# Finite dimensional local rings with infinitely many minimal prime ideals

Is there a finite dimensional local ring with infinitely many minimal prime ideals?

Equivalent formulation:

Is there a ring with a prime ideal $$\mathfrak p$$ of finite height such that the set of minimal prime sub-ideals of $$\mathfrak p$$ is infinite?

Here "ring" means "commutative ring with one", "dimension" means "Krull dimension", and "local ring" means "ring with exactly one maximal ideal" (warning: some authors call "quasi-local ring" a ring with exactly one maximal ideal, and "local ring" a noetherian ring with exactly one maximal ideal; it is well known that a noetherian ring has only finitely many minimal prime ideals).

• @Pierre-YvesGaillard Full disclosure: I have not thought this through, and it might be way off track.... but if I was going to try to come up with an example the first place I would look would be $R[x]$ where $R$ is a one-dimensional Pruefer domain with infinitely many primes and non-zero jacobson radical. e.g. appropriate localization of algebraic integers. I suspect you might find max (height 2) ideals containing infinitely many height one primes not contracting to $0$ in $R$, which would guarantee an element $f \in R[x]$ such that $R[x]/f$ has a max ideal as you desire. Jun 9 '20 at 15:01
• Not exactly the same question, but your question is also answered by my example at math.stackexchange.com/questions/1542104/…. Jun 9 '20 at 16:50
Let $$R$$ be the quotient of a polynomial ring $$k[x_1,x_2,\dots]$$ in infinitely many variables over a field by the ideal generated by all products $$x_ix_j$$ for $$i\neq j$$. Geometrically, $$\operatorname{Spec} R$$ looks like infinitely many copies of $$\mathbb{A}^1_k$$ with their origins identified. So, $$R$$ is a $$1$$-dimensional ring with infinitely many minimal prime ideals, each of which are contained in the maximal ideal generated by all the $$x_i$$. See this answer for more details about this ring.