# Noetherian ring with infinite Krull dimension (Nagata's example).

I just started to read about the Krull dimension (definition and basic theory), at first when I thought about the Krull dimension of a noetherian ring my idea was that it must be finite, however this turned out to be wrong.

I am looking for an example of a commutative noetherian ring that has infinite Krull dimension.

I read that there is a famous example due to Nagata, but I was unable to find it.

The example of Nagata:

Let $$A=k[x_{\mathbf N}]$$ be the polynomial ring over a field $$k$$ in countably many indeterminates. Let $$m_1,m_2,\ldots$$ be an increasing sequence of positive integers such that $$m_{i+1}-m_i>m_i-m_{i-1}\forall i>1$$. Let $$\mathfrak p_i=(x_{m_i+1},\ldots,x_{m_{i+1}})$$ and let $$S$$ be the complement in $$A$$ of the union of the ideals $$\mathfrak p_i$$. Each $$\mathfrak p_i$$ is a prime ideal and so $$S$$ is multiplicatively closed. Each $$S^{-1}\mathfrak p_i$$ has height equal to $$m_{i+1}-m_i$$, hence $$\operatorname{dim} S^{-1}A=\infty$$.

Claim. $$S^{-1}A$$ is noetherian.

Lemma. Let $$A$$ be a ring such that
(1) for each maximal ideal $$\mathfrak m$$ of $$A$$, the local ring $$A_{\mathfrak m}$$ is noetherian;
(2) for each $$0\ne x\in A$$, the set of maximal ideals of $$A$$ which contain $$x$$ is finite.
Then $$A$$ is noetherian. (Proof e.g. in Atiyah-Macdonald ex. 7.9).

The claim will be deduced from the following remarks.

Remark 1. The $$S^{-1}\mathfrak p_i$$ are maximal.
Proof. Suppose $$\alpha\in S^{-1}A$$, $$\alpha\not\in S^{-1}\mathfrak p_i$$ and $$\alpha\not\in k$$. Then $$\alpha$$ includes a monomial not including any of the generators of $$\mathfrak p_i$$ as a factor (we may assume $$\alpha\in A$$ after clearing denominators). Removing monomials in $$\alpha$$ belonging to $$\mathfrak p_i$$, we may assume $$\alpha$$ contains no monomial such as $$x_{m_i+1}$$ with nonzero coefficient; then $$\alpha+x_{m_i+1}\in S$$ hence is a unit.

Remark 2. Any $$0\ne x\in A$$ can be in only finitely many $$S^{-1}\mathfrak p_i$$.
Proof. Can be checked for $$x\in k[x_{\mathbf N}]$$, where it is obvious.

Remark 3. (generalized prime avoidance) Any ideal $$I\subset k[x_{\mathbf N}]$$ contained in $$\bigcup_i\mathfrak p_i$$ is contained in $$\mathfrak p_i$$ for some $$i$$.

Note that Remark 3 and 1 show that the maximal ideals of $$S^{-1}A$$ are precisely the $$S^{-1}\mathfrak p_i$$. Remark 2 then satisfies condition (2) of the lemma. To see that $$S^{-1}A_{S^{-1}\mathfrak p_i}$$ is noetherian, note that it coincides with $$A_{\mathfrak p_i}\cong k(x_j)[x_{m_i+1},\ldots,x_{m_{i+1}}]_{(x_{m_i+1},\ldots,x_{m_{i+1}})}$$, a localization of a noetherian ring, where the index $$j$$ runs over all $$\mathbf N$$ except for $$m_i+1,\ldots,m_{i+1}$$. This will satisfy (1), meaning it suffices to prove Remark 3.

Proof of Remark 3. Suppose $$I\subset A$$, $$I\subset\bigcup_{i\in L}\mathfrak p_i$$. If $$|L|<\infty$$, the result follows from usual (finite) prime avoidance. Hence assume $$|L|=|\mathbf N|$$ and that $$I$$ is not contained in $$\bigcup_{k\in K}\mathfrak p_k$$ for $$K\subset\mathbf N$$ finite. For $$f\in A$$, put $$D(f):=\{i\in\mathbf N \text{ s.t. } S^{-1}\mathfrak p_i\ni f\}.$$ Let $$f\in I$$, then if $$\nexists g\in I$$ s.t. $$D(f)\cap D(g)\neq\emptyset$$, then $$I\subset\bigcup_{i\in D(f)}\mathfrak p_i$$, and $$D(f)$$ is a finite set. Hence $$\exists g\in I$$ s.t. $$D(f)\cap D(g)=\emptyset$$. Note that if $$D(f)=\emptyset$$ or $$D(g)=\emptyset$$ then one or the other lies outside of $$\bigcup_i\mathfrak p_i$$, contradicting $$I\subset\bigcup_{i\in L}\mathfrak p_i$$. Hence we may assume $$D(f)\ne\emptyset\ne D(g)$$. Let $$\sigma\in D(g)$$, $$d:=\deg f$$. Then the claim is that $$D(f+x_{m_\sigma+1}^{d+1}g)=\emptyset$$, providing a contradiction. Clearly $$D(x_{m_\sigma+1}^{d+1}g)=D(g)$$, and since $$D(f)\cap D(g)=\emptyset$$, $$f+x_{m_\sigma+1}^{d+1}g$$ is not contained in any $$\mathfrak p_\ell$$ for $$\ell\in D(f)\cup D(g)$$. At the same time, since the term of lowest degree of $$x_{m_\sigma+1}^{d+1}g$$ is of greater degree than the term of highest degree of $$f$$, there can be no cancellation among the monomials, so, fixing an index $$\ell$$, if $$f\not\in\mathfrak p_\ell$$, $$f$$ has a nonzero monomial not in $$\mathfrak p_\ell$$, and that monomial persists with the same nonzero coefficient in $$f+x_{m_\sigma+1}^{d+1}g$$, hence $$f+x_{m_\sigma+1}^{d+1}g$$ cannot lie in $$\mathfrak p_\ell$$, since for a pol. to lie in $$\mathfrak p_\ell$$, every monomial with nonzero coefficient must lie in $$(x_{m_\ell+1},\ldots,x_{m_{\ell+1}})$$. This proves Remark 3.

• Why is there an $S^{-1}$ in the definition of $D(f)$? Is this a typo?
– john
Feb 17 at 14:04
• It is a distinction without a difference: as in the proof of Remark 2, $S^{-1}\mathfrak p_i\cap A=\mathfrak p_i$.
– Tomo
Feb 17 at 19:43

Take a polynomial ring $A$ in infinitely many variables over a field, and consider the infinite family of prime ideals formed by disjoint subsets of the variables, the number of variables increasing in lenght. Then localise $A$ by the complement of the union of these prime ideals. Thanks who ? Thanks Nagata ! ;-)