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?