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Let $Y$ be a variety. Let $\mathcal{O}_{P, Y} = \mathcal{O}_{P}$ be the ring of germs of regular functions on $Y$ near $P$. That is, an element of $\mathcal{O}_P$ is pair $\langle U, f \rangle$ where $U$ is an open subset of $Y$ containing $P$ and $f$ is a regular function on $U$.

How to show that

(1) $\mathcal{O}_P$ is a local ring (has only one maximal ideal $\mathfrak{m}$) and

(2) $\mathcal{O}_{P}/\mathfrak{m} \cong k$?

These questions comes from page 16 of the book Algebraic Geometry by Hartshorne.

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Take $\mathfrak{m}=\{ f \in \mathcal{O}_P \mid f(P) = 0 \}$. Then $\mathfrak{m} \subseteq \mathcal{O}_P$. It is clear that $\mathfrak{m}$ is an ideal of $\mathcal{O}_P$. Suppose that $f(P) \neq 0$ and $f \in \mathcal{O}_P$. Then $1/f$ is regular in some neighborhood of $P$ and hence $1/f \in \mathcal{O}_P$. How could we show that $1/f \in \mathfrak{m}$? I think that if $1/f \in \mathfrak{m}$, then $1=f \cdot 1/f \in \mathfrak{m}$ and hence $\mathfrak{m} = \mathcal{O}_P$. Is this true? Thank you very much.

Edit: I made a mistake. I should consider the ideal $\mathfrak{n}$ such that $\mathfrak{m} \subsetneq \mathfrak{n} \subseteq \mathcal{O}_P$ and show that $\mathfrak{n} = \mathcal{O}_P$.

LJR
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2 Answers2

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Define a map $\varphi : \mathcal{O}_P \to k$ that sends a regular function $f$ at $P$ to $f(P) \in k$. This is just the evaluation homomorphism and clearly $\mathfrak{m} = \ker \varphi$. Now this map is also surjective because for any element $a \in k$ I can just define the constant function on $Y$ that sends every $y \in Y$ to $a$. Such a function is clearly regular at $P$. Thus it follows that

$$\overline{\varphi} : \mathcal{O}_P/\mathfrak{m} \stackrel{\cong}{\longrightarrow} k$$

is an isomorphism.

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When you ask “How can we show that $1/f \in \mathfrak{m}$?” it is possible that there is some confusion about notation. If we take $A(Y)_{\mathfrak{m}_p}$, then what actually gets put in the denominator are functions which are non-vanishing at $p$. Compare for instance Hartshorne’s Theorem 3.4 and notation for the graded homogeneous setting: if $S$ is a graded ring and $\mathfrak{p}$ a homogeneous prime ideal in $S$, then we denote by $S_{(\mathfrak{p})}$ the subring of elements of degree $0$ in the localization of $S$ with respect to the multiplicative subset $T$ consisting of the homogeneous elements of $S$ not in $\mathfrak{p}$.

locally trivial
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