Why the ring of regular functions on a variety at a point is a local ring

Question

In the Hartshorne's book on Algebraic Geometry one reads the following, '' Let Y be a variety. Denote by $O(Y) $ the ring of all regular functions on $Y. $ For $p\in Y, $ we define the local ring of $P$ on $Y, O_P, $ to be the ring of germs of regular functions on $Y$ near $P$''. He then gives an argument, why $O_P $ is a local ring: '' its maximal ideal $m$ is the set of germs of regular functions which vanish at $P. $ For if $f(P) \neq 0,$ then $1/f$ is regular in some neighborhood of $P$ ''. I unfortunately don't understand the argument, why $O_P$ contains a maximal ideal. Can somebody give some more explanations on that? Many thanks for your comment.

Answer 1

The ideal $\mathfrak{M}_P = \{f \in \mathcal{O}_X(U): f(p) = 0\}$ is maximal as it is the kernel of the evaluation map at $P$, which gives an isomorphism of $\mathcal{O}_{X, P}/\mathfrak{M}_P \cong k$.

Answer 2

Following the comment, every element not in the maximal ideal is a unit. All functions where $f(P) \neq 0$ is invertible and therefore a unit. So that gives us that all functions $f(P) = 0$ must be in some maximal ideal, implying that $\mathfrak{m}_P$ is maximal. Further, it is the unique maximal ring in $\mathcal{O}_{P.Y}$ meaning that $\mathcal{O}_{P.Y}$ is local.