# Krull-dimension of local domain

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?

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.

• I am deeply impressed by your example. Thanks! – Pierre MATSUMI Jun 27 '16 at 7:57

Let $A=k[xy_1,xy_2,\dots,xy_n,\dots]\subseteq k[x,y_1,y_2,\dots,y_n,\dots]$, $\mathfrak m=(x)$ and $R=A_{\mathfrak m}$ with $\mathfrak m_R=\mathfrak m_{\mathfrak m}$.

This $R$ satisfies the conditions, but $\dim R=+\infty$, because (for instance) if $I=(x-1)$, then $R/I\simeq k[y_1,y_2,\dots,y_n,\dots]$

• I am a little confused by your example. Since $x \not \in A$, I assume you mean $\mathfrak m = (x) \cap A$. But this is not finitely generated... – R. van Dobben de Bruyn Jun 27 '16 at 5:11
• @R. van Dobben de Bruyn: I think there is simply an x missing: $A=k[x,xy_1,xy_2,\ldots]$ should work. – tj_ Jun 27 '16 at 20:43
• @tj_ But then $I = (x-1) \subseteq A$ does not pass to $A_{\mathfrak m}$. Also, in both cases $A$ is abstractly isomorphic to $k[x_1,x_2,\ldots]$, so I don't see the advantage of writing it like this. – R. van Dobben de Bruyn Jun 27 '16 at 20:51
• I don't think $k[x,xy_1,xy_2,\ldots]$ is isomorphic to $k[x_1,x_2,\ldots]$ because $(x)$ is a maximal ideal of the former and the latter doesn't have a f.g. max. ideal. – tj_ Jun 27 '16 at 21:04
• It's not a maximal ideal, because $xy_1$ is not divisible by $x$ in $A$. Specifically, it seems to me that the map $k[x_0,x_1,\ldots] \to k[x,xy_1,\ldots]$ given by $x_0 \mapsto x$ and $x_i \mapsto xy_i$ is an isomorphism (the only content of this statement is that the $xy_i$ are algebraically independent). – R. van Dobben de Bruyn Jun 27 '16 at 21:24

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. $W = \omega + 1$.

• 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. – R. van Dobben de Bruyn Jun 27 '16 at 18:06
• 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! – R. van Dobben de Bruyn Jun 27 '16 at 19:15