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Why is it true that prime divisors on nonsingular curves (curve: integral separated scheme of finite type over an algebraically closed field, of dimension 1 with all local rings regular) are only the closed points? In particular, if $C$ is curve, then $dim(C)=1$. Now i can see that if $p$ is a closed point, then it is a prime divisor, since it is irreducible and it has codimension 1 (the codimension could not be higher than 1, otherwise we have a violation of $dim(C)=1$). But why can we not have a closed irreducible proper subset $Z$, which is not a point, such that $Z$ is a prime divisor?

Intuition: if every closed, irreducible proper subset $Z$ of a curve $C$, contains a closed point $p$ of $C$, then $Z=\left\{p\right\}$, otherwise we would have a chain $\left\{p\right\} \subsetneq Z \subsetneq C$. Is that the case?

Is it true that all points of a curve are closed?

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up vote 3 down vote accepted

An irreducible scheme with more than one point always has a non-closed point, namely the generic point. If $X$ is a $1$-dimensional irreducible scheme, then it has closed points, necessarily of codimension $1$, and a unique generic point, necessarily of codimension zero. Here's an argument. Note that a proper closed subset of an irreducible space of finite dimension has dimension strictly less than that of the ambient space. So let $x\in X$. Then $\overline{\{x\}}$ is a closed irreducible subset of $X$. If it has dimension $1$, then it equals $X$, whence $x$ is the generic point. Otherwise this space has dimension $0$. Take any $y\in\overline{\{x\}}$. Then $\overline{\{y\}}\subseteq\overline{\{x\}}$, and this inclusion cannot be strict because the latter space is zero-dimensional. So $y$ and $x$ have the same closure. Since schemes are Kolmogorov (distinct points have distinct closures), we conclude that $x=y$. Thus $\overline{\{x\}}=\{x\}$ is a closed point.

So the irreducible closed subsets of $X$ are closed points and $X$ itself. A closed point has codimension $1$ because we always have the chain $\{x\}\subsetneq X$.

If $X$ is not assumed irreducible, then there can be closed points of codimension $0$. For example, take $\mathbf{A}_k^1\coprod\mathrm{Spec}(k)$ has a closed point which is an irreducible component, hence of codimension zero, although the ambient scheme is $1$-dimensional.

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