TITLE: Reduced scheme and closed points
QUESTION [24 upvotes]: In The Geometry of Schemes by Eisenbud and Harris, Exercise I-32 asks one to show that a scheme $X$ is reduced if and only if every local ring $\mathcal{O}_{X,p}$ is reduced for closed points $p \in X$.  However, this does not seem to work in general, since $X$ may not have enough closed points.  What additional hypotheses on $X$ do I need for such an assertion to hold?

REPLY [4 votes]: Brunoh: 
1) If $X$ is a quasi-compact scheme such that $\mathscr O_{X,x}$ is reduced for every closed point $x$, then $X$ is reduced. Indeed, let $y\in X$. The scheme $\overline{\{y\}}$ is a closed subscheme of $X$, hence is quasi-compact, and non-empty because it contains $y$. It thus has a closed point $x$, which is closed in $X$ as well. Now $\mathscr O_{X,y}$ is a localization of $\mathscr O_{X,x}$, hence is reduced because so is $\mathscr O_{X,x}$ by assumption. 
2) Let $k$ be a field and let $v$ be the valuation on $k(X_i)_{i\in \mathbb Z_{> 0}}$ defined by the composition of the successive discrete valuations provided by the $X_i$'s. Let $X$ be the spectrum of the corresponding valuation ring. Then topologically, $X=\{x_0,\ldots, x_n,\ldots\}\bigcup \{x_\infty\}$ where every $x_i$ specializes to $x_{i+1}$, and where $x_\infty$ is the unique closed point (the point $x_0$ is the generic one, and $x_i$ corresponds to the prime ideal generated by $X_i$). Now if you remove $x_\infty$ you get an open subscheme $U$ of $X$, without any closed point. 
Of course, $U$ is reduced, but $U\times_k \mathrm{Spec}\; k[\epsilon]$ (with $\epsilon\neq 0$ and $\epsilon^2=0$) is not reduced, and homeomorphic to $U$.