Reduced scheme and closed points 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?
 A: 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 $U$. 
A: There do exist schemes without a closed point, yes. (Liu, exercises 3.3.26/27) 
But under some very reasonable additional conditions - I think quasi-compactness will be sufficient, if you are happy with using Zorn's lemma - the result holds. Use/prove the existence of a closed point, and the fact that localizing a reduced ring still gives you a reduced ring. 
A: It seems to me that looking at closed points only is not sufficient since they are not always a dense set of X ...
A: I think that any quasi-compact, non-empty sober topological space has a closed point (a topological space $X$ is sober if every irreducible closed subset of $X$ has a unique generic point; any scheme is sober). 
So, let $X$ be such a space. Let $\mathscr F$ be the set of irreducible closed subsets of $X$, ordered by reverse inclusion. 
The set $\mathscr F$ is inductive; indeed, let $(F_i)_{i\in I}$  be a totally ordered family of irreducible closed subsets of $X$. Let's first prove that its intersection is non-empty. If it were empty then by quasi-compactness (in its dual version, involving intersection of closed subsets), they would exist a finite subset $J$ of $I$ such that $\bigcap _{i\in J}F_i=\varnothing$. But then we get a contradiction: if $J=\varnothing$, then $\bigcap _{i\in J}F_i=X$ and the latter is non-empty by assumption; and if $J$ is  non-empty, $\bigcap _{i\in J}F_i=F_i$ for some $i\in J$, hence is non-empty. 
Now if $x$ is a point in $\bigcap F_i$, then $\overline{\{x\}}$ is an irreducible closed subset of $X$ contained in all the $F_i$'s. Therefore $\mathscr F$ is inductive. 
It thus has a maximal element $G$. Since $X$ is sober, its closed irreducible subset $G$ has a unique generic point $\eta$. Let $x\in G$. By maximality of $G$ one has $\overline{\{x\}}=G$, and $x=\eta$. As a consequence, $G=\{\eta\}$ and $\eta$ is closed. 
A: I don't think quasi-compactness is enough,for Noether scheme it is true. in a noether scheme, every point P has a closed point in its closure, so .....
but i don't find a necessary and sufficient condition 
