In a quasicompact scheme every nonreduced point has a closed nonreduced point in its closure

Question

Let $x\in X$ be a non reduced point. Then I can find a nilpotent $f\in \Gamma(U, \mathcal O_X)$, where $U$ is some open neighborhood of $x$. I also know that when $X$ is assumed to be quasicompact, every point has a closed point in its closure. Now, if the $\operatorname{cl} (x) \subset U$, I'm done. But how do I proceed if this is not the case? Can I find a $W\supset \bar U$, and $\tilde f$ such that $\operatorname{res_{W\rightarrow U}} \tilde f = f$?

Answer 1

Let $x$ be a non-reduced point and let $y$ be a closed point lying in the closure of $\{x\}$. Let $U=\mathrm{Spec}(A)$ be an affine open around $y$, so that we necessarily also have $x\in U$. Let $\mathfrak{p}_x$ and $\mathfrak{p}_y$ be the corresponding primes of $A$. Then $\mathfrak{p}_x\subseteq\mathfrak{p}_y$, and $A_{\mathfrak{p}_x}=\mathscr{O}_{U,x}=\mathscr{O}_{X,x}$ is a localization of $A_{\mathfrak{p}_y}=\mathscr{O}_{U,y}=\mathscr{O}_{X,y}$. Since localizations of reduced rings are reduced, $y$ must be a non-reduced point.

Answer 2

You can try to induct on the number of components in a cover of $X$ with open affine subschemes $X=\cup_{i=1}^n X_i$. Your argument works in the case $n=1$. Now suppose $x\in X_1$ and consider a point $\tilde{x}$ in the closure $\bar x$ of $x$ inside $X_1$. Note that $\tilde{x}$ is a non-reduced point. If this is a closed point of $X$ then you are done. Otherwise, the closure of $\tilde{x}$ is not contained in $X\backslash X_1$ (i.e., a specialization of $\bar{x}$ cannot land inside $X_1$). So we can replace $X$ by $X\backslash X_1$ and $x$ by $\tilde{x}$. Note that $X\backslash X_1$ is quasi-compact and use induction to complete the proof.