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It is 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 whose 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 $\emptyset$, $X$, $\{x_1\}$, $\{x_1,x_2\}$, ..., $\{x_1,...,x_n\}$. Then $X$ with this topology is a spectral space, which means that there exists 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.

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up vote 5 down vote accepted

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.

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