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I am a math graduate student, and I'm working through Fulton. This is my first exposure to algebraic geometry. I'm having trouble with problem 2.18:

Let $\mathcal{O}_P(V)$ be the local ring of a variety $V$ at a point $P$. Show that there is a natural one-to-one correspondence between the prime ideals in $\mathcal{O}_P(V)$ and the subvarieties of $V$ that pass through $P$. (Hint: If $I$ is prime in $\mathcal{O}_P(V)$, $I\cap\Gamma(V)$ is prime in $\Gamma(V)$, and $I$ is generated by $I\cap\Gamma(V)$; use Problem 2.2.)

And problem 2.2 reads: Let $V\in\mathbb{A}^n$ be a variety. A subvariety of $V$ is a variety $W\in\mathbb{A}^n$ that is contained in $V$. Show that there is a natural one-to-one correspondence between algebraic subsets (resp. subvarieties, resp. points) of $V$ and radical ideals (resp. prime ideals, resp. maximal ideals) of $\Gamma(V)$.

The solution to 2.2 is simple: Since $\Gamma(V)=k[x_1,\dots,x_n]/I(V)$, there is a one-to-one correspondence between prime ideals of $\Gamma(V)$ and prime ideals of $k[x_1,\dots,x_n]$ containing $I(V)$. And there is also a one-to-one correspondence between prime ideals of $k[x_1,\dots,x_n]$ containing $I(V)$ and subvarieties of $V$.

Here's what I have so far: If $I$ is a prime ideal in $\mathcal{O}_P(V)$, $J=I\cap\Gamma(V)$ is a prime ideal in $\Gamma(V)$, and by problem 2.2 there is a corresponding subvariety $W$. But I need to show that $P\in{W}$, and the information about $P$ is contained in the denominators of functions in I, which I threw away when I intersected $I$ with $\Gamma(V)$.

The other direction is easy: If $W$ is a subvariety of $V$, then there is a corresponding prime ideal $J=I_V(W)$ of $\Gamma(V)$. And the ideal generated by $J$ in $\mathcal{O}_P(V)$ is a prime ideal.

What am I missing in the "hard" direction?

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    With the "easy" direction, how do you know that the ideal generated by J in $O_{p}(V)$ is prime?2019-04-27

2 Answers 2

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Let $f \in I \cap \Gamma(V)$. We need to show that $f(P) = 0$. If not, then $1/f \in O_P(V)$. Since $f \in I$, and $I$ is an ideal of $O_P(V)$, we would have $1 \in I$, a contradiction since a prime ideal cannot equal $(1)$.

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    @Seven Yes, we are required to show that, but that is part of the proof of the general result. See e.g. http://www2.gsu.edu/~matfxe/commalglectures/lect6.pdf Proposition 1.102015-06-21
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Alternatively, since $I$ is a proper ideal $\mathcal{O}_P (V)$ is it contained in the maximal ideal $m_P (V)$ which is the ideal of rational functions on $V$ that vanish at $P.$