The stalk $O_{X,x}$ is an integral domain implies $x$ is in the closure of only one associated point

Question

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 $y \in U$.

Answer 1

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.)