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).

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    $\begingroup$ @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. $\endgroup$ Jun 9 '20 at 15:01
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    $\begingroup$ Not exactly the same question, but your question is also answered by my example at math.stackexchange.com/questions/1542104/…. $\endgroup$ Jun 9 '20 at 16:50
  • $\begingroup$ @EricWofsey --- Thanks! If you posted a short answer linking to your other answer, I'd be glad to accept it. $\endgroup$ Jun 9 '20 at 18:17

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.


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