I was reading Section I.6 from Hartshorne's Algebraic geometry book. He claims that the ring $\mathcal{O}_p$ of regular functions at a point $p$ in some nonsingular curve $Y$ is has dimension 1. I understand this holds when $Y$ is an affine variety since $\dim\mathcal{O}_p=\dim Y=1$ in affine case. However, what if $Y$ is projective?
In affine case, let $m_p$ be the maximal ideal corresponding to point $p$, then $\mathcal{O}_p\cong A(Y)_{m_p}$, so $\dim \mathcal{O}_p=\operatorname{ht}(m_p)=\dim A(Y)-\dim A(Y)/m_p$ where $\dim A(Y)=\dim Y$ and $\dim A(Y)/m_p=0$ since it's a field. If we do the same thing in projective case, then most claims hold true except $\dim S(Y)=\dim Y+1$, so I'm expecting $\dim\mathcal{O}_p=\dim Y+1=2$ when $Y$ is a projective curve, which doesn't match with Hartshorne's claim.
Is there anything I missed while coming to the projective case or is Hartshorne assuming the curve is affine here?
Best Answer
Hartshorne's claim is true when $Y$ is projective. The key fact is that every point in a variety has an affine open neighborhood (see proposition I.4.3, for instance) - combining this with the fact that the local ring of a point can be calculated from any open neighborhood, we have that the result for the affine case carries over to the projective case. (The key error in your calculation is that you haven't justified why you should be able to replace $A(Y)$ with $S(Y)$ and have everything else work.)