$ is notherian. I was sure that it is the key to check my curiosity. But, unfortunately, this fails because the polynomial is $k[x,y,z]$ is not a PID.(More precisely, rings of polynomials in only one variable with coefficients in a field is PID)

Hanwoong Cho
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    A localization of a Noetherian ring is again Noetherian: https://math.stackexchange.com/questions/110735/why-is-the-localization-of-a-commutative-noetherian-ring-still-noetherian – Sergey Guminov Jul 12 '22 at 10:52
  • To close the loop, note that the local ring of a point in projective space is the same as the local ring of that point considered as a point in one of the standard affine opens $D_+(x_i)\cong\Bbb A^n$. Now you may apply the fact that the coordinate ring of $\Bbb A^n$ is noetherian and the linked duplicate, which shows that the localization of a noetherian ring is again noetherian, to conclude that the local ring of a point in $\Bbb P^n$ is noetherian. – KReiser Jul 23 '22 at 21:33

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