Let $(R,{\frak m}_R)$ be a local domain (not necessarily Noetherian). That is, $R$ is integral and ${\frak m}_R$ is the unique maximal ideal of $R$. Suppose that ${\frak m}_R$ is finitely generated.

Question. Does ${\frak m}_R$ have finite height? That is, is the Krull dimension of $R$ necessarily finite?


2 Answers 2


Non-discrete valuation rings form a great class of examples.

Remark. Recall that if $(\Gamma,\geq)$ is a totally ordered abelian group, then we can obtain a valuation ring in the following manner. Let $M = \{m \in \Gamma\ \big|\ m \geq 0\}$ be the submonoid of nonnegative elements, and let $k[M]$ be the monoid algebra on $M$. Then we obtain a nonarchimedean valuation on $k[M]$ by \begin{align*} v \colon k[M] &\to M \cup \{\infty\}\\ \sum a_m m &\mapsto \min\left\{m\ \big|\ a_m \neq 0\right\}, \end{align*} where the minimum is taken with respect to the ordering of $\Gamma$, and by convention $v(0) = \infty$.

The set $\mathfrak m = \{x \in k[M]\ \big|\ v(x) > 0\}$ is a maximal ideal, and the ring $R = k[M]_\mathfrak m$ is local domain with a valuation taking values in $M$. (Prime) ideals in $M$ correspond to (prime) ideals in $R$.

Moreover, $\mathfrak m$ is finitely generated if and only if the corresponding ideal $\{m \in M\ |\ m > 0\}$ in $M$ is finitely generated. We will denote the latter set by $\mathfrak m$ as well, as we will be working exclusively in ordered group language.

So for valuation rings, the question is equivalent to the following:

Question. Does there exist a totally ordered group $\Gamma$ such that $\mathfrak m$ is finitely generated, but $M$ has an infinite chain of prime ideals?

My example is motivated by, but not identical to, this example of a valuation ring of infinite Krull dimension (but I don't think the maximal ideal is finitely generated in that example).

Example. Let $\Gamma = \mathbb Z[x]$, with the total ordering given by $$p \succeq q \Longleftrightarrow p(x) \geq q(x) \text{ as } x\to \infty.$$ It is a total ordering because $p-q$ is eventually positive or eventually negative for $p \neq q$. The explicit description for $p = \sum_{i = 0}^m a_i x^i$ and $q = \sum_{j = 0}^n b_j x^j$ is: we have $p \succeq q$ if and only if either

  • $m > n$ and $a_m > 0$;
  • $m = n$ and $a_m \geq b_m$;
  • $m < n$ and $b_n < 0$.

Now I claim that $\mathfrak m$ is finitely generated. Indeed, it is generated by $1$: if $p \succ 0$, then $p - 1 \succeq 0$, i.e. $p - 1 \in M$, which means that $p = (p-1) + 1$ is in the ideal generated by $1$.

Finally, I have to exhibit an infinite chain of prime ideals. For each $n \in \mathbb N$, let $I_n$ be the set of polynomials $p$ of degree $\geq n$. If $p, q \succeq 0$, then $$\deg(p+q) = \max(\deg(p),\deg(q)),\tag{1}$$ because there is no cancellation of leading terms (both are positive). In the case that $\deg p \geq n$, we find $\deg(p+q) \geq n$. This shows that $I_n$ is an ideal. The formula (1) also shows that $I_n$ is prime. This gives an infinite chain of prime ideals \begin{equation} I_0 \supseteq I_1 \supseteq I_2 \supseteq \ldots\tag*{$\square$} \end{equation}

Remark. By identifying $k\left[\mathbb Z[x]\right]$ with $k\left[x_0^{\pm 1},x_1^{\pm 1},\ldots\right]$, we can write $R$ as a suitable localisation of the ring $$k\left[\left\{x_0^{a_0} \cdots x_n^{a_n}\ \big|\ a_n > 0\right\}\right] \subseteq k[x_0^{\pm 1},x_1^{\pm 1}, \ldots].$$ I don't see a neater way to write down more explicitly what this ring is.

  • $\begingroup$ I am deeply impressed by your example. Thanks! $\endgroup$ Jun 27, 2016 at 7:57

Here's a different exposition of some non-discrete valuation ring examples such as in the answer by R. van Dobben de Bruyn.

Let $W$ be a partially ordered set, $\{x_i: i \in W\}$ indeterminate variables, $R=k[x_i/x_j^n: i,j \in W, i<j, n \geq 0]_{(x_i:i \in W)}$. The primes of $R$ are $(0)$ and $(x_i/x_j^n: i \in V, i<j, n \geq 0)$ where $V$ is an initial segment of $W$.

If $W$ contains finitely many maximal elements (Edit: and every element is bounded by a maximal element) and contains chains of unbounded length then $m_R$ is finitely generated and $dim R = \infty$ e.g. $W = \omega + 1$.

  • $\begingroup$ That's a very clean way of saying it. I like the strong emphasis on the order structure; the ring theory and even the ordered group theory is not so important. $\endgroup$ Jun 27, 2016 at 18:06
  • $\begingroup$ I guess my example corresponds to the ordered set $\mathbb Z_{\leq 0}$, whereas the one I linked to corresponds to $\mathbb Z_{\geq 0}$ (which does not have a maximal element, so $\mathfrak m$ is not principal). This also explains why mine has an infinite decreasing chain but not an increasing one. Thanks for helping me understand the bigger picture! $\endgroup$ Jun 27, 2016 at 19:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.