TITLE: Do codimension 1 subsets of a scheme cover it?
QUESTION [6 upvotes]: Let $X$ be an irreducible scheme. A point $p\in X$ is said to have codimension $n\in\mathbb{Z}_{\geq 0}\cup \{\infty\}$ if $\overline{\{p\}}$ has codimension $n$. Is it true that any point of positive codimension lies in the closure of a point of codimension 1?

REPLY [3 votes]: This is false, already for spectra of valuation rings. Recall that ideals in a valuation ring $R$ correspond to ideals in the value group $\Gamma$, which can be any totally ordered abelian group. Moreover, since the ideals are totally ordered, so are the prime ideals, so it suffices to construct a totally ordered abelian group that does not have a height $1$ prime. This is easy: 
Example. Let $\Gamma$ be the group $\mathbf Z[x]$, with the ordering defined by $f \geq g$ if and only if $f(n) \geq g(n)$ for $n \gg 0$. This is equivalent to the lexicographic ordering with $1 < x < x^2 < \ldots$.
Suppose $I \subseteq \Gamma$ is a height $1$ prime ideal, i.e. $I+\Gamma_{\geq 0} \subseteq I$, if $i + j \in I$ then $i \in I$ or $j \in I$, and if $J \subsetneq I$ is another prime ideal, then $J = 0$. Let $f \in I$ be a nonzero element, and assume $\deg f = n$. Then $x^{n+1} > f$, so $(x^{n+1}) = x^{n+1} + \Gamma_{\geq 0} \subsetneq I$ is a strictly smaller nonzero prime ideal inside $I$, contradicting the height $1$ hypothesis. $\square$
Remark. An example of a valuation ring with this value group can be constructed as follows: let $K = k(x_0, x_1, \ldots)$ be a rational function field in (countably) infinitely many variables. It has a valuation $K \to \mathbf Z[x]$ defined on monomials in $k[x_0,\ldots]$ by
$$\prod_i x_i^{n_i} \mapsto \sum_i n_ix^i,$$
then on polynomials by mapping to the leading term (with respect to the lexicographic ordering with $x_0 < x_1 < \ldots$), and finally on $K$ by extending multiplicatively. Then one may take $R$ to be the value ring for this valuation.
Remark. Obviously the result is true if every point has finite height; for example if $X$ is Noetherian.