# Theorem 1 in Fulton's algebraic curves section 3.2

I am reading section 3.2 in Fulton's Algebraic Curves.

Theorem 1. Let $$F$$ be an irreducible plane curve, $$P\in F$$. $$P$$ is a simple point of F if and only if $$O_p(F)$$ is a discrete valuation ring.

Theorem 2. Let $$P$$ be a point on an irreducible curve F. Then for all sufficiently large n, $$m=dim_k(m_p(F)^n/m_p(F)^{n+1})$$ where $$m$$ is multiplicity of F at p and $$m_p(F)$$ is a unique maximal ideal of $$O_p(F)$$.

It proves $$\gets$$ direction of Theorem 1 by using Theorem 2 and problem 2.50 in this book. But it needs the condition, which $$k\to O_p(F)\to O_p(F)/m_p(F)$$ is isomorphism to apply problem 2.50. I was trying to prove $$O_p(F)/m_p(F)$$ is ring finite over $$k$$(algebraically closed field) to show that homomorphism is an isomorphism, but I failed. I need help or some hints. I appreciate any help you can provide.

It comes from Nullstellensatz: the quotient $$O_P(F)/m_P(F)$$ is a field which is a finite type $$k$$-algebra hence is a finite extension. It is then algebraic over $$k$$ so, from the closure of $$k$$, equal to $$k$$.

• Thanks for your answer! I know the coordinate ring of F is finitely generated k algebra, but I don't know how to deal with the denominator part of its local ring at P. Could you give me more details? Commented Dec 4, 2022 at 10:41
• I guess that $O_P(F)/m_P(F)$ is the localization of $O(F)/m_P(F)$. The last is $k$, so for the first. Isn't? Commented Dec 4, 2022 at 11:21
• What is $O(F)$ ? Commented Dec 6, 2022 at 19:08
• It is the function ring. If $F$ is affine, of equation $f(x,y)=0$ then $O(F)=k[x,y]/(f(x,y))$. Commented Dec 8, 2022 at 14:17
• I see. But I still need clarification about what $O(F)/m_p(F)$ is since $m_p(F)$ isn't contained in $O(F)$ but in $O_p(F)$. Commented Dec 8, 2022 at 16:59

This is from section $$2.4$$:

The ideal $$m_P(V ) = \{f ∈ \mathcal O_P(V ) | f (P) =0\}$$ is called the maximal ideal of $$V$$ at $$P$$. It is the kernel of the evaluation homomorphism $$f \mapsto f (P)$$ of $$\mathcal O_P(V )$$ onto $$k$$, so $$\mathcal O_P(V ) \mathbin{/}m_P(V)$$ is isomorphic to $$k$$.

You can easily verify that the isomorphism mentioned above is the inverse of $$k\to O_p(F)\to O_p(F)/m_p(F)$$.