I am looking for a counterexample to the following version of the Krull height theorem.
Let $R$ be a commutative Noetherian ring with $1$. Let $I := (x_1, \ldots, x_n)$ be a finitely generated ideal with minimal generating set $\{x_1, \ldots, x_n\}$ and let $P \in \operatorname{Spec}(R)$ with $I \subsetneq P$ minimal among all primes in $R$. Then $\operatorname{ht}(P) = n$.
Here "minimal generating set" is meant in the sense that $\forall i \in \{1, \ldots, n\}: (x_1, \ldots, x_{i-1}, x_{i+1}, \ldots, x_n) \subsetneq (x_1, \ldots, x_n)$.
i.e. I am looking for a prime $P$, which is minimal over $I$, whose height is strictly smaller than $n$.
The standard version of the Krull height theorem I am refering to is
Let $R$ be a commutative Noetherian ring with $1$. Let $I := (x_1, \ldots, x_n)$ be a finitely generated ideal and let $P \in \operatorname{Spec}(R)$ with $I \subseteq P$ minimal among all primes in $R$. Then $\operatorname{ht}(P) \leq n$.
Also, under what more conditions does the theorem hold? or under what conditions can one give a lower bound $m$ for $\operatorname{ht}(P)$.
Comments:
If we leave out the condition for $P$ to be strictly larger than $I$, then any minimally prime principal ideal will do since it is minimally prime over itself (having one generator), but has height $0$.
If we leave out the condition of the generating set being minimal, then any $(x_1, \ldots, x_n, a)$ with $a \in (x_1, \ldots, x_n)$ gives a counterexample since $P$ minimal over $(x_1, \ldots, x_n, a)$ implies $\operatorname{ht}(P) \leq n+1$, but $(x_1, \ldots, x_n) = (x_1, \ldots, x_n, a) \subsetneq P$ minimally, so $\operatorname{ht}(P) \leq n < n+1$.
Under the stronger assumption of minimalism of the generating set meaning that there does not exist a set with strictly fewer elements over which $P$ is minimal then the statement is correct. Assume that $P$ is minimal over $I$, but has height strictly smaller than $n$. Then that is a contradiction to minimality of the generating set by the converse of Krull height theorem.