Question

Its clear that for any $n$ there is a ring $R$ with exactly $n$ prime ideals (e.g., consider a product of $n$ fields). Can one construct such a ring for which its prime ideals form a chain ??

I can show that such rings do exist: Let $X$ be the set with $n$ points $x_1$, ..., $x_n$ and define a topology on $X$ with open sets $\emptyset$, $X$, $(X_1)$, $(x_1,x_2)$ ..., $(x_1,...,x_n)$ (I mean the set containig $x_1,..,x_n$). Then $X$ with this topology is an specteral space which means that there exixts a ring $R$ such that $Spec(R)$ is homeomorphic to $X$. Then clearly this $R$ has exactly $n$ prime ideals and they form a chain.

Answer 1

First, construct a totally ordered group $G$ of height $n$ by iterating the construction in Bourbaki's Algèbre commutative, VI.4.2 Exemple 1. Second, construct a valuation ring $A$ with group of values $G$ following the recipe in loc.cit., VI.3.4 Exemple 6. Then, the spectrum of $A$ has cardinality $n$ and is totally ordered by inclusion.