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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.

Jyrki Lahtonen
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G. Chiusole
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  • I think [tag:dimension-theory] is intended to be used for questions about topological and/or fractal dimensions only. This is an ok question about commutative algebra, though. – Jyrki Lahtonen Sep 24 '19 at 13:15
  • @JyrkiLahtonen That’s fair. How about the krull-dimension tag? I’d think it’s closest to questions about height. – G. Chiusole Sep 24 '19 at 13:19
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    There are one dimensional local Noetherian rings $R_n$ with maximal ideal generated by any given $n>0$. This will give you easy counterexamples. – Mohan Sep 24 '19 at 13:31
  • @Mohan that is helpful, but I don’t think it gives an example in which the prime is strictly greater – G. Chiusole Sep 24 '19 at 13:40
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    It does. Take such a ring and take $n-1$ elements of the $n$ generators. The ideal generated is clearly strictly contained in the maximal ideal, which is the only non-zero prime and thus minimal. – Mohan Sep 24 '19 at 14:18
  • @Mohan I’m sorry, how does one know that the maximal ideal is the only non $0$ prime ideal? If the ring were an integral domain $(0)$ were prime and dimension would give the claim. – G. Chiusole Sep 24 '19 at 14:35
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    You can find domains with the property. Here is a standard example. $R_n=k[[t^n, t^{n+1},\ldots, t^{2n-1}]]$. – Mohan Sep 24 '19 at 15:27
  • Nice example. Thank you very much – G. Chiusole Sep 30 '19 at 11:23

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