Rings with given "dimension" (in a specific sense, related to generators of ideals)

Question

Let $R$ be a noetherian ring. If $I \trianglelefteq R$ is an ideal of $R$, we define $\mu(I)$ to be the minimal$^{(1)}$ number of generators of $I$ (this is well-defined since $R$ is noetherian). Then we define the "dimension" of $R$ as $$d(R) := \sup\limits_{I \trianglelefteq R} \;\mu(I).$$

My question is:

Given a positive integer $n>0$, is there a ring $R$ such that $d(R)=n$?

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Obviously, $R$ is a PID iff $d(R)=1$. Dedekind domains satisfy $d(R)≤2$. For instance the non-UFD (hence non-PID) $\Bbb Z[i\sqrt 5] = \mathscr{O}_{\mathbb Q(i\sqrt 5)}\;$ has "dimension" $2$. Moreover, I know that $d(\Bbb Z[X]) = \infty$.

But I don't have an example of a ring with "dimension" $3$, for instance. Possibly something like $\Bbb Z[\sqrt 2, \sqrt 3]$ could work, but we have to avoid ring of integers of number fields…

My notion of "dimension" is maybe related to the notion of Krull dimension, probably by taking the supremum over $\text{Spec}(R)$ instead of all the ideals $I \trianglelefteq R$, i.e., $d_p(R) := \sup\limits_{P\in \text{Spec}(R)} \;\mu(P)\;$ (similar notations are here; moreover we have $d_p(\Bbb Z[X]) =2$ which is the Krull dimension). I don't really know the notion of "height" of an ideal, however, but I think that the inequality $\mu(I) \ge \text{ht}(I)$ holds.

Anyway, any hint would be greatly appreciated!

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$^{(1)}$ minimal in the following sense: if $m<\mu(I)$ and $x_1,\dots,x_m \in I$ then the ideal generated by $x_1,\dots,x_n$ is strictly contained in $I$, i.e. $\langle x_1,\dots,x_m \rangle ≠ I$. Literally: $\mu(I) = \min\{\#E \mid \langle E \rangle = I\}$.

Answer 1

How about $R = k[x_1, \ldots, x_n]/\mathfrak m^2$ where $\mathfrak m = (x_1, \ldots, x_n)$? This is a local ring where $\mathfrak m$ is minimally generated by $n$ elements. Take a proper ideal $I$. Then $I$ is a $k$-subspace of $\mathfrak m$ viewed as a $k$-vector space, and a $k$-basis of $I$ is also a generating set of $I$. So $I$ is generated by at most $\dim_k I \le \dim_k \mathfrak m = n$ elements.

Answer 2

A comment. In this answer I assume that $\langle x_1,\dots,x_m \rangle$ denotes the subring generated by $x_1,\dots,x_m$, not the ideal generated by them. As pointed out by @user26857 in the comments, Watson may have meant the latter case, but I decided to leave my answer here for its own sake.

The example appears somewhere in the sequence $\mathbb{F}_2, \mathbb{F}_2^2, \mathbb{F}_2^3,\ldots$. In fact, every ideal of $\mathbb{F}_2^m$ has the form $\mathbb{F}_2^k$, and we have $$d(\mathbb{F}_2^{m})\leqslant d(\mathbb{F}_2^{m+1})\leqslant d(\mathbb{F}_2^{m})+1.$$ This condition guarantees that every value between $1$ and $d(\mathbb{F}_2^{m})$ appears as the dimension of some $\mathbb{F}_2^{k}$ with $k\leqslant m$. It remains to prove that $d(\mathbb{F}_2^{t})\to\infty$ as $t\to\infty$; to this end, note that $d(\mathbb{F}_2^{t})\geqslant \log_2t$ because every set of elements in $\mathbb{F}_2^{t}$ of cardinality less than $\log_2t$ has a pair of coordinates in which all elements coincide.