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Let $X$ be a scheme of finite type over an algebraically closed field $k$. Suppose stalk $O_{X,x}$ is an integral domain. Then it implies that $x$ is in the closure of only one associated point of $O_X$, which I am not sure how to deduce. How can I prove this? Thank you.

The definition of associated point I have is: $z \in X$ is an associated point of $O_X$ if there exists an open neighbourhood $U$ of $z$ and $s \in \Gamma(U, O_X)$ such that $s_y \neq 0$ if and only if $y \in \overline{ \{ z \} }$ for all $y \in U$.

Johnny T.
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  • Hi. I’m a bit confused by your definition. Can you tell me what are the associated points of Spec ( C x C ), where C is the field of complex numbers? – Youngsu Jun 29 '20 at 09:16
  • This is the definition from Mumford's red book III.2, and to be honest I don't have a good understanding of it.. – Johnny T. Jun 29 '20 at 10:00
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    This follows from the correspondence between irreducible subvarieties of $X$ through $x$ and prime ideals of $O_{X,x}$ as linked in the duplicates. Associated points of $O_X$ correspond exactly to generic points of irreducible components of $X$ modulo some niceness assumptions. – KReiser Jun 29 '20 at 20:00
  • @KReiser I am not quite seeing how to deduce the conclusion from your links, thank you for the links.. I get that $x$ can not lie on the intersection of the closures of two generic points of $X$ (the generic points of the components of $X$ are associated points of $O_X$). However, how does one deal with the non-generic associated points? – Johnny T. Jun 30 '20 at 06:50
  • @KReiser I believe I'm just talking about III.2 Proposition 3 of Mumford (unless I am misunderstanding something). It states that if $X$ is noetherian then the generic points of the components of $X$ are associated points of $O_X$. If $X$ is reduced, then the generic points of $X$ are the only associated points of $O_X$. And he gives an example of non-generic points of $O_X$ after the proof. And the example seems to be for a scheme of finite type over $k$, namely $\operatorname{Spec} k[x,y]/(x^2, xy)$. So I thought I would have to consider non-generic associated points as well? – Johnny T. Jun 30 '20 at 07:26
  • Fine, fine, you've convinced me the duplicated did not address the whole of your problem. Reopened and answer posted. – KReiser Jun 30 '20 at 07:40
  • @KReiser Thank you, I was just trying to understand what I was missing or misunderstanding. My intention was not to try to convince you to reopen the problem etc. Thank you for your help – Johnny T. Jun 30 '20 at 07:45

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I would recommend proving it in the forward direction so you don't have to think too hard about what associated points are. Hint: if the stalk is reduced, there is an open neighborhood which is reduced too, and any noetherian scheme has finitely many irreducible components, so you should be able to reduce to the case that $X$ is affine integral over a field, where the proof is straightforwards. Details under the spoilers.

Step 1: reduce to the case where $X$ is affine and integral. Since reducedness is stalk-local and open, we know there is an open neighborhood of $x$ where $X$ is reduced. By the correspondence between minimal primes of $\mathcal{O}_{X,x}$ and subvarieties through $x$, we have that $x$ lies on only one irreducible component. Since $X$ is noetherian, it has finitely many irreducible components, so the set of points just on the irreducible component that $x$ is on is open.

Step 2: Prove the result for affine integral schemes over a field. Look at the generic point, and bask in its light. (To be precise, show that if $s_y\neq 0$, then $s_{y'}\neq 0$ whenever $y$ is a specialization of $y'$, and conclude from there.)

KReiser
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